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Factoring Polynomials Worksheets

Factoring is a process of splitting the algebraic expressions into factors that can be multiplied. Included here are factoring worksheets to factorize linear expressions, quadratic expressions, monomials, binomials and polynomials using a variety of methods like grouping, synthetic division and box method. These factoring polynomials worksheet pdfs with answer keys offer high school students immense practice. Begin your practice with our free worksheets!

Factors of Monomials

Factors of Monomials

These printable two-part worksheets encompass ten monomials each. Choose the factors of the monomials in the multiple response questions in Part A and list out all possible factors of the given monomial in Part B.

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Factoring Linear Expressions

Factoring Linear Expressions

Familiarize students with the topic by employing these factoring linear expression worksheets. Figure out the common factor of each linear expression and express in factor form.

Factoring Quadratic Expressions - Easy

Factoring Quadratic Expressions

A quadratic expression involves a squared term, in ax 2 +bx+c format. Factor each second degree polynomial into two first degree polynomials in these factoring quadratic expression pdf worksheets.

Factoring Binomials - Easy

Factoring Binomials

Determine the factors of the individual terms and then track down the common factor to factorize the given binomial expressions. Worksheets are available in two levels of difficulties.

Factoring Polynomials by Grouping

Factoring Polynomials by Grouping

When a polynomial expression involves four terms with no common factors, then grouping method comes handy. Organize the terms and then factorize the polynomials by applying the grouping method.

Factorize Polynomial using Synthetic Division Method

Factorize Polynomial using Synthetic Division Method

This stack of factorize using synthetic division method worksheets provides an alternate way to factorize the polynomials. Also, determine if the given expression is a factor of the polynomial.

Factoring Polynomials - Additional Practice

Factoring Polynomials - Additional Practice

Employ these printable high school worksheets on factoring polynomials to intensify your practice. Recapitulate and test comprehension with these worksheets consisting of ten unique polynomial expressions.

Factoring Quadratic Trinomials - Box method

Factoring Quadratic Trinomials - Box method

The box method is an alternative to the grouping method. It makes factoring trinomials easier by dividing the process into simpler steps. Arrange the terms in the 2x2 boxes and get factoring!

Factorization using Algebraic Identities

Factorization using Algebraic Identities

These factorization worksheets involving algebraic identities feature exercises on factorization of algebraic expressions using identities like (a+b) 2 , (a-b) 2 , algebraic identities involving substitution method and more!

(45 Worksheets)

GCF of Polynomials

GCF of Polynomials

Walk through these Greatest Common Factor of polynomials pdf worksheets to find the GCF of two or three monomials, find the GCF of polynomials, available in easy and moderate levels, find the GCF using the division method and more!

LCM of Polynomials

LCM of Polynomials

Implement these LCM of polynomials printable worksheets to hone your skills in finding the LCM of two, three monomials and polynomials categorized into different levels. Find the other polynomial in linear, quadratic expression and more.

(35 Worksheets)

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» Multiplying Polynomials

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Factoring Polynomials Worksheets

The factoring polynomials worksheets require a child to take the common factor or GCF out. The expressions deal with single as well as multiple variables.

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Factoring Polynomial Worksheets

Factoring is a process of splitting. Factoring polynomial worksheets help students understand the factorization of linear expressions, quadratic expressions, monomials, binomials, and polynomials using different types of methods like grouping, synthetic division, and box method.

Benefits of Factoring Polynomial Worksheets

Polynomials have numerous applications in everyday life, in diverse areas from shopping to engineering. Hence understanding polynomials is beneficial for students to get a broader perspective of algebra in real life.

The benefit of factoring polynomials worksheets is that they will engage students and help them practice different theories that revolve around the polynomial. These math worksheets follow a stepwise approach and help students learn how to factorize polynomials.

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Students get equipped with the factorization of polynomials with the help of these worksheets. These worksheets can be downloaded for free in PDF formats.

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Mathematics LibreTexts

4.4: Solve Polynomial Equations by Factoring

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Learning Objectives

  • Review general strategies for factoring.
  • Solve polynomial equations by factoring.
  • Find roots of a polynomial function.
  • Find polynomial equations given the solutions.

Reviewing General Factoring Strategies

We have learned various techniques for factoring polynomials with up to four terms. The challenge is to identify the type of polynomial and then decide which method to apply. The following outlines a general guideline for factoring polynomials.

general guidelines for factoring polynomials

Step 1: Check for common factors. If the terms have common factors, then factor out the greatest common factor (GCF).

Step 2: Determine the number of terms in the polynomial.

  • Factor four-term polynomials by grouping.
  • Factor trinomials (3 terms) using “trial and error” or the AC method.
  • Difference of squares :\(a^{2}−b^{2}=(a+b)(a−b)\)
  • Sum of squares : \(a^{2}+b^{2}\)  no general formula
  • Difference of cubes : \(a^{3}−b^{3}=(a−b)(a^{2}+ab+b^{2})\)
  • Sum of cubes : \(a^{3}+b^{3}=(a+b)(a^{2}−ab+b^{2})\)

Step 3: Look for factors that can be factored further.

Step 4: Check by multiplying.

If a binomial is both a difference of squares and a difference cubes, then first factor it as difference of squares. This will result in a more complete factorization. In addition, not all polynomials with integer coefficients factor. When this is the case, we say that the polynomial is prime.

If an expression has a GCF, then factor this out first. Doing so is often overlooked and typically results in factors that are easier to work with. Furthermore, look for the resulting factors to factor further; many factoring problems require more than one step. A polynomial is completely factored when none of the factors can be factored further.

Example \(\PageIndex{1}\):

Factor \(54 x ^ { 4 } - 36 x ^ { 3 } - 24 x ^ { 2 } + 16 x\).

This four-term polynomial has a GCF of \(2x\). Factor this out first.

\(54 x ^ { 4 } - 36 x ^ { 3 } - 24 x ^ { 2 } + 16 x = 2 x \left( 27 x ^ { 3 } - 18 x ^ { 2 } - 12 x + 8 \right)\)

Now factor the resulting four-term polynomial by grouping and look for resulting factors to factor further.

434605097e320aada4a22b964d70fe07.png

\(2 x ( 3 x - 2 ) ^ { 2 } ( 3 x + 2 )\). The check is left to the reader.

Example \(\PageIndex{2}\):

Factor: \(x ^ { 4 } - 3 x ^ { 2 } - 4\).

This trinomial does not have a GCF.

\(\begin{aligned} x ^ { 4 } - 3 x ^ { 2 } - 4 & = \left( x ^ { 2 } \quad\right) \left( x ^ { 2 }\quad \right) \\ & = \left( x ^ { 2 } + 1 \right) \left( x ^ { 2 } - 4 \right) \quad\color{Cerulean} { Difference\: of\: squares } \\ & = \left( x ^ { 2 } + 1 \right) ( x + 2 ) ( x - 2 ) \end{aligned}\)

The factor \(\left( x ^ { 2 } + 1 \right)\) is prime and the trinomial is completlely factored.

\(\left( x ^ { 2 } + 1 \right) ( x + 2 ) ( x - 2 )\)

Example \(\PageIndex{3}\):

Factor: \(x ^ { 6 } + 6 x ^ { 3 } - 16\).

Begin by factoring \(x ^ { 6 } = x ^ { 3 } \cdot x ^ { 3 }\) and look for the factors of \(16\) that add to \(6\).

\(\begin{aligned} x ^ { 6 } + 6 x ^ { 3 } - 16 & = \left( x ^ { 3 }\quad \right) \left( x ^ { 3 }\quad \right) \\ & = \left( x ^ { 3 } - 2 \right) \left( x ^ { 3 } + 8 \right)\quad \color{Cerulean} { sum\:of\:cubes } \\ & = \left( x ^ { 3 } - 2 \right) ( x + 2 ) \left( x ^ { 2 } - 2 x + 4 \right) \end{aligned}\)

The factor \(\left( x ^ { 3 } - 2 \right)\) cannot be factored any further using integers and the factorization is complete.

\(\left( x ^ { 3 } - 2 \right) ( x + 2 ) \left( x ^ { 2 } + 2 x + 4 \right)\)

Exercise \(\PageIndex{1}\)

Factor: \(9 x ^ { 4 } + 17 x ^ { 2 } - 2\)

\(( 3 x + 1 ) ( 3 x - 1 ) \left( x ^ { 2 } + 2 \right)\)

www.youtube.com/v/lsAP8_UgUx0

Solving Polynomial Equations by Factoring

In this section, we will review a technique that can be used to solve certain polynomial equations. We begin with the zero-product property 20 :

\(a⋅b=0\)   if and only if   \(a=0\) or \(b=0\)

The zero-product property is true for any number of factors that make up an equation. In other words, if any product is equal to zero, then at least one of the variable factors must be equal to zero. If an expression is equal to zero and can be factored into linear factors, then we will be able to set each factor equal to zero and solve for each equation.

Example \(\PageIndex{4}\)

Solve: \(2 x ( x - 4 ) ( 5 x + 3 ) = 0\).

Set each variable factor equal to zero and solve.

\(\begin{aligned} 2 x & = 0 \\ \frac { 2 x } { \color{Cerulean}{2} } & = \frac { 0 } {\color{Cerulean}{ 2} } \\ x & = 0 \end{aligned}\) or \(\begin{aligned} x - 4 & = 0 \\ x & = 4 \end{aligned}\) or \(\begin{aligned} 5 x + 3 & = 0 \\ \frac { 5 x } { \color{Cerulean}{5} } & = \frac { - 3 } {\color{Cerulean}{ 5} } \\ x & = - \frac { 3 } { 5 } \end{aligned}\)

To check that these are solutions we can substitute back into the original equation to see if we obtain a true statement. Note that each solution produces a zero factor. This is left to the reader.

The solutions are \(0, 4\), and \(\frac{−3}{5}\).

Of course, most equations will not be given in factored form.

Example \(\PageIndex{5}\)

Solve: \(4 x ^ { 3 } - x ^ { 2 } - 100 x + 25 = 0\).

Begin by factoring the left side completely.

\(\begin{aligned} 4 x ^ { 3 } - x ^ { 2 } - 100 x + 25 & = 0 \quad\color{Cerulean}{Factor\:by\:grouping.} \\ x ^ { 2 } ( 4 x - 1 ) - 25 ( 4 x - 1 ) & = 0 \\ ( 4 x - 1 ) \left( x ^ { 2 } - 25 \right) & = 0\quad\color{Cerulean}{Factor\:as\:a\:difference\:of\:squares.} \\ ( 4 x - 1 ) ( x + 5 ) ( x - 5 ) & = 0 \end{aligned}\)

Set each factor equal to zero and solve.

\(\begin{array} { r } { 4 x - 1 = 0 } \\ { 4 x = 1 } \\ { x = \frac { 1 } { 4 } } \end{array}\) or \(\begin{aligned} x + 5 & = 0 \\ x & = - 5 \end{aligned}\) or \(\begin{aligned} x - 5 & = 0 \\ x & = 5 \end{aligned}\)

The solutions are \(\frac{1}{4}, −5\), and \(5\).

Using the zero-product property after factoring an equation that is equal to zero is the key to this technique. However, the equation may not be given equal to zero, and so there may be some preliminary steps before factoring. The steps required to solve by factoring 21 are outlined in the following example.

Example \(\PageIndex{6}\)

Solve: \(15 x ^ { 2 } + 3 x - 8 = 5 x - 7\).

Step 1: Express the equation in standard form, equal to zero. In this example, subtract \(5x\) from and add \(7\) to both sides.

\(\begin{array} { l } { 15 x ^ { 2 } + 3 x - 8 = 5 x - 7 } \\ { 15 x ^ { 2 } - 2 x - 1 = 0 } \end{array}\)

Step 2: Factor the expression.

\((3x−1)(5x+1)=0\)

Step 3: Apply the zero-product property and set each variable factor equal to zero.

\(3x−1=0\)        or        \(5x+1=0\)

Step 4: Solve the resulting linear equations.

\(\begin{array} { r } { 3 x - 1 = 0 } \\ { 3 x = 1 } \\ { x = \frac { 1 } { 3 } } \end{array}\) or \(\begin{aligned} 5 x + 1 & = 0 \\ 5 x & = - 1 \\ x & = - \frac { 1 } { 5 } \end{aligned}\)

The solutions are \(\frac{1}{3}\) and \(\frac{−1}{5}\). The check is optional.

Example \(\PageIndex{7}\):

Solve: \((3x+2)(x+1)=4\).

This quadratic equation appears to be factored; hence it might be tempting to set each factor equal to \(4\). However, this would lead to incorrect results. We must rewrite the equation equal to zero, so that we can apply the zero-product property.

\(\begin{array} { r } { ( 3 x + 2 ) ( x + 1 ) = 4 } \\ { 3 x ^ { 2 } + 3 x + 2 x + 2 = 4 } \\ { 3 x ^ { 2 } + 5 x + 2 = 4 } \\ { 3 x ^ { 2 } + 5 x - 2 = 0 } \end{array}\)

Once it is in standard form, we can factor and then set each factor equal to zero.

        \(\begin{array} { c } { ( 3 x - 1 ) ( x + 2 ) = 0 } \\ { 3 x - 1 = 0 \quad \text { or } \quad x + 2 = 0 } \\ \quad\quad{ 3 x = 1 }\quad\quad\quad\:\:\quad x=-2\\ { x = \frac { 1 } { 3 } }\quad\quad\quad\quad\quad\: \end{array}\)

The solutions are \(\frac{1}{3}\) and \(−2\).

Finding Roots of Functions

Recall that any polynomial with one variable is a function and can be written in the form,

\(f ( x ) = a _ { n } x ^ { n } + a _ { n - 1 } x ^ { n - 1 } + \cdots + a _ { 1 } x + a _ { 0 }\)

A root 22 of a function is a value in the domain that results in zero. In other words, the roots occur when the function is equal to zero, \(f(x)=0\).

Example \(\PageIndex{8}\)

Find the roots: \(f ( x ) = ( x + 2 ) ^ { 2 } - 4\).

To find roots we set the function equal to zero and solve.

\(\begin{aligned} f ( x ) & = 0 \\ ( x + 2 ) ^ { 2 } - 4 & = 0 \\ x ^ { 2 } + 4 x + 4 - 4 & = 0 \\ x ^ { 2 } + 4 x & = 0 \\ x ( x + 4 ) & = 0 \end{aligned}\)

Next, set each factor equal to zero and solve.

\(\begin{aligned} x = 0 \quad \text { or } \quad x + 4 = 0 \\ x = - 4 \end{aligned}\)

We can show that these \(x\)-values are roots by evaluating.

\(\begin{aligned} f ( 0 ) & = ( 0 + 2 ) ^ { 2 } - 4 \\ & = 4 - 4 \\ & = 0 \color{Cerulean}{✓}\end{aligned}\)   \(\begin{aligned} f ( - 4 ) & = ( - 4 + 2 ) ^ { 2 } - 4 \\ & = ( - 2 ) ^ { 2 } - 4 \\ & = 4 - 4 \\ & = 0 \color{Cerulean}{✓}\end{aligned}\)

The roots are \(0\) and \(−4\).

If we graph the function in the previous example we will see that the roots correspond to the \(x\)-intercepts of the function. Here the function \(f\) is a basic parabola shifted \(2\) units to the left and \(4\) units down.

b7624df66849d5be91434b044cf0d510.png

Example \(\PageIndex{9}\)

Find the roots: \(f ( x ) = x ^ { 4 } - 5 x ^ { 2 } + 4\).

\(\begin{aligned} f ( x ) & = 0 \\ x ^ { 4 } - 5 x ^ { 2 } + 4 & = 0 \\ \left( x ^ { 2 } - 1 \right) \left( x ^ { 2 } - 4 \right) & = 0 \\ ( x + 1 ) ( x - 1 ) ( x + 2 ) ( x - 2 ) & = 0 \end{aligned}\)

\(\begin{aligned} x + 1 & = 0 \\ x & = - 1 \end{aligned}\) or \(\begin{array} { r } { x - 1 = 0 } \\ { x = 1 } \end{array}\) or \(\begin{aligned} x + 2 & = 0 \\ x & = - 2 \end{aligned}\) or \(\begin{aligned} x - 2 & = 0 \\ x & = 2 \end{aligned}\)

The roots are \(−1, 1, −2\), and \(2\).

Graphing the previous function is not within the scope of this course. However, the graph is provided below:

6aaf3a5ab540885474d58855068b64ce.png

Notice that the degree of the polynomial is \(4\) and we obtained four roots. In general, for any polynomial function with one variable of degree \(n\), the fundamental theorem of algebra 23 guarantees \(n\) real roots or fewer. We have seen that many polynomials do not factor. This does not imply that functions involving these unfactorable polynomials do not have real roots. In fact, many polynomial functions that do not factor do have real solutions. We will learn how to find these types of roots as we continue in our study of algebra.

Example \(\PageIndex{10}\)

Find the roots: \(f ( x ) = - x ^ { 2 } + 10 x - 25\).

\(\begin{aligned} f ( x ) & = 0 \\ - x ^ { 2 } + 10 x - 25 & = 0 \\ - \left( x ^ { 2 } - 10 x + 25 \right) & = 0 \\ - ( x - 5 ) ( x - 5 ) & = 0 \end{aligned}\)

Next, set each variable factor equal to zero and solve.

\(\begin{aligned} x - 5 & = 0 \quad\quad\text { or }& x - 5 = 0 \\ & = 5 \quad &x= 5 \end{aligned}\)

A solution that is repeated twice is called a double root 24 . In this case, there is only one solution.

The root is \(5\).

The previous example shows that a function of degree \(2\) can have one root. From the factoring step, we see that the function can be written

\(f ( x ) = - ( x - 5 ) ^ { 2 }\)

In this form, we can see a reflection about the \(x\)-axis and a shift to the right \(5\) units. The vertex is the \(x\)-intercept, illustrating the fact that there is only one root.

252fe9f0928cc6d78892b3f06550a0b5.png

Exercise \(\PageIndex{2}\)

Find the roots of \(f ( x ) = x ^ { 3 } + 3 x ^ { 2 } - x - 3\).

\(±1, −3\)

www.youtube.com/v/t1ShqqhoaVE

Example \(\PageIndex{11}\)

Assuming dry road conditions and average reaction times, the safe stopping distance in feet is given by \(d ( x ) = \frac { 1 } { 20 } x ^ { 2 } + x\), where \(x\) represents the speed of the car in miles per hour. Determine the safe speed of the car if you expect to stop in \(40\) feet.

We are asked to find the speed \(x\) where the safe stopping distance \(d(x)=40\) feet.

\(\begin{array} { c } { d ( x ) = 40 } \\ { \frac { 1 } { 20 } x ^ { 2 } + x = 40 } \end{array}\)

To solve for \(x\), rewrite the resulting equation in standard form. In this case, we will first multiply both sides by \(20\) to clear the fraction.

\(\begin{aligned} \color{Cerulean}{20}\color{black}{ \left( \frac { 1 } { 20 } x ^ { 2 } + x \right)} & = \color{Cerulean}{20}\color{black}{ (} 40 ) \\ x ^ { 2 } + 20 x & = 800 \\ x ^ { 2 } + 20 x - 800 & = 0 \end{aligned}\)

Next factor and then set each factor equal to zero.

\(\begin{aligned} x ^ { 2 } + 20 x - 800 & = 0 \\ ( x + 40 ) ( x - 20 ) & = 0 \\ x + 40 & = 0\quad or\quad \:x-20=0 \\ x & = - 40\quad \quad\quad\quad x=20 \end{aligned}\)

The negative answer does not make sense in the context of this problem. Consider \(x=20\) miles per hour to be the only solution.

\(20\) miles per hour

Finding Equations with Given Solutions

We can use the zero-product property to find equations, given the solutions. To do this, the steps for solving by factoring are performed in reverse.

Example \(\PageIndex{12}\)

Find a quadratic equation with integer coefficients, given solutions \(\frac{−3}{2}\) and \(\frac{1}{3}\).

Given the solutions, we can determine two linear factors. To avoid fractional coefficients, we first clear the fractions by multiplying both sides by the denominator.

\(\begin{aligned} x & = - \frac { 3 } { 2 } \\ 2 x & = - 3 \\ 2 x + 3 & = 0 \end{aligned}\) or \(\begin{aligned} x & = \frac { 1 } { 3 } \\ 3 x & = 1 \\ 3 x - 1 & = 0 \end{aligned}\)

The product of these linear factors is equal to zero when \(x=\frac{−3}{2}\) or \(x=\frac{1}{3}\).

\((2x+3)(3x−1)=0\)

Multiply the binomials and present the equation in standard form.

\(\begin{array} { r } { 6 x ^ { 2 } - 2 x + 9 x - 3 = 0 } \\ { 6 x ^ { 2 } + 7 x - 3 = 0 } \end{array}\)

We may check our equation by substituting the given answers to see if we obtain a true statement. Also, the equation found above is not unique and so the check becomes essential when our equation looks different from someone else’s. This is left as an exercise.

\(6 x ^ { 2 } + 7 x - 3 = 0\)

Example \(\PageIndex{13}\)

Find a polynomial function with real roots \(1, −2\), and \(2\).

Given solutions to \(f(x)=0\) we can find linear factors.

\(\begin{array} { r } { x = 1 } \\ { x - 1 = 0 } \end{array}\) or \(\begin{aligned} x & = - 2 \\ x + 2 & = 0 \end{aligned}\) or \(\begin{aligned} x & = 2 \\ x - 2 & = 0 \end{aligned}\)

Apply the zero-product property and multiply.

\(\begin{aligned} ( x - 1 ) ( x + 2 ) ( x - 2 ) & = 0 \\ ( x - 1 ) \left( x ^ { 2 } - 4 \right) & = 0 \\ x ^ { 3 } - 4 x - x ^ { 2 } + 4 & = 0 \\ x ^ { 3 } - x ^ { 2 } - 4 x + 4 & = 0 \end{aligned}\)

\(f ( x ) = x ^ { 3 } - x ^ { 2 } - 4 x + 4\)

Exercise \(\PageIndex{3}\)

Find a polynomial equation with integer coefficients, given solutions \(\frac{1}{2}\) and \(\frac{−3}{4}\).

\(8 x ^ { 2 } + 2 x - 3 = 0\)

www.youtube.com/v/o4cuUWWEGdU

Key Takeaways

  • Factoring and the zero-product property allow us to solve equations.
  • To solve a polynomial equation, first write it in standard form. Once it is equal to zero, factor it and then set each variable factor equal to zero. The solutions to the resulting equations are the solutions to the original.
  • Not all polynomial equations can be solved by factoring. We will learn how to solve polynomial equations that do not factor later in the course.
  • A polynomial function can have at most a number of real roots equal to its degree. To find roots of a function, set it equal to zero and solve.
  • To find a polynomial equation with given solutions, perform the process of solving by factoring in reverse.

Exercise \(\PageIndex{4}\)

Factor completely.

  • \(50 x ^ { 2 } - 18\)
  • \(12 x ^ { 3 } - 3 x\)
  • \(10 x ^ { 3 } + 65 x ^ { 2 } - 35 x\)
  • \(15 x ^ { 4 } + 7 x ^ { 3 } - 4 x ^ { 2 }\)
  • \(6 a ^ { 4 } b - 15 a ^ { 3 } b ^ { 2 } - 9 a ^ { 2 } b ^ { 3 }\)
  • \(8 a ^ { 3 } b - 44 a ^ { 2 } b ^ { 2 } + 20 a b ^ { 3 }\)
  • \(36 x ^ { 4 } - 72 x ^ { 3 } - 4 x ^ { 2 } + 8 x\)
  • \(20 x ^ { 4 } + 60 x ^ { 3 } - 5 x ^ { 2 } - 15 x\)
  • \(3 x ^ { 5 } + 2 x ^ { 4 } - 12 x ^ { 3 } - 8 x ^ { 2 }\)
  • \(10 x ^ { 5 } - 4 x ^ { 4 } - 90 x ^ { 3 } + 36 x ^ { 2 }\)
  • \(x ^ { 4 } - 23 x ^ { 2 } - 50\)
  • \(2 x ^ { 4 } - 31 x ^ { 2 } - 16\)
  • \(- 2 x ^ { 5 } - 6 x ^ { 3 } + 8 x\)
  • \(- 36 x ^ { 5 } + 69 x ^ { 3 } + 27 x\)
  • \(54 x ^ { 5 } - 78 x ^ { 3 } + 24 x\)
  • \(4 x ^ { 6 } - 65 x ^ { 4 } + 16 x ^ { 2 }\)
  • \(x ^ { 6 } - 7 x ^ { 3 } - 8\)
  • \(x ^ { 6 } - 25 x ^ { 3 } - 54\)
  • \(3 x ^ { 6 } + 4 x ^ { 3 } + 1\)
  • \(27 x ^ { 6 } - 28 x ^ { 3 } + 1\)

1. \(2 ( 5 x + 3 ) ( 5 x - 3 )\)

3. \(5 x ( x + 7 ) ( 2 x - 1 )\)

5. \(3 a ^ { 2 } b ( 2 a + b ) ( a - 3 b )\)

7. \(4 x ( x - 2 ) ( 3 x + 1 ) ( 3 x - 1 )\)

9. \(x ^ { 2 } ( 3 x + 2 ) ( x + 2 ) ( x - 2 )\)

11. \(\left( x ^ { 2 } + 2 \right) ( x + 5 ) ( x - 5 )\)

13. \(- 2 x \left( x ^ { 2 } + 4 \right) ( x - 1 ) ( x + 1 )\)

15. \(6 x ( x + 1 ) ( x - 1 ) ( 3 x + 2 ) ( 3 x - 2 )\)

17. \(( x + 1 ) \left( x ^ { 2 } - x + 1 \right) ( x - 2 ) \left( x ^ { 2 } + 2 x + 4 \right)\)

19. \(\left( 3 x ^ { 3 } + 1 \right) ( x + 1 ) \left( x ^ { 2 } - x + 1 \right)\)

Exercise \(\PageIndex{5}\)

  • \(( 6 x - 5 ) ( x + 7 ) = 0\)
  • \(( x + 9 ) ( 3 x - 8 ) = 0\)
  • \(5 x ( 2 x - 5 ) ( 3 x + 1 ) = 0\)
  • \(4 x ( 5 x - 1 ) ( 2 x + 3 ) = 0\)
  • \(( x - 1 ) ( 2 x + 1 ) ( 3 x - 5 ) = 0\)
  • \(( x + 6 ) ( 5 x - 2 ) ( 2 x + 9 ) = 0\)
  • \(( x + 4 ) ( x - 2 ) = 16\)
  • \(( x + 1 ) ( x - 7 ) = 9\)
  • \(( 6 x + 1 ) ( x + 1 ) = 6\)
  • \(( 2 x - 1 ) ( x - 4 ) = 39\)
  • \(x ^ { 2 } - 15 x + 50 = 0\)
  • \(x ^ { 2 } + 10 x - 24 = 0\)
  • \(3 x ^ { 2 } + 2 x - 5 = 0\)
  • \(2 x ^ { 2 } + 9 x + 7 = 0\)
  • \(\frac { 1 } { 10 } x ^ { 2 } - \frac { 7 } { 15 } x - \frac { 1 } { 6 } = 0\)
  • \(\frac { 1 } { 4 } - \frac { 4 } { 9 } x ^ { 2 } = 0\)
  • \(6 x ^ { 2 } - 5 x - 2 = 30 x + 4\)
  • \(6 x ^ { 2 } - 9 x + 15 = 20 x - 13\)
  • \(5 x ^ { 2 } - 23 x + 12 = 4 ( 5 x - 3 )\)
  • \(4 x ^ { 2 } + 5 x - 5 = 15 ( 3 - 2 x )\)
  • \(( x + 6 ) ( x - 10 ) = 4 ( x - 18 )\)
  • \(( x + 4 ) ( x - 6 ) = 2 ( x + 4 )\)
  • \(4 x ^ { 3 } - 14 x ^ { 2 } - 30 x = 0\)
  • \(9 x ^ { 3 } + 48 x ^ { 2 } - 36 x = 0\)
  • \(\frac { 1 } { 3 } x ^ { 3 } - \frac { 3 } { 4 } x = 0\)
  • \(\frac { 1 } { 2 } x ^ { 3 } - \frac { 1 } { 50 } x = 0\)
  • \(- 10 x ^ { 3 } - 28 x ^ { 2 } + 48 x = 0\)
  • \(- 2 x ^ { 3 } + 15 x ^ { 2 } + 50 x = 0\)
  • \(2 x ^ { 3 } - x ^ { 2 } - 72 x + 36 = 0\)
  • \(4 x ^ { 3 } - 32 x ^ { 2 } - 9 x + 72 = 0\)
  • \(45 x ^ { 3 } - 9 x ^ { 2 } - 5 x + 1 = 0\)
  • \(x ^ { 3 } - 3 x ^ { 2 } - x + 3 = 0\)
  • \(x ^ { 4 } - 5 x ^ { 2 } + 4 = 0\)
  • \(4 x ^ { 4 } - 37 x ^ { 2 } + 9 = 0\)

1. \(- 7 , \frac { 5 } { 6 }\)

3. \(0 , \frac { 5 } { 2 } , - \frac { 1 } { 3 }\)

5. \(- \frac { 1 } { 2 } , 1 , \frac { 5 } { 3 }\)

7. \(- 6,4\)

9. \(- \frac { 5 } { 3 } , \frac { 1 } { 2 }\)

11. \(5,10\)

13. \(- \frac { 5 } { 3 } , 1\)

15. \(- \frac { 1 } { 3 } , 5\)

17. \(- \frac { 1 } { 6 } , 6\)

19. \(\frac { 3 } { 5 } , 8\)

21. \(2,6\)

23. \(0 , - \frac { 3 } { 2 } , 5\)

25. \(0 , \pm \frac { 3 } { 2 }\)

27. \(- 4,0 , \frac { 6 } { 5 }\)

29. \(\pm 6 , \frac { 1 } { 2 }\)

31. \(\pm \frac { 1 } { 3 } , \frac { 1 } { 5 }\)

33. \(\pm 1 , \pm 2\)

Exercise \(\PageIndex{6}\)

Find the roots of the given functions.

  • \(f ( x ) = x ^ { 2 } + 10 x - 24\)
  • \(f ( x ) = x ^ { 2 } - 14 x + 48\)
  • \(f ( x ) = - 2 x ^ { 2 } + 7 x + 4\)
  • \(f ( x ) = - 3 x ^ { 2 } + 14 x + 5\)
  • \(f ( x ) = 16 x ^ { 2 } - 40 x + 25\)
  • \(f ( x ) = 9 x ^ { 2 } - 12 x + 4\)
  • \(g ( x ) = 8 x ^ { 2 } + 3 x\)
  • \(g ( x ) = 5 x ^ { 2 } - 30 x\)
  • \(p ( x ) = 64 x ^ { 2 } - 1\)
  • \(q ( x ) = 4 x ^ { 2 } - 121\)
  • \(f ( x ) = \frac { 1 } { 5 } x ^ { 3 } - 1 x ^ { 2 } - \frac { 1 } { 20 } x + \frac { 1 } { 4 }\)
  • \(f ( x ) = \frac { 1 } { 3 } x ^ { 3 } + \frac { 1 } { 2 } x ^ { 2 } - \frac { 4 } { 3 } x - 2\)
  • \(g ( x ) = x ^ { 4 } - 13 x ^ { 2 } + 36\)
  • \(g ( x ) = 4 x ^ { 4 } - 13 x ^ { 2 } + 9\)
  • \(f ( x ) = ( x + 5 ) ^ { 2 } - 1\)
  • \(g ( x ) = - ( x + 5 ) ^ { 2 } + 9\)
  • \(f ( x ) = - ( 3 x - 5 ) ^ { 2 }\)
  • \(g ( x ) = - ( x + 2 ) ^ { 2 } + 4\)

1. \(2 , - 12\)

3. \(- \frac { 1 } { 2 } , 4\)

5. \(\frac { 5 } { 4 }\)

7. \(- \frac { 3 } { 8 } , 0\)

9. \(\pm \frac { 1 } { 8 }\)

11. \(\pm \frac { 1 } { 2 } , 5\)

13. \(\pm 2 , \pm 3\)

15. \(- 6 , - 4\)

17. \(\frac { 5 } { 3 }\)

Exercise \(\PageIndex{7}\)

Given the graph of a function, determine the real roots.

a787931cf5e210ca86255385e54c378b.png

5. The sides of a square measure \(x − 2\) units. If the area is \(36\) square units, then find \(x\).

6. The sides of a right triangle have lengths that are consecutive even integers. Find the lengths of each side. (Hint: Apply the Pythagorean theorem)

7. The profit in dollars generated by producing and selling n bicycles per week is given by the formula \(P ( n ) = - 5 n ^ { 2 } + 400 n - 6000\). How many bicycles must be produced and sold to break even?

8. The height in feet of an object dropped from the top of a \(64\)-foot building is given by \(h ( t ) = - 16 t ^ { 2 } + 64\) where \(t\) represents the time in seconds after it is dropped. How long will it take to hit the ground?

9. A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box of height 2 inches is given.

af33d445388f22bf636c5f708b3e7020.png

What is the length of each side of the cardboard sheet if the volume of the box is to be \(98\) cubic inches?

10. The height of a triangle is \(4\) centimeters less than twice the length of its base. If the total area of the triangle is \(48\) square centimeters, then find the lengths of the base and height.

11. A uniform border is to be placed around an \(8 × 10\) inch picture.

017977d6b65606ad4e19f5ee4f12abd4.png

If the total area including the border must be \(168\) square inches, then how wide should the border be?

12. The area of a picture frame including a \(3\)-inch wide border is \(120\) square inches.

a6e4cd33b2fc5b3201cf8055d0a50607.png

If the width of the inner area is \(2\) inches less than its length, then find the dimensions of the inner area.

13. Assuming dry road conditions and average reaction times, the safe stopping distance in feet is given by \(d ( x ) = \frac { 1 } { 20 } x ^ { 2 } + x\) where \(x\) represents the speed of the car in miles per hour. Determine the safe speed of the car if you expect to stop in \(75\) feet.

14. A manufacturing company has determined that the daily revenue in thousands of dollars is given by the formula \(R ( n ) = 12 n - 0.6 n ^ { 2 }\) where \(n\) represents the number of palettes of product sold \((0 ≤ n < 20)\). Determine the number of palettes sold in a day if the revenue was \(45\) thousand dollars.

1. \(- 3 , - 1,0,2\)

3. \(- 2,3\)

5. \(8\) units

7. \(20\) or \(60\) bicycles

9. \(11\) in

11. \(2\) inches

13. \(30\) miles per hour

Exercise \(\PageIndex{8}\)

Find a polynomial equation with the given solutions.

  • \(2, \frac{1}{3}\)
  • \(- \frac { 3 } { 4 } , 5\)
  • \(- 3,1,3\)
  • \(- 5 , - 1,1\)

1. \(x ^ { 2 } - 2 x - 15 = 0\)

3. \(3 x ^ { 2 } - 7 x + 2 = 0\)

5. \(x ^ { 2 } + 4 x = 0\)

7. \(x ^ { 2 } - 49 = 0\)

9. \(x ^ { 3 } - x ^ { 2 } - 9 x + 9 = 0\)

Exercise \(\PageIndex{9}\)

Find a function with the given roots.

  • \(\frac { 1 } { 2 } , \frac { 2 } { 3 }\)
  • \(\frac { 2 } { 5 } , - \frac { 1 } { 3 }\)
  • \(\pm \frac { 3 } { 4 }\)
  • \(\pm \frac { 5 } { 2 }\)
  • \(5\) double root
  • \(-3\) double root

1. \(f ( x ) = 6 x ^ { 2 } - 7 x + 2\)

3. \(f ( x ) = 16 x ^ { 2 } - 9\)

5. \(f ( x ) = x ^ { 2 } - 10 x + 25\)

7. \(f ( x ) = x ^ { 3 } - 2 x ^ { 2 } - 3 x\)

Exercise \(\PageIndex{10}\)

Recall that if \(| X | = p\), then \(X=-p\) or \(X=p\). Use this to solve the following absolute value equations.

  • \(\left| x ^ { 2 } - 8 \right| = 8\)
  • \(\left| 2 x ^ { 2 } - 9 \right| = 9\)
  • \(\left| x ^ { 2 } - 2 x - 1 \right| = 2\)
  • \(\left| x ^ { 2 } - 8 x + 14 \right| = 2\)
  • \(\left| 2 x ^ { 2 } - 4 x - 7 \right| = 9\)
  • \(\left| x ^ { 2 } - 3 x - 9 \right| = 9\)

1. \(\pm 4,0\)

3. \(\pm 1,3\)

5. \(- 2,1,4\)

Exercise \(\PageIndex{11}\)

  • Explain to a beginning algebra student the difference between an equation and an expression.
  • What is the difference between a root and an \(x\)-intercept? Explain.
  • Create a function with three real roots of your choosing. Graph it with a graphing utility and verify your results. Share your function on the discussion board.
  • Research and discuss the fundamental theorem of algebra.

1. Answer may vary

3. Answer may vary

20 A product is equal to zero if and only if at least one of the factors is zero.

21 The process of solving an equation that is equal to zero by factoring it and then setting each variable factor equal to zero.

22 A value in the domain of a function that results in zero.

23 Guarantees that there will be as many (or fewer) roots to a polynomial function with one variable as its degree.

24 A root that is repeated twice.

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Factoring Polynomials Worksheets

Created: October 31, 2022

Last updated: June 6, 2023

The process of factoring entails division, albeit a more complex division. The factoring polynomials worksheet with answers is an excellent resource to teach your graders how to factor quadratic expressions, linear expressions, binomials, monomials, and polynomials. Below are more things you need to know about these worksheets.

Understanding how to factor polynomials

Factoring polynomials is the opposite approach of multiplying the factors of polynomials. In a polynomial expression, all variables and constants are separated by mathematical signs like subtraction or addition.

When expressed as another linear polynomial, the zeros of polynomials are referred to as factors. After a polynomial has been factored, you will get a remainder of zero if you divide it by its factors. Also, you can factor polynomials by their GCF (Greatest Common Factor).

Polynomial Factoring Worksheets

Polynomial Factoring Worksheets

Factor Polynomials Worksheet

Factor Polynomials Worksheet

Polynomial Factoring Worksheet

Polynomial Factoring Worksheet

Factoring Polynomials Worksheets

Polynomials And Factoring Worksheet

Algebra 1 Factoring Polynomials Worksheets

Algebra 1 Factoring Polynomials Worksheets

Factorization can be defined as the method of finding the factors of a particular mathematical expression or value. Multiplication of fractions brings an original number. For instance, 18, 9, 6, 3, and 2 are factors of 18.

Like with fractions, the factors of a polynomial are the polynomials that are multiplied to obtain the original. For instance, x² + 5x + 6 has its factors as (x + 2) (x + 3). Getting the original polynomial involves multiplying x +2 and x +3.

Factoring polynomials involves several methods:

  • GCF (Greatest Common Factor)
  • Difference or sum in two cubes
  • Trinomials method
  • General trinomials
  • The difference in two squares approach

Free factoring polynomials worksheet with answers

It is no secret that the equation aspects of mathematics are usually the most complex for students. However, finding a suitable approach to aid learning and understanding is pertinent irrespective of the problems’ difficulty. One of the most appropriate resources is factoring polynomials by grouping worksheet with answers.

Math for Kids

There are several perks of using Brighterly’s factoring polynomial worksheets. You can use them online on the Brighterly’s website. Excellent tutors design the worksheets with a perfect understanding of the concept and the ways to simplify them. When you use these learning tools, your kids will learn faster.

The polynomial factoring worksheets may not be suitable for kids below third grade. The worksheets are excellent for higher-grade kids and prepare them for subsequent phases. For instance, offering the worksheets to kids in grade five will make them more equipped and prepared for sixth grade.

The factor polynomials worksheet is designed in a stepwise way. Therefore, there is no chance of the questions overwhelming your kids. They will proceed from easy to more complex tasks, which makes understanding seamless.

Printable factoring polynomials worksheet PDF

Factoring polynomials worksheets pdf.

Factoring Polynomials Worksheet

Factoring Polynomials Worksheet With Answers

Factoring Polynomials By Grouping Worksheet With Answers

Factoring Polynomials Worksheet PDF

The printable factoring polynomials worksheet PDF is available on the Brighterly website. You have to download and print these sheets. With the paper format, the worksheets become easier to use as you don’t have to go online whenever your kids have to study.

You can keep your kids occupied anytime when you have printable versions of worksheets. Also, since the worksheet contains answers, your kids will know when they are wrong, which enables self-learning.

More Factoring Worksheets

  • Factoring Trinomials Worksheets
  • Solving Quadratic Equations By Factoring Worksheets
  • Factoring Quadratics Worksheets
  • Factoring By Grouping Worksheets
  • Factoring Worksheets

Jessica is a a seasoned math tutor with over a decade of experience in the field. With a BSc and Master’s degree in Mathematics, she enjoys nurturing math geniuses, regardless of their age, grade, and skills. Apart from tutoring, Jessica blogs at Brighterly. She also has experience in child psychology, homeschooling and curriculum consultation for schools and EdTech websites.

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  • 1.5 Factoring Polynomials
  • Introduction to Prerequisites
  • 1.1 Real Numbers: Algebra Essentials
  • 1.2 Exponents and Scientific Notation
  • 1.3 Radicals and Rational Exponents
  • 1.4 Polynomials
  • 1.6 Rational Expressions
  • Key Equations
  • Key Concepts
  • Review Exercises
  • Practice Test
  • Introduction to Equations and Inequalities
  • 2.1 The Rectangular Coordinate Systems and Graphs
  • 2.2 Linear Equations in One Variable
  • 2.3 Models and Applications
  • 2.4 Complex Numbers
  • 2.5 Quadratic Equations
  • 2.6 Other Types of Equations
  • 2.7 Linear Inequalities and Absolute Value Inequalities
  • Introduction to Functions
  • 3.1 Functions and Function Notation
  • 3.2 Domain and Range
  • 3.3 Rates of Change and Behavior of Graphs
  • 3.4 Composition of Functions
  • 3.5 Transformation of Functions
  • 3.6 Absolute Value Functions
  • 3.7 Inverse Functions
  • Introduction to Linear Functions
  • 4.1 Linear Functions
  • 4.2 Modeling with Linear Functions
  • 4.3 Fitting Linear Models to Data
  • Introduction to Polynomial and Rational Functions
  • 5.1 Quadratic Functions
  • 5.2 Power Functions and Polynomial Functions
  • 5.3 Graphs of Polynomial Functions
  • 5.4 Dividing Polynomials
  • 5.5 Zeros of Polynomial Functions
  • 5.6 Rational Functions
  • 5.7 Inverses and Radical Functions
  • 5.8 Modeling Using Variation
  • Introduction to Exponential and Logarithmic Functions
  • 6.1 Exponential Functions
  • 6.2 Graphs of Exponential Functions
  • 6.3 Logarithmic Functions
  • 6.4 Graphs of Logarithmic Functions
  • 6.5 Logarithmic Properties
  • 6.6 Exponential and Logarithmic Equations
  • 6.7 Exponential and Logarithmic Models
  • 6.8 Fitting Exponential Models to Data
  • Introduction to Systems of Equations and Inequalities
  • 7.1 Systems of Linear Equations: Two Variables
  • 7.2 Systems of Linear Equations: Three Variables
  • 7.3 Systems of Nonlinear Equations and Inequalities: Two Variables
  • 7.4 Partial Fractions
  • 7.5 Matrices and Matrix Operations
  • 7.6 Solving Systems with Gaussian Elimination
  • 7.7 Solving Systems with Inverses
  • 7.8 Solving Systems with Cramer's Rule
  • Introduction to Analytic Geometry
  • 8.1 The Ellipse
  • 8.2 The Hyperbola
  • 8.3 The Parabola
  • 8.4 Rotation of Axes
  • 8.5 Conic Sections in Polar Coordinates
  • Introduction to Sequences, Probability and Counting Theory
  • 9.1 Sequences and Their Notations
  • 9.2 Arithmetic Sequences
  • 9.3 Geometric Sequences
  • 9.4 Series and Their Notations
  • 9.5 Counting Principles
  • 9.6 Binomial Theorem
  • 9.7 Probability

Learning Objectives

In this section, you will:

  • Factor the greatest common factor of a polynomial.
  • Factor a trinomial.
  • Factor by grouping.
  • Factor a perfect square trinomial.
  • Factor a difference of squares.
  • Factor the sum and difference of cubes.
  • Factor expressions using fractional or negative exponents.

Imagine that we are trying to find the area of a lawn so that we can determine how much grass seed to purchase. The lawn is the green portion in Figure 1 .

The area of the entire region can be found using the formula for the area of a rectangle.

The areas of the portions that do not require grass seed need to be subtracted from the area of the entire region. The two square regions each have an area of A = s 2 = 4 2 = 16 A = s 2 = 4 2 = 16 units 2 . The other rectangular region has one side of length 10 x − 8 10 x − 8 and one side of length 4 , 4 , giving an area of A = l w = 4 ( 10 x − 8 ) = 40 x − 32 A = l w = 4 ( 10 x − 8 ) = 40 x − 32 units 2 . So the region that must be subtracted has an area of 2 ( 16 ) + 40 x − 32 = 40 x 2 ( 16 ) + 40 x − 32 = 40 x units 2 .

The area of the region that requires grass seed is found by subtracting 60 x 2 − 40 x 60 x 2 − 40 x units 2 . This area can also be expressed in factored form as 20 x ( 3 x − 2 ) 20 x ( 3 x − 2 ) units 2 . We can confirm that this is an equivalent expression by multiplying.

Many polynomial expressions can be written in simpler forms by factoring. In this section, we will look at a variety of methods that can be used to factor polynomial expressions.

Factoring the Greatest Common Factor of a Polynomial

When we study fractions, we learn that the greatest common factor (GCF) of two numbers is the largest number that divides evenly into both numbers. For instance, 4 4 is the GCF of 16 16 and 20 20 because it is the largest number that divides evenly into both 16 16 and 20 20 The GCF of polynomials works the same way: 4 x 4 x is the GCF of 16 x 16 x and 20 x 2 20 x 2 because it is the largest polynomial that divides evenly into both 16 x 16 x and 20 x 2 . 20 x 2 .

When factoring a polynomial expression, our first step should be to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables.

Greatest Common Factor

The greatest common factor (GCF) of polynomials is the largest polynomial that divides evenly into the polynomials.

Given a polynomial expression, factor out the greatest common factor.

  • Identify the GCF of the coefficients.
  • Identify the GCF of the variables.
  • Combine to find the GCF of the expression.
  • Determine what the GCF needs to be multiplied by to obtain each term in the expression.
  • Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by.

Factoring the Greatest Common Factor

Factor 6 x 3 y 3 + 45 x 2 y 2 + 21 x y . 6 x 3 y 3 + 45 x 2 y 2 + 21 x y .

First, find the GCF of the expression. The GCF of 6, 45, and 21 is 3. The GCF of x 3 , x 2 x 3 , x 2 , and x x is x x . (Note that the GCF of a set of expressions in the form x n x n will always be the exponent of lowest degree.) And the GCF of y 3 , y 2 y 3 , y 2 , and y y is y y . Combine these to find the GCF of the polynomial, 3 x y 3 x y .

Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that 3 x y ( 2 x 2 y 2 ) = 6 x 3 y 3 , 3 x y ( 15 x y ) = 45 x 2 y 2 3 x y ( 2 x 2 y 2 ) = 6 x 3 y 3 , 3 x y ( 15 x y ) = 45 x 2 y 2 , and 3 x y ( 7 ) = 21 x y . 3 x y ( 7 ) = 21 x y .

Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by.

After factoring, we can check our work by multiplying. Use the distributive property to confirm that ( 3 x y ) ( 2 x 2 y 2 + 15 x y + 7 ) = 6 x 3 y 3 + 45 x 2 y 2 + 21 x y . ( 3 x y ) ( 2 x 2 y 2 + 15 x y + 7 ) = 6 x 3 y 3 + 45 x 2 y 2 + 21 x y .

Factor x ( b 2 − a ) + 6 ( b 2 − a ) x ( b 2 − a ) + 6 ( b 2 − a ) by pulling out the GCF.

Factoring a Trinomial with Leading Coefficient 1

Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial x 2 + 5 x + 6 x 2 + 5 x + 6 has a GCF of 1, but it can be written as the product of the factors ( x + 2 ) ( x + 2 ) and ( x + 3 ) . ( x + 3 ) .

Trinomials of the form x 2 + b x + c x 2 + b x + c can be factored by finding two numbers with a product of c c and a sum of b . b . The trinomial x 2 + 10 x + 16 , x 2 + 10 x + 16 , for example, can be factored using the numbers 2 2 and 8 8 because the product of those numbers is 16 16 and their sum is 10. 10. The trinomial can be rewritten as the product of ( x + 2 ) ( x + 2 ) and ( x + 8 ) . ( x + 8 ) .

A trinomial of the form x 2 + b x + c x 2 + b x + c can be written in factored form as ( x + p ) ( x + q ) ( x + p ) ( x + q ) where p q = c p q = c and p + q = b . p + q = b .

Can every trinomial be factored as a product of binomials?

No. Some polynomials cannot be factored. These polynomials are said to be prime.

Given a trinomial in the form x 2 + b x + c , x 2 + b x + c , factor it.

  • List factors of c . c .
  • Find p p and q , q , a pair of factors of c c with a sum of b . b .
  • Write the factored expression ( x + p ) ( x + q ) . ( x + p ) ( x + q ) .

Factor x 2 + 2 x − 15. x 2 + 2 x − 15.

We have a trinomial with leading coefficient 1 , b = 2 , 1 , b = 2 , and c = −15. c = −15. We need to find two numbers with a product of −15 −15 and a sum of 2. 2. In the table below, we list factors until we find a pair with the desired sum.

Now that we have identified p p and q q as −3 −3 and 5 , 5 , write the factored form as ( x − 3 ) ( x + 5 ) . ( x − 3 ) ( x + 5 ) .

We can check our work by multiplying. Use FOIL to confirm that ( x − 3 ) ( x + 5 ) = x 2 + 2 x − 15. ( x − 3 ) ( x + 5 ) = x 2 + 2 x − 15.

Does the order of the factors matter?

No. Multiplication is commutative, so the order of the factors does not matter.

Factor x 2 − 7 x + 6. x 2 − 7 x + 6.

Factoring by Grouping

Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can factor by grouping by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial 2 x 2 + 5 x + 3 2 x 2 + 5 x + 3 can be rewritten as ( 2 x + 3 ) ( x + 1 ) ( 2 x + 3 ) ( x + 1 ) using this process. We begin by rewriting the original expression as 2 x 2 + 2 x + 3 x + 3 2 x 2 + 2 x + 3 x + 3 and then factor each portion of the expression to obtain 2 x ( x + 1 ) + 3 ( x + 1 ) . 2 x ( x + 1 ) + 3 ( x + 1 ) . We then pull out the GCF of ( x + 1 ) ( x + 1 ) to find the factored expression.

Factor by Grouping

To factor a trinomial in the form a x 2 + b x + c a x 2 + b x + c by grouping, we find two numbers with a product of a c a c and a sum of b . b . We use these numbers to divide the x x term into the sum of two terms and factor each portion of the expression separately, then factor out the GCF of the entire expression.

  • List factors of a c . a c .
  • Find p p and q , q , a pair of factors of a c a c with a sum of b . b .
  • Rewrite the original expression as a x 2 + p x + q x + c . a x 2 + p x + q x + c .
  • Pull out the GCF of a x 2 + p x . a x 2 + p x .
  • Pull out the GCF of q x + c . q x + c .
  • Factor out the GCF of the expression.

Factoring a Trinomial by Grouping

Factor 5 x 2 + 7 x − 6 5 x 2 + 7 x − 6 by grouping.

We have a trinomial with a = 5 , b = 7 , a = 5 , b = 7 , and c = −6. c = −6. First, determine a c = −30. a c = −30. We need to find two numbers with a product of −30 −30 and a sum of 7. 7. In the table below, we list factors until we find a pair with the desired sum.

So p = −3 p = −3 and q = 10. q = 10.

We can check our work by multiplying. Use FOIL to confirm that ( 5 x − 3 ) ( x + 2 ) = 5 x 2 + 7 x − 6. ( 5 x − 3 ) ( x + 2 ) = 5 x 2 + 7 x − 6.

  • ⓐ 2 x 2 + 9 x + 9 2 x 2 + 9 x + 9
  • ⓑ 6 x 2 + x − 1 6 x 2 + x − 1

Factoring a Perfect Square Trinomial

A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.

We can use this equation to factor any perfect square trinomial.

Perfect Square Trinomials

A perfect square trinomial can be written as the square of a binomial:

Given a perfect square trinomial, factor it into the square of a binomial.

  • Confirm that the first and last term are perfect squares.
  • Confirm that the middle term is twice the product of a b . a b .
  • Write the factored form as ( a + b ) 2 . ( a + b ) 2 .

Factor 25 x 2 + 20 x + 4. 25 x 2 + 20 x + 4.

Notice that 25 x 2 25 x 2 and 4 4 are perfect squares because 25 x 2 = ( 5 x ) 2 25 x 2 = ( 5 x ) 2 and 4 = 2 2 . 4 = 2 2 . Then check to see if the middle term is twice the product of 5 x 5 x and 2. 2. The middle term is, indeed, twice the product: 2 ( 5 x ) ( 2 ) = 20 x . 2 ( 5 x ) ( 2 ) = 20 x . Therefore, the trinomial is a perfect square trinomial and can be written as ( 5 x + 2 ) 2 . ( 5 x + 2 ) 2 .

Factor 49 x 2 − 14 x + 1. 49 x 2 − 14 x + 1.

Factoring a Difference of Squares

A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.

We can use this equation to factor any differences of squares.

Differences of Squares

A difference of squares can be rewritten as two factors containing the same terms but opposite signs.

Given a difference of squares, factor it into binomials.

  • Write the factored form as ( a + b ) ( a − b ) . ( a + b ) ( a − b ) .

Factor 9 x 2 − 25. 9 x 2 − 25.

Notice that 9 x 2 9 x 2 and 25 25 are perfect squares because 9 x 2 = ( 3 x ) 2 9 x 2 = ( 3 x ) 2 and 25 = 5 2 . 25 = 5 2 . The polynomial represents a difference of squares and can be rewritten as ( 3 x + 5 ) ( 3 x − 5 ) . ( 3 x + 5 ) ( 3 x − 5 ) .

Factor 81 y 2 − 100. 81 y 2 − 100.

Is there a formula to factor the sum of squares?

No. A sum of squares cannot be factored.

Factoring the Sum and Difference of Cubes

Now, we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial.

Similarly, the difference of cubes can be factored into a binomial and a trinomial, but with different signs.

We can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: S ame O pposite A lways P ositive. For example, consider the following example.

The sign of the first 2 is the same as the sign between x 3 − 2 3 . x 3 − 2 3 . The sign of the 2 x 2 x term is opposite the sign between x 3 − 2 3 . x 3 − 2 3 . And the sign of the last term, 4, is always positive .

Sum and Difference of Cubes

We can factor the sum of two cubes as

We can factor the difference of two cubes as

Given a sum of cubes or difference of cubes, factor it.

  • Confirm that the first and last term are cubes, a 3 + b 3 a 3 + b 3 or a 3 − b 3 . a 3 − b 3 .
  • For a sum of cubes, write the factored form as ( a + b ) ( a 2 − a b + b 2 ) . ( a + b ) ( a 2 − a b + b 2 ) . For a difference of cubes, write the factored form as ( a − b ) ( a 2 + a b + b 2 ) . ( a − b ) ( a 2 + a b + b 2 ) .

Factoring a Sum of Cubes

Factor x 3 + 512. x 3 + 512.

Notice that x 3 x 3 and 512 512 are cubes because 8 3 = 512. 8 3 = 512. Rewrite the sum of cubes as ( x + 8 ) ( x 2 − 8 x + 64 ) . ( x + 8 ) ( x 2 − 8 x + 64 ) .

After writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check.

Factor the sum of cubes: 216 a 3 + b 3 . 216 a 3 + b 3 .

Factoring a Difference of Cubes

Factor 8 x 3 − 125. 8 x 3 − 125.

Notice that 8 x 3 8 x 3 and 125 125 are cubes because 8 x 3 = ( 2 x ) 3 8 x 3 = ( 2 x ) 3 and 125 = 5 3 . 125 = 5 3 . Write the difference of cubes as ( 2 x − 5 ) ( 4 x 2 + 10 x + 25 ) . ( 2 x − 5 ) ( 4 x 2 + 10 x + 25 ) .

Just as with the sum of cubes, we will not be able to further factor the trinomial portion.

Factor the difference of cubes: 1,000 x 3 − 1. 1,000 x 3 − 1.

Factoring Expressions with Fractional or Negative Exponents

Expressions with fractional or negative exponents can be factored by pulling out a GCF. Look for the variable or exponent that is common to each term of the expression and pull out that variable or exponent raised to the lowest power. These expressions follow the same factoring rules as those with integer exponents. For instance, 2 x 1 4 + 5 x 3 4 2 x 1 4 + 5 x 3 4 can be factored by pulling out x 1 4 x 1 4 and being rewritten as x 1 4 ( 2 + 5 x 1 2 ) . x 1 4 ( 2 + 5 x 1 2 ) .

Factoring an Expression with Fractional or Negative Exponents

Factor 3 x ( x + 2 ) −1 3 + 4 ( x + 2 ) 2 3 . 3 x ( x + 2 ) −1 3 + 4 ( x + 2 ) 2 3 .

Factor out the term with the lowest value of the exponent. In this case, that would be ( x + 2 ) − 1 3 . ( x + 2 ) − 1 3 .

Factor 2 ( 5 a − 1 ) 3 4 + 7 a ( 5 a − 1 ) − 1 4 . 2 ( 5 a − 1 ) 3 4 + 7 a ( 5 a − 1 ) − 1 4 .

Access these online resources for additional instruction and practice with factoring polynomials.

  • Identify GCF
  • Factor Trinomials when a Equals 1
  • Factor Trinomials when a is not equal to 1
  • Factor Sum or Difference of Cubes

1.5 Section Exercises

If the terms of a polynomial do not have a GCF, does that mean it is not factorable? Explain.

A polynomial is factorable, but it is not a perfect square trinomial or a difference of two squares. Can you factor the polynomial without finding the GCF?

How do you factor by grouping?

For the following exercises, find the greatest common factor.

14 x + 4 x y − 18 x y 2 14 x + 4 x y − 18 x y 2

49 m b 2 − 35 m 2 b a + 77 m a 2 49 m b 2 − 35 m 2 b a + 77 m a 2

30 x 3 y − 45 x 2 y 2 + 135 x y 3 30 x 3 y − 45 x 2 y 2 + 135 x y 3

200 p 3 m 3 − 30 p 2 m 3 + 40 m 3 200 p 3 m 3 − 30 p 2 m 3 + 40 m 3

36 j 4 k 2 − 18 j 3 k 3 + 54 j 2 k 4 36 j 4 k 2 − 18 j 3 k 3 + 54 j 2 k 4

6 y 4 − 2 y 3 + 3 y 2 − y 6 y 4 − 2 y 3 + 3 y 2 − y

For the following exercises, factor by grouping.

6 x 2 + 5 x − 4 6 x 2 + 5 x − 4

2 a 2 + 9 a − 18 2 a 2 + 9 a − 18

6 c 2 + 41 c + 63 6 c 2 + 41 c + 63

6 n 2 − 19 n − 11 6 n 2 − 19 n − 11

20 w 2 − 47 w + 24 20 w 2 − 47 w + 24

2 p 2 − 5 p − 7 2 p 2 − 5 p − 7

For the following exercises, factor the polynomial.

7 x 2 + 48 x − 7 7 x 2 + 48 x − 7

10 h 2 − 9 h − 9 10 h 2 − 9 h − 9

2 b 2 − 25 b − 247 2 b 2 − 25 b − 247

9 d 2 −73 d + 8 9 d 2 −73 d + 8

90 v 2 −181 v + 90 90 v 2 −181 v + 90

12 t 2 + t − 13 12 t 2 + t − 13

2 n 2 − n − 15 2 n 2 − n − 15

16 x 2 − 100 16 x 2 − 100

25 y 2 − 196 25 y 2 − 196

121 p 2 − 169 121 p 2 − 169

4 m 2 − 9 4 m 2 − 9

361 d 2 − 81 361 d 2 − 81

324 x 2 − 121 324 x 2 − 121

144 b 2 − 25 c 2 144 b 2 − 25 c 2

16 a 2 − 8 a + 1 16 a 2 − 8 a + 1

49 n 2 + 168 n + 144 49 n 2 + 168 n + 144

121 x 2 − 88 x + 16 121 x 2 − 88 x + 16

225 y 2 + 120 y + 16 225 y 2 + 120 y + 16

m 2 − 20 m + 100 m 2 − 20 m + 100

25 p 2 − 120 p + 144 25 p 2 − 120 p + 144

36 q 2 + 60 q + 25 36 q 2 + 60 q + 25

For the following exercises, factor the polynomials.

x 3 + 216 x 3 + 216

27 y 3 − 8 27 y 3 − 8

125 a 3 + 343 125 a 3 + 343

b 3 − 8 d 3 b 3 − 8 d 3

64 x 3 −125 64 x 3 −125

729 q 3 + 1331 729 q 3 + 1331

125 r 3 + 1,728 s 3 125 r 3 + 1,728 s 3

4 x ( x − 1 ) − 2 3 + 3 ( x − 1 ) 1 3 4 x ( x − 1 ) − 2 3 + 3 ( x − 1 ) 1 3

3 c ( 2 c + 3 ) − 1 4 − 5 ( 2 c + 3 ) 3 4 3 c ( 2 c + 3 ) − 1 4 − 5 ( 2 c + 3 ) 3 4

3 t ( 10 t + 3 ) 1 3 + 7 ( 10 t + 3 ) 4 3 3 t ( 10 t + 3 ) 1 3 + 7 ( 10 t + 3 ) 4 3

14 x ( x + 2 ) − 2 5 + 5 ( x + 2 ) 3 5 14 x ( x + 2 ) − 2 5 + 5 ( x + 2 ) 3 5

9 y ( 3 y − 13 ) 1 5 − 2 ( 3 y − 13 ) 6 5 9 y ( 3 y − 13 ) 1 5 − 2 ( 3 y − 13 ) 6 5

5 z ( 2 z − 9 ) − 3 2 + 11 ( 2 z − 9 ) − 1 2 5 z ( 2 z − 9 ) − 3 2 + 11 ( 2 z − 9 ) − 1 2

6 d ( 2 d + 3 ) − 1 6 + 5 ( 2 d + 3 ) 5 6 6 d ( 2 d + 3 ) − 1 6 + 5 ( 2 d + 3 ) 5 6

Real-World Applications

For the following exercises, consider this scenario:

Charlotte has appointed a chairperson to lead a city beautification project. The first act is to install statues and fountains in one of the city’s parks. The park is a rectangle with an area of 98 x 2 + 105 x − 27 98 x 2 + 105 x − 27 m 2 , as shown in the figure below. The length and width of the park are perfect factors of the area.

Factor by grouping to find the length and width of the park.

A statue is to be placed in the center of the park. The area of the base of the statue is 4 x 2 + 12 x + 9   m 2 . 4 x 2 + 12 x + 9   m 2 . Factor the area to find the lengths of the sides of the statue.

At the northwest corner of the park, the city is going to install a fountain. The area of the base of the fountain is 9 x 2 − 25   m 2 . 9 x 2 − 25   m 2 . Factor the area to find the lengths of the sides of the fountain.

For the following exercise, consider the following scenario:

A school is installing a flagpole in the central plaza. The plaza is a square with side length 100 yd. as shown in the figure below. The flagpole will take up a square plot with area x 2 − 6 x + 9 x 2 − 6 x + 9 yd 2 .

Find the length of the base of the flagpole by factoring.

For the following exercises, factor the polynomials completely.

16 x 4 − 200 x 2 + 625 16 x 4 − 200 x 2 + 625

81 y 4 − 256 81 y 4 − 256

16 z 4 − 2,401 a 4 16 z 4 − 2,401 a 4

5 x ( 3 x + 2 ) − 2 4 + ( 12 x + 8 ) 3 2 5 x ( 3 x + 2 ) − 2 4 + ( 12 x + 8 ) 3 2

( 32 x 3 + 48 x 2 − 162 x − 243 ) −1 ( 32 x 3 + 48 x 2 − 162 x − 243 ) −1

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Access for free at https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
  • Authors: Jay Abramson
  • Publisher/website: OpenStax
  • Book title: College Algebra 2e
  • Publication date: Dec 21, 2021
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
  • Section URL: https://openstax.org/books/college-algebra-2e/pages/1-5-factoring-polynomials

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Polynomial Worksheets

These worksheets focus on the topics typically covered in Algebra I

  • Multiplying Monomials Worksheet
  • Multiplying and Dividing Monomials Sheet
  • Adding and Subtracting Polynomials worksheet
  • Multiplying Monomials with Polynomials Worksheet
  • Multiplying Binomials Worksheet
  • Multiplying Polynomials
  • Simplifying Polynomials
  • Factoring Trinomials
  • Operations with Polynomials Worksheet

Algebra 2 Polynomial Worksheets

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Factoring Polynomials

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Factoring Polynomials

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  • Writing variable expressions
  • Order of operations
  • Evaluating expressions
  • Number sets
  • Adding rational numbers
  • Adding and subtracting rational numbers
  • Multiplying and dividing rational numbers
  • The distributive property
  • Combining like terms
  • Percent of change
  • One-step equations
  • Two-step equations
  • Multi-step equations
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  • Percent problems
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  • Work word problems
  • Literal Equations
  • Graphing one-variable inequalities
  • One-step inequalities
  • Two-step inequalities
  • Multi-step inequalities
  • Compound inequalities
  • Absolute value inequalities
  • Discrete relations
  • Continuous relations
  • Evaluating and graphing functions
  • Finding slope from a graph
  • Finding slope from two points
  • Finding slope from an equation
  • Graphing lines using slope-intercept form
  • Graphing lines using standard form
  • Writing linear equations
  • Graphing linear inequalities
  • Graphing absolute value equations
  • Direct variation
  • Solving systems of equations by graphing
  • Solving systems of equations by elimination
  • Solving systems of equations by substitution
  • Systems of equations word problems
  • Graphing systems of inequalities
  • Discrete exponential growth and decay word problems
  • Exponential functions and graphs
  • Writing numbers in scientific notation
  • Operations with scientific notation
  • Addition and subtraction with scientific notation
  • Naming polynomials
  • Adding and subtracting polynomials
  • Multiplying polynomials
  • Multiplying special case polynomials
  • Factoring special case polynomials
  • Factoring by grouping
  • Dividing polynomials
  • Graphing quadratic inequalities
  • Completing the square
  • By taking square roots
  • By factoring
  • With the quadratic formula
  • By completing the square
  • Simplifying radicals
  • Adding and subtracting radical expressions
  • Multiplying radicals
  • Dividing radicals
  • Using the distance formula
  • Using the midpoint formula
  • Simplifying rational expressions
  • Finding excluded values / restricted values
  • Multiplying rational expressions
  • Dividing rational expressions
  • Adding and subtracting rational expressions
  • Finding trig. ratios
  • Finding angles of triangles
  • Finding side lengths of triangles
  • Visualizing data
  • Center and spread of data
  • Scatter plots
  • Using statistical models

Factoring Polynomials Worksheets

Tips for Factoring Polynomials - Factoring polynomials is one of the most basic calculations students learn in their quest for mathematical excellence. However, as polynomials become tougher, so is factorization practices. Thus, it's essential that you learn the tips for factoring polynomials before you commence with the actual process. Let us highlight those tips below. For any polynomial, no matter how many terms it has, you need to check for the greatest common factor first. In other words, in an expression, you search for a common term that occurs in the entire expression. For instance, you have an expression of 5x - 15. Here, the greatest common factor is 5. So, you can write the equation as 5(x-3). Remember, if you wish to resolve these equations, you have to equal it to zero. Thus, 5(x-3)=0. Now, you need to divide the GCF number throughout the expression. Like (5(x-3))/5=0. The equation then becomes (x-3) = 0. Now, you get the term x = 3. However, if the equation is trinomial (three terms), you can use the FOIL method for multiplying binomials backwards. Moreover, if it's a binomial, then you need to look for the difference of squares, the difference of cubes as well as the sum of cubes.

Basic Lesson

Guides students solving equations that involve an Factoring Polynomials. Demonstrates answer checking. Factor Polynomial: 8x+24 For solving the above equation we have to take the common factor 8. Place the common factor outside as a coefficient = 8(x+3).

Intermediate Lesson

Demonstrates how to solve more difficult problems. We have, Coefficient of x 2 = 2, Coefficient of x= 5, Constant term = 3. Now, try to split the coefficient of x that is 5 into two parts whose sum is 5 and product equals to the coefficient of x 2 and constant term that is 2 x 3 = 6.

Independent Practice 1

A really great activity for allowing students to understand the concept of Factoring Polynomials.

Independent Practice 2

Students find the Factoring Polynomials in assorted problems. The answers can be found below.

Homework Worksheet

Students are provided with problems to achieve the concepts of Factoring Polynomials.

This tests the students ability to evaluate Factoring Polynomials.

Answers for math worksheets, quiz, homework, and lessons.

Remember Who He Ran Against?

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    Factor Polynomial: 8x+24 For solving the above equation we have to take the common factor 8. Place the common factor outside as a coefficient = 8 (x+3). View worksheet Intermediate Lesson Demonstrates how to solve more difficult problems. We have, Coefficient of x 2 = 2, Coefficient of x= 5, Constant term = 3.