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Factoring Polynomials

Factor the greatest common factor of a polynomial.

Factoring Polynomials

Master a proven technique for note taking: The Cornell Method.
Use the Cornell Process to study from your notes.

The Cornell Way. Using your Cornell notes to study is referred to as the Cornell Way.
The process includes Cornell note taking (presented in Objective 1), note making, note interacting, and note reflecting.
After taking notes in class, follow up with these study methods.

Note making: This is where you will fill in any gaps left in your Cornell notes meaning add in any missed details. Fill in your question column with sample test questions, any formulas used, and highlight, circle or star important any important ideas. Complete your summary row with a few sentences summarizing the important ideas. This should be completed right after class or within the next day. Be creative and add some color, make these notes into something you enjoy working with.

Note interacting: This is the ongoing process of studying from your notes. Fold your left question column back and ask yourself the practice test questions. Include note interactions in the review for your exams. These note interactions can be as short as 5 minutes in length but need to happen regularly and at least daily for the week before exams.

Note reflecting: This is where you assess how helpful your notes were. Do this right after you get back your graded exam. Did the regular note interactions help you to perform better on the exam? Were there problems similar to those you predicted on the exam? Enhance your notes with any important ideas you had initially left out. Then use what you learn from this assessment to improve your note taking in the future.

Try it.

Complete the Cornell Process by revisiting your factoring of polynomials summary notes. For now, follow the Note Making, and Note Interacting steps outlined above. Keep these Cornell notes in your divided binder for the course so you can easily refer back when you encounter an exercise requiring factoring throughout the term.

Try it.

Share these completed Cornell notes with your classmates. Fill in any gaps in your work. Work on each other’s practice problems.

Condensed — the full section is in OpenStax College Algebra 2e.

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, \(4\) is the GCF of \(16\) and \(20\) because it is the largest number that divides evenly into both \(16\) and \(20\) The GCF of polynomials works the same way: \(4x\) is the GCF of \(16x\) and \(20{x}^{2}\) because it is the largest polynomial that divides evenly into both \(16x\) and \(20{x}^{2}.\)

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.

Example

Try it.

Factor \(6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy.\)

Solution

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

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

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

\[(3xy)(2{x}^{2}{y}^{2}+15xy+7)\]

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 \({x}^{2}+5x+6\) has a GCF of 1, but it can be written as the product of the factors \((x+2)\) and \((x+3).\)

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

Example

Try it.

Factor \({x}^{2}+2x-15.\)

Solution

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

Factors of \(-15\)	Sum of Factors
\(1,-15\)	\(-14\)
\(-1,15\)	14
\(3,-5\)	\(-2\)
\(-3,5\)	2

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

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 \(2{x}^{2}+5x+3\) can be rewritten as \((2x+3)(x+1)\) using this process. We begin by rewriting the original expression as \(2{x}^{2}+2x+3x+3\) and then factor each portion of the expression to obtain \(2x(x+1)+3(x+1).\) We then pull out the GCF of \((x+1)\) to find the factored expression.

Example

Try it.

Factor \(5{x}^{2}+7x-6\) by grouping.

Solution

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

Factors of \(-30\)	Sum of Factors
\(1,-30\)	\(-29\)
\(-1,30\)	29
\(2,-15\)	\(-13\)
\(-2,15\)	13
\(3,-10\)	\(-7\)
\(-3,10\)	7

So \(p=-3\) and \(q=10.\)

\[\begin{array}{ll}5{x}^{2}-3x+10x-6 & \ \text{Rewrite the original expression as }a{x}^{2}+px+qx+c. \\ x(5x-3)+2(5x-3) & \ \text{Factor out the GCF of each part}. \\ (5x-3)(x+2) & \ \text{Factor out the GCF}\text{}\ \text{of the expression}.\end{array}\]

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.

\[\begin{array}{lll}{a}^{2}+2ab+{b}^{2} & = & {(a+b)}^{2} \\ & \text{and} & \\ {a}^{2}-2ab+{b}^{2} & = & {(a-b)}^{2}\end{array}\]

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

Example

Try it.

Factor \(25{x}^{2}+20x+4.\)

Solution

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

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.

\[{a}^{2}-{b}^{2}=(a+b)(a-b)\]

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

Example

Try it.

Factor \(9{x}^{2}-25.\)

Solution

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

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.

\[{a}^{3}+{b}^{3}=(a+b)({a}^{2}-ab+{b}^{2})\]

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

\[{a}^{3}-{b}^{3}=(a-b)({a}^{2}+ab+{b}^{2})\]

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: Same Opposite Always Positive. For example, consider the following example.

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

The sign of the first 2 is the same as the sign between \({x}^{3}-{2}^{3}.\) The sign of the \(2x\) term is opposite the sign between \({x}^{3}-{2}^{3}.\) And the sign of the last term, 4, is always positive.

Example

Try it.

Factor \({x}^{3}+512.\)

Solution

Notice that \({x}^{3}\) and \(512\) are cubes because \({8}^{3}=512.\) Rewrite the sum of cubes as \((x+8)({x}^{2}-8x+64).\)

Example

Try it.

Factor \(8{x}^{3}-125.\)

Solution

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

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, \(2{x}^{\frac{1}{4}}+5{x}^{\frac{3}{4}}\) can be factored by pulling out \({x}^{\frac{1}{4}}\) and being rewritten as \({x}^{\frac{1}{4}}(2+5{x}^{\frac{1}{2}}).\)

Example

Try it.

Factor \(3x{(x+2)}^{\frac{-1}{3}}+4{(x+2)}^{\frac{2}{3}}.\)

Solution

Factor out the term with the lowest value of the exponent. In this case, that would be \({(x+2)}^{-\frac{1}{3}}.\)

\[\begin{array}{ll}{(x+2)}^{-\frac{1}{3}}(3x+4(x+2)) & \ \text{Factor out the GCF}. \\ {(x+2)}^{-\frac{1}{3}}(3x+4x+8) & \ \text{Simplify}. \\ {(x+2)}^{-\frac{1}{3}}(7x+8) & \end{array}\]

Key Equations

difference of squares	\({a}^{2}-{b}^{2}=(a+b)(a-b)\)
perfect square trinomial	\({a}^{2}+2ab+{b}^{2}={(a+b)}^{2}\)
sum of cubes	\({a}^{3}+{b}^{3}=(a+b)({a}^{2}-ab+{b}^{2})\)
difference of cubes	\({a}^{3}-{b}^{3}=(a-b)({a}^{2}+ab+{b}^{2})\)

Factoring Polynomials

The greatest common factor, or GCF, can be factored out of a polynomial. Checking for a GCF should be the first step in any factoring problem. See .
Trinomials with leading coefficient 1 can be factored by finding numbers that have a product of the third term and a sum of the second term. See .
Trinomials can be factored using a process called factoring by grouping. See .
Perfect square trinomials and the difference of squares are special products and can be factored using equations. See and .
The sum of cubes and the difference of cubes can be factored using equations. See and .
Polynomials containing fractional and negative exponents can be factored by pulling out a GCF. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

The following book section includes a variety of methods for factoring polynomials. Read the section carefully and complete Cornell notes for the process of factoring. This work will help you throughout the semester because the ability to factor polynomials is one of those linchpin topics that will continue to emerge throughout the term. Remember to write practice factoring examples in the left question or cues column.
Need some inspiration? Search for videos on Cornell note taking on the web. There are some interesting videos of students showing off their beautiful notes that have been enhanced with highlighters and sketches. Many of the students talk about the difference this technique has made in their learning and attribute their success to this Cornell format. Another topic to search is sketchnoting, which helps to visually enhance your Cornell notes and truly tell a story.
Complete the Cornell Process by revisiting your factoring of polynomials summary notes. For now, follow the Note Making, and Note Interacting steps outlined above. Keep these Cornell notes in your divided binder for the course so you can easily refer back when you encounter an exercise requiring factoring throughout the term.
Share these completed Cornell notes with your classmates. Fill in any gaps in your work. Work on each other’s practice problems.
Factor \(6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy.\)

Otkrij odgovor

First, find the GCF of the expression. The GCF of 6, 45, and 21 is 3. The GCF of \({x}^{3},{x}^{2}\) , and \(x\) is \(x\) . (Note that the GCF of a set of expressions in the form \({x}^{n}\) will always be the exponent of lowest degree.) And the GCF of \({y}^{3},{y}^{2}\) , and \(y\) is \(y\) . Combine these to find the GCF of the polynomial, \(3xy\) .
Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that \(3xy(2{x}^{2}{y}^{2})=6{x}^{3}{y}^{3},3xy(15xy)=45{x}^{2}{y}^{2}\) , and \(3xy(7)=21xy.\)
Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by.
\[(3xy)(2{x}^{2}{y}^{2}+15xy+7)\]
Factor \(x({b}^{2}-a)+6({b}^{2}-a)\) by pulling out the GCF.

Otkrij odgovor

\(({b}^{2}-a)(x+6)\)
Factor \({x}^{2}+2x-15.\)

Otkrij odgovor

We have a trinomial with leading coefficient \(1,b=2,\) and \(c=-15.\) We need to find two numbers with a product of \(-15\) and a sum of \(2.\) In the table below, we list factors until we find a pair with the desired sum.
Factors of \(-15\) Sum of Factors
\(1,-15\) \(-14\)
\(-1,15\) 14
\(3,-5\) \(-2\)
\(-3,5\) 2
Now that we have identified \(p\) and \(q\) as \(-3\) and \(5,\) write the factored form as \((x-3)(x+5).\)
Factor \({x}^{2}-7x+6.\)

Otkrij odgovor

\((x-6)(x-1)\)
Factor \(5{x}^{2}+7x-6\) by grouping.

Otkrij odgovor

We have a trinomial with \(a=5,b=7,\) and \(c=-6.\) First, determine \(ac=-30.\) We need to find two numbers with a product of \(-30\) and a sum of \(7.\) In the table below, we list factors until we find a pair with the desired sum.
Factors of \(-30\) Sum of Factors
\(1,-30\) \(-29\)
\(-1,30\) 29
\(2,-15\) \(-13\)
\(-2,15\) 13
\(3,-10\) \(-7\)
\(-3,10\) 7
So \(p=-3\) and \(q=10.\)
\[\begin{array}{ll}5{x}^{2}-3x+10x-6 & \ \text{Rewrite the original expression as }a{x}^{2}+px+qx+c. \\ x(5x-3)+2(5x-3) & \ \text{Factor out the GCF of each part}. \\ (5x-3)(x+2) & \ \text{Factor out the GCF}\text{}\ \text{of the expression}.\end{array}\]
Factor
1. ⓐ \(2{x}^{2}+9x+9\)
2. ⓑ \(6{x}^{2}+x-1\)
Otkrij odgovor
1. ⓐ \((2x+3)(x+3)\)
2. ⓑ \((3x-1)(2x+1)\)
Factor \(25{x}^{2}+20x+4.\)

Otkrij odgovor

Notice that \(25{x}^{2}\) and \(4\) are perfect squares because \(25{x}^{2}={(5x)}^{2}\) and \(4={2}^{2}.\) Then check to see if the middle term is twice the product of \(5x\) and \(2.\) The middle term is, indeed, twice the product: \(2(5x)(2)=20x.\) Therefore, the trinomial is a perfect square trinomial and can be written as \({(5x+2)}^{2}.\)
Factor \(49{x}^{2}-14x+1.\)

Otkrij odgovor

\({(7x-1)}^{2}\)
Factor \(9{x}^{2}-25.\)

Otkrij odgovor

Notice that \(9{x}^{2}\) and \(25\) are perfect squares because \(9{x}^{2}={(3x)}^{2}\) and \(25={5}^{2}.\) The polynomial represents a difference of squares and can be rewritten as \((3x+5)(3x-5).\)
Factor \(81{y}^{2}-100.\)

Otkrij odgovor

\((9y+10)(9y-10)\)
Factor \({x}^{3}+512.\)

Otkrij odgovor

Notice that \({x}^{3}\) and \(512\) are cubes because \({8}^{3}=512.\) Rewrite the sum of cubes as \((x+8)({x}^{2}-8x+64).\)
Factor the sum of cubes: \(216{a}^{3}+{b}^{3}.\)

Otkrij odgovor

\((6a+b)(36{a}^{2}-6ab+{b}^{2})\)
Factor \(8{x}^{3}-125.\)

Otkrij odgovor

Notice that \(8{x}^{3}\) and \(125\) are cubes because \(8{x}^{3}={(2x)}^{3}\) and \(125={5}^{3}.\) Write the difference of cubes as \((2x-5)(4{x}^{2}+10x+25).\)
Factor the difference of cubes: \(1,000{x}^{3}-1.\)

Otkrij odgovor

\((10x-1)(100{x}^{2}+10x+1)\)
Factor \(3x{(x+2)}^{\frac{-1}{3}}+4{(x+2)}^{\frac{2}{3}}.\)

Otkrij odgovor

Factor out the term with the lowest value of the exponent. In this case, that would be \({(x+2)}^{-\frac{1}{3}}.\)
\[\begin{array}{ll}{(x+2)}^{-\frac{1}{3}}(3x+4(x+2)) & \ \text{Factor out the GCF}. \\ {(x+2)}^{-\frac{1}{3}}(3x+4x+8) & \ \text{Simplify}. \\ {(x+2)}^{-\frac{1}{3}}(7x+8) & \end{array}\]
Factor \(2{(5a-1)}^{\frac{3}{4}}+7a{(5a-1)}^{-\frac{1}{4}}.\)

Otkrij odgovor

\({(5a-1)}^{-\frac{1}{4}}(17a-2)\)
If the terms of a polynomial do not have a GCF, does that mean it is not factorable? Explain.

Otkrij odgovor

The terms of a polynomial do not have to have a common factor for the entire polynomial to be factorable. For example, \(4{x}^{2}\) and \(-9{y}^{2}\) don’t have a common factor, but the whole polynomial is still factorable: \(4{x}^{2}-9{y}^{2}=(2x+3y)(2x-3y).\)
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?

Otkrij odgovor

Divide the \(x\) term into the sum of two terms, factor each portion of the expression separately, and then factor out the GCF of the entire expression.
\(14x+4xy-18x{y}^{2}\)
\(49m{b}^{2}-35{m}^{2}ba+77m{a}^{2}\)

Otkrij odgovor

\(7m\)
\(30{x}^{3}y-45{x}^{2}{y}^{2}+135x{y}^{3}\)
\(200{p}^{3}{m}^{3}-30{p}^{2}{m}^{3}+40{m}^{3}\)

Otkrij odgovor

\(10{m}^{3}\)
\(36{j}^{4}{k}^{2}-18{j}^{3}{k}^{3}+54{j}^{2}{k}^{4}\)
\(6{y}^{4}-2{y}^{3}+3{y}^{2}-y\)

Otkrij odgovor

\(y\)
\(6{x}^{2}+5x-4\)
\(2{a}^{2}+9a-18\)

Otkrij odgovor

\((2a-3)(a+6)\)
\(6{c}^{2}+41c+63\)
\(6{n}^{2}-19n-11\)

Otkrij odgovor

\((3n-11)(2n+1)\)
\(20{w}^{2}-47w+24\)
\(2{p}^{2}-5p-7\)

Otkrij odgovor

\((p+1)(2p-7)\)
\(7{x}^{2}+48x-7\)
\(10{h}^{2}-9h-9\)

Otkrij odgovor

\((5h+3)(2h-3)\)
\(2{b}^{2}-25b-247\)
\(9{d}^{2}-73d+8\)

Otkrij odgovor

\((9d-1)(d-8)\)
\(90{v}^{2}-181v+90\)
\(12{t}^{2}+t-13\)

Otkrij odgovor

\((12t+13)(t-1)\)

Symbols used here

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Factoring Polynomials

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.
Master a proven technique for note taking: The Cornell Method.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Pokušaj sam.

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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