Factoring Polynomials

Useful Formulas

Computing factors of polynomials requires knowledge of different formulas and some experience to find out which formula to be applied. Below, we give some important formulas:

${x^{2}-y^{2}=(x+y)(x-y)}$

${x^{2}+2xy+y^{2}=(x+y)^{2}}$

${x^{2}-2xy+y^{2}=(x-y)^{2}}$

${x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})}$

${x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})}$

${a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-ac-bc)}$

Methods

Given the expression $x^{2}+3x+2$ , one may ask "what are the values of $x$ that make this expression 0?" If we factor we obtain

$x^{2}+3x+2=(x+2)(x+1)$

.

If $x=-1,-2$ , then one of the factors on the right becomes zero. Therefore, the whole must be zero. So, by factoring we have discovered the values of $x$ that render the expression zero. These values are termed "roots." In general, given a quadratic polynomial $px^{2}+qx+r$ that factors as

$px^{2}+qx+r=(ax+c)(bx+d)$

then we have that $x=-{\frac {c}{a}}$ and $x=-{\frac {d}{b}}$ are roots of the original polynomial.

A special case to be on the look out for is the difference of two squares, $a^{2}-b^{2}$ . In this case, we are always able to factor as

$a^{2}-b^{2}=(a+b)(a-b)$

For example, consider $4x^{2}-9$ . On initial inspection we would see that both $4x^{2}$ and $9$ are squares of $2x$ and $3$, respectively. Applying the previous rule we have

$4x^{2}-9=(2x+3)(2x-3)$

The AC method

There is a way of simplifying the process of factoring using the AC method. Suppose that a quadratic polynomial has a formula of

$ax^{2}+bx+c$

If there are numbers $r$ and $q$ that satisfy both

$r\cdot q=ac$ and $r+q=b$

Then, we can rewrite the polynomial as

$ax^{2}+bx+c=ax^{2}+rx+qx+c$

and factor out a common term from $(ax^{2}+rx)$ and $(qx+c)$ to factor the polynomial.

For example, take the polynomial $2x^{2}+7x-15$.

$ac=2(-15)=-30$. $10(-3)=-30=ac$ and $10+(-3)=7=b$, so we can rewrite the polynomial as $2x^{2}+10x-3x-15$. From here, we can factor $2x$ from the first two terms, and $-3$ from the last two, which gives us $2x(x+5)+(-3)(x+5)=(2x-3)(x+5)$. Thus, the factored form of $2x^{2}+7x-15$ is $(2x-3)(x+5)$. Note that if we reverse the order of $10x$ and $-3x$, we get $2x^{2}-3x+10x-15=x(2x-3)+5(2x-3)=(x+5)(2x-3)$, which is the same factored form that we got previously. Thus, the order of $rx$ and $qx$ should not matter when using the AC method to factor a quadratic polynomial.

The quadratic formula

Given any quadratic equation $ax^{2}+bx+c=0\ ,\ a\neq 0$, all solutions of the equation are given by the quadratic formula:

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$

Note that the value of $b^{2}-4ac$ will affect the number of real solutions of the equation.

If	Then
$b^{2}-4ac>0$	There are two real solutions for the equation
$b^{2}-4ac=0$	There are only one real solutions for the equation
$b^{2}-4ac<0$	There are no real solutions for the equation

Remainder and Factor Theorem

The polynomial division algorithm is as follows: suppose $d(x)$ and $p(x)$ are nonzero polynomials where the degree of $p(x)$ is greater than or equal to the degree of $d(x)$. Then there exist two unique polynomials, $q(x)$ and $r(x)$, such that $p(x)=d(x)q(x)+r(x)$, where either $r(x)=0$ or the degree of $r(x)$ is strictly less than the degree of $d(x)$.

Remainder Theorem

Suppose $p(x)$ is a polynomial of degree at least 1 and c is a real number. When $p(x)$ is divided by $(x-c)$ the remainder is $p(c)$.

Proof: By the division algorithm, $p(x)=(x-c)q(x)+r$, where r must be a constant since $d(x)=x-c$ has a degree of 1. $p(x)=(x-c)q(x)+r$ must hold for all values of $x$, so we can set $x=c$ and get that $p(c)=(c-c)q(x)+r=r$. Thus the remainder $r=p(c)$.

Factor Theorem

Suppose $p(x)$ is a nonzero polynomial. The real number $c$ is a zero of $p(x)$ if and only if $(x-c)$ is a factor of $p(x)$.

By the division algorithm, $x-c$ is a factor of $p(x)$ if and only if $r=0$. So, since $p(c)=r$ when $p(x)$ is divided by $x-c$, $x-c$ is a factor of $p(x)$ if and only if $p(c)=0$; that is, if $c$ is a zero of $p(x)$.

Factor Theorem Example Problem

Determine if x + 2 is a factor of $2x^{2}+3x-2$.

Since c is positive instead of negative we need to use this basic identity:

$x+2=x-\left(-2\right)$

Now we can use the factor theorem.

$2\left(-2\right)^{2}+3\left(-2\right)-2=8-6-2=0$.

Since the resultant is 0, $(x+2)$ is a factor of $2x^{2}+3x-2$.

This means it is possible to re-state the polynomial in the form (x+2)( some linear expression of x).

So $2x^{2}+3x-2=(x+2)(ax+b)$

Expanding the right hand side we get :

$2x^{2}+3x-2=ax^{2}+x(2a+b)+2b$

Equating like terms we get :

$2=a$

$2a+b=3$, and

$2b=-2$

Giving $a=2$, $b=-1$ from the first and third equations and this works in the second, so

$2x^{2}+3x-2=(x+2)(2x-1)$

Example Problems

EXAMPLE 1: Find all the roots of $4x^{2}+7x-2$

Finding the roots is equivalent to solving the equation $4x^{2}+7x-2=0$ . Applying the quadratic formula with $a=4\ ,\ b=7\ ,\ c=-2$ , we have:

$x={\frac {-7\pm {\sqrt {7^{2}-4(4)(-2)}}}{2(4)}}$

$x={\frac {-7\pm {\sqrt {49+32}}}{8}}$

$x={\frac {-7\pm {\sqrt {81}}}{8}}$

$x={\frac {-7\pm 9}{8}}$

$x={\frac {2}{8}}\ ,\ x={\frac {-16}{8}}$

$x={\frac {1}{4}}\ ,\ x=-2$

The quadratic formula can also help with factoring, as the next example demonstrates.

EXAMPLE 2: Factor the polynomial $4x^{2}+7x-2$

We already know from the previous example that the polynomial has roots $x={\frac {1}{4}}$ and $x=-2$ . Our factorization will take the form
$C(x+2)\left(x-{\tfrac {1}{4}}\right)$
.

All we have to do is set this expression equal to our polynomial and solve for the unknown constant C:

$C(x+2)\left(x-{\tfrac {1}{4}}\right)=4x^{2}+7x-2$

$C\left(x^{2}+\left(-{\tfrac {1}{4}}+2\right)x-{\tfrac {2}{4}}\right)=4x^{2}+7x-2$

$C\left(x^{2}+{\tfrac {7}{4}}x-{\tfrac {1}{2}}\right)=4x^{2}+7x-2$

You can see that $C=4$ solves the equation. So the factorization is

$4x^{2}+7x-2=4(x+2)\left(x-{\tfrac {1}{4}}\right)=(x+2)(4x-1)$

EXAMPLE 3: Factor the polynomial $6x^{2}+7x+2$.

For this problem, let's use the AC method. $a=6$, $b=7$, and $c=2$. So, $ac=6(2)=12$, and we need to find two numbers whose product equals 12, and whose sum equals 7. Both ac and b are positive, so we are only concerned with the positive factors of 12, which are $1,2,3,4,6,$and $12$. $3(4)=12=ac$ and $3+4=7=b$, so we can rewrite the polynomial as $6x^{2}+3x+4x+2=(3x)(2x+1)=2(2x+1)=(3x+2)(2x+1)$.

Resources

• Factoring Polynomials with Example Problems, Paul's Online Notes
• Factoring Polynomials, Lumen Learning
• Difference of Squares. Produced by TA Catherine Sporer, UTSA
• 6 Methods of Factoring with Practive Problems, AlgebraLAB
• AC Factoring Method, Texas Wesleyan University

Licensing

Content obtained and/or adapted from:

• Algebra, Wikibooks: Calculus under a CC BY-SA license
• Factoring Polynomials, Wikibooks: Algebra under a CC BY-SA license