# 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:

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

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

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

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

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

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

## Methods

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

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

.

If ${\displaystyle x=-1,-2}$ , 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 ${\displaystyle x}$ that render the expression zero. These values are termed "roots." In general, given a quadratic polynomial ${\displaystyle px^{2}+qx+r}$ that factors as

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

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

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

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

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

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

### 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

${\displaystyle ax^{2}+bx+c}$

If there are numbers ${\displaystyle r}$ and ${\displaystyle q}$ that satisfy both

${\displaystyle r\cdot q=ac}$ and ${\displaystyle r+q=b}$

Then, we can rewrite the polynomial as

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

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

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

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

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

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

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

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

## Remainder and Factor Theorem

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

### Remainder Theorem

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

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

### Factor Theorem

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

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

#### Factor Theorem Example Problem

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

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

${\displaystyle x+2=x-\left(-2\right)}$

Now we can use the factor theorem.

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

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

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

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

Expanding the right hand side we get :

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

Equating like terms we get :

${\displaystyle 2=a}$

${\displaystyle 2a+b=3}$, and

${\displaystyle 2b=-2}$

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

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

## Example Problems

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

Finding the roots is equivalent to solving the equation ${\displaystyle 4x^{2}+7x-2=0}$ . Applying the quadratic formula with ${\displaystyle a=4\ ,\ b=7\ ,\ c=-2}$ , we have:
${\displaystyle x={\frac {-7\pm {\sqrt {7^{2}-4(4)(-2)}}}{2(4)}}}$
${\displaystyle x={\frac {-7\pm {\sqrt {49+32}}}{8}}}$
${\displaystyle x={\frac {-7\pm {\sqrt {81}}}{8}}}$
${\displaystyle x={\frac {-7\pm 9}{8}}}$
${\displaystyle x={\frac {2}{8}}\ ,\ x={\frac {-16}{8}}}$
${\displaystyle x={\frac {1}{4}}\ ,\ x=-2}$
The quadratic formula can also help with factoring, as the next example demonstrates.

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

We already know from the previous example that the polynomial has roots ${\displaystyle x={\frac {1}{4}}}$ and ${\displaystyle x=-2}$ . Our factorization will take the form
${\displaystyle C(x+2)\left(x-{\tfrac {1}{4}}\right)}$
.
All we have to do is set this expression equal to our polynomial and solve for the unknown constant C:
${\displaystyle C(x+2)\left(x-{\tfrac {1}{4}}\right)=4x^{2}+7x-2}$
${\displaystyle C\left(x^{2}+\left(-{\tfrac {1}{4}}+2\right)x-{\tfrac {2}{4}}\right)=4x^{2}+7x-2}$
${\displaystyle C\left(x^{2}+{\tfrac {7}{4}}x-{\tfrac {1}{2}}\right)=4x^{2}+7x-2}$
You can see that ${\displaystyle C=4}$ solves the equation. So the factorization is
${\displaystyle 4x^{2}+7x-2=4(x+2)\left(x-{\tfrac {1}{4}}\right)=(x+2)(4x-1)}$

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

For this problem, let's use the AC method. ${\displaystyle a=6}$, ${\displaystyle b=7}$, and ${\displaystyle c=2}$. So, ${\displaystyle ac=6(2)=12}$, 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 ${\displaystyle 1,2,3,4,6,}$and ${\displaystyle 12}$. ${\displaystyle 3(4)=12=ac}$ and ${\displaystyle 3+4=7=b}$, so we can rewrite the polynomial as ${\displaystyle 6x^{2}+3x+4x+2=(3x)(2x+1)=2(2x+1)=(3x+2)(2x+1)}$.