How to factor polynomials

The methods that are used to factor a polynomial depend on how many terms the polynomial has.
Note: This page assumes that the polynomials are being factored on the real field (that the coefficients of the factors are real numbers).

Any polynomial

The first step to factor any polynomial is always to factor out the Greatest Common Factor, commonly referred to as the GCF.

For example:

${\displaystyle 2a^{2}b^{3}+4a^{4}b^{4}c^{3}-2a^{3}b^{3}=2a^{2}b^{3}(1+2a^{2}bc^{3}-a),}$ 

or

${\displaystyle 3x^{3}-6x+9x^{4}=3x(x^{2}-2+3x^{3}),}$ 

or

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

Binomial (two terms)

Again, the first step is to factor out the GCF. If there is no GCF, then there are three possibilities: Difference of squares, sum of cubes, or difference of cubes.

Difference of squares

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

For example:

${\displaystyle y^{2}-9=(y+3)(y-3),}$ 

or

${\displaystyle 16a^{2}-49b^{2}=(4a+7b)(4a-7b).}$ 

Sum of cubes

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

For example:

${\displaystyle z^{3}+27=(z+3)(z^{2}-3z+9),}$ 

or

${\displaystyle 8x^{3}+125=(2x)^{3}+(5)^{3}=(2x+5)[(2x)^{2}-(5)(2x)+(5)^{2}]=(2x+5)(4x^{2}-10x+25).}$ 

Difference of cubes

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

For example:
${\displaystyle z^{3}-27=(z-3)(z^{2}+3z+9),}$ 

or

${\displaystyle 8x^{3}-125=(2x)^{3}-(5)^{3}=(2x-5)[(2x)^{2}+(5)(2x)+(5)^{2}]=(2x-5)(4x^{2}+10x+25).}$ 

Trinomial (three terms)

There are three possibilities for factoring a trinomial depending on which type of trinomial it is.

Monic trinomials

A monic trinomial has 1 as the leading coefficient.
${\displaystyle x^{2}+bx+c=(x+d)(x+e),}$       where ${\displaystyle (e)(d)=c}$  and ${\displaystyle d+e=b}$ .

For example:
${\displaystyle x^{2}-x-6=(x-3)(x+2)}$  because ${\displaystyle (-3)(2)=-6}$  and ${\displaystyle -3+2=-1}$ 

or

${\displaystyle x^{2}-11x+24=(x-3)(x-8)}$  because ${\displaystyle (-3)(-8)=24}$  and ${\displaystyle -3+-8=-11}$ 

Non-monic trinomials

A non-monic trinomial has a constant other than 1 as the leading coefficient.

${\displaystyle ax^{2}+bx+c=(mx+p)(nx+q)}$ 
where ${\displaystyle mn=a}$ , ${\displaystyle pq=c}$ , and ${\displaystyle mq+pn=b}$ 

Many times students are taught that to factor a non-monic trinomials, they must guess different combinations of m,n,p,and q and then FOIL the factors to see if they had guessed correctly. There is a method of factoring that, while not often taught, will work.

Example:
Factor ${\displaystyle 6x^{2}+7x-3}$ 

Step 1   Multiply a and c. (Multiply the number in front of ${\displaystyle x^{2}}$  and the constant)
Multipy 6 and -3, ${\displaystyle 6(-3)=-18}$ 

Step 2   Find factors of ac.
Find factors of -18: -1(18), 1(-18), -2(9), 2(-9), -3(6), and 3(-6).

Step 3   Decide which factors of ac that when added together will give b.
The combination of -2 and 9 is the one needed since -2+9=7.

Step 4   Rewrite the middle term of bx using the factors found in step 3.
Instead of ${\displaystyle 7x}$ , write ${\displaystyle 6x^{2}-2x+9x-3}$ 

Step 5   Factor by grouping.
${\displaystyle (6x^{2}-2x)+(9x-3)}$ 
${\displaystyle 2x(3x-1)+3(3x-1)}$ 
${\displaystyle (3x-1)(2x+3)}$ 

Therefore ${\displaystyle 6x^{2}+7x-3=(3x-1)(2x+3)}$ 

Perfect square trinomials

Perfect square trinomials are of the form ${\displaystyle a^{2}+2ab+b^{2}}$  or ${\displaystyle a^{2}-2ab+b^{2}}$ 
${\displaystyle a^{2}+2ab+b^{2}=(a+b)^{2}}$  and
${\displaystyle a^{2}-2ab+b^{2}=(a-b)^{2}}$ 

For example:
${\displaystyle 4h^{2}-20h+25=(2h)^{2}-2(2h)(5)+(5)^{2}=(2h-5)^{2}}$ 

or

${\displaystyle 100x^{2}+180xy+81y^{2}=(10x)^{2}+2(10x)(9y)+(9y)^{2}=(10x+9y)^{2}}$ 

Polynomials with four terms

Some polynomials with four terms can be factored by some form of grouping. There are special groupings but the most common form is referred to as factoring by grouping and is described step by step below.

Example:
${\displaystyle a^{2}-3ab+4ac-12bc}$ 

Step 1   Split the polynomial into groups of 2 terms.
${\displaystyle (a^{2}-3ab)+(4ac-12bc)}$ 

Step 2   Find the GCF (greatest common factor) of each group.
${\displaystyle a(a-3b)+4c(a-3b)}$ 

Step 3   If the 'leftovers match' factor them out.
Since there is a (a-3b) in each term, factor out (a-3b) from each term.
${\displaystyle a\mathbf {(a-3b)} +4c\mathbf {(a-3b)} =(a-3b)(a+4c)}$ 

Therefore ${\displaystyle a^{2}-3ab+4ac-12bc=(a-3b)(a+4c)}$ 

When nothing works

If the polynomial can't be factored, then it is considered prime.

References

Lial, Margaret L.; Hornsby, John; McGinnis, Terry (2008). Intermediate Algebra 10th edition. Boston: Addison Wesley. pp. 346–388. ISBN 978-0321443625.

Tussy, Alan S.; Gustafson, R. David (2008). Intermediate Algebra 4th edition. Brooks Cole. ISBN 978-0495389736. {{cite book}}: Unknown parameter |month= ignored (help)

External links

• Factoring Quadratics: The Simple Case