Factoring Polynomials

To understand how to factor polynomials, it is first helpful to understand something about factoring numbers.  Factoring numbers is process whereby we determine what numbers are multiplied together to form a given quantity.  For example

18 = (6)(3)

18 = (-6)(-3)

18 = (2)(3)(3)

The last example is considered to be completely factored, since it is broken down into it’s prime factors (2 and 3 are prime numbers).  When factoring polynomials it is done in a similar fashion.  We are trying to determine what factors multiplied together equal the given polynomial.

Methods of Factoring

  1. Given a polynomial ax2+bx+c, where a, b, and c are constants it can be factored by finding all possible factors of c, and then determining which pair of factors add up to b (given that a=1).  And finding inner product + outer product that add to give b (when a ≠ 1).
  2. a2 – b2 = (a+b)(a-b)     Difference of perfect squares.
  3. a3 – b3 = (a-b)(a2+ab+b2)    Difference of perfect cubes.
  4. a3 + b3 = (a+b)(a2 – ab+b2)    Sum of perfect cubes.
  5. Factoring by grouping, this is useful when a≠1 from method 1
  6. Using “-b” formula to first get roots of the quadratic equation and then form factors.
  7. Find the roots by trial and error and then use them to form the factors.

Examples of Factoring

Factor polynomial of form ax2+bx+c using method 1

1) x2+5x+4

first start by listing all factors of 4:

4 = (1)(4)

4 = (2)(2)

4 = (-1)(-4)

4 = (-2)(-2)

now we need to determine which set of these factors adds up to the value of b, which is 5

1+4 = 5

it appears that the first group of factors satisfies that condition, hence the factored form of the polynomial is:

x2+5x+4 = (x+1)(x+4)

Factor the following polynomial using method 5

2) 3x2+16x+5

in this case a = 3.  So we will use the method of factoring by grouping.  That means we will now find all possible factors of (a)(c) = (3)(5) = 15

So we need to list all possible factors of 15 which add up to b = 16

15 = (1)(15)

15 = (3)(5)

15 = (-1)(-15)

15 = (-3)(-5)

now we need to determine which set of these factors adds up to the value of b, which is 16

1+15 = 16

So it appears our first group of factors satisfies that condition, now we rewrite the polynomial in the following form

3x2+16x+5 = 3x2+1x+15x+5

note that the polynomial are equivalent.  However, the way in which the new polynomial is written allows us to factor it by grouping

3x2+1x+15x+5 = x(3x+1) + 5(3x+1)

notice that this is the sum of two terms x(3x+1) and 5(3x+1), which both have a similar factor of (3x+1).  That means we can factor out the (3x+1) and we are left with,

x(3x+1)+5(3x+1) = (x+5)(3x+1)


3x2+16x+5 = (x+5)(3x+1)

Factor the difference of perfect squares using method 2

3) 9x2-81

first we need to write this in the form a2-b2

9x2-81 = (3x)2 – 92

where a = 3x, and b=9.  Since

a2 – b2 = (a+b)(a-b)


(3x)2 – 92 = (3x+9)(3x-9)

and we can see that

9x2-81 = (3x+3)(3x-3)

Factoring the difference of perfect squares can be regarded as a special case of method 1 above where the coefficient of the x-term is zero.

When given something like x2 – 9, we can rewrite it as x2+0x-9.
To factor this, we will need two numbers that add up to 0 and multiply to -9. These numbers are 3 and -3. Therefore when we factor this, we will get (x+3)(x-3).

Generally, whenever we have x2-b2   we will always be able to factor it as (x+b)(x-b).

Or (x2 – number) will always be factored as (x+number)(x-number).

For expressions more complex, such as 4x2-12y4;
First, take the 
square root of the first term and place it in the front of each set of parentheses, and then take the square root of the second term and place it in the two parentheses with a plus or minus sign. This will be factored as (2x-11y2)(2x-11y2).

Factor the difference of perfect cubes using method 3

4)  27x3– 64y3

first write this in the form a3-b3

27x3– 64y3 = (3x)3 – (4y)3

where a = 3x, and b = 4y.  Since

a3 – b3 = (a-b)(a2+ab+b2)


(3x)3 – (4y)3 = (3x-4y)(9x2+12xy+16y2)

and we can see that

27x3– 64y3 = (3x-4y)(9x2+12xy+16y2)

5) Factoring by grouping

Factoring by grouping is a method of factoring that works on four-term polynomials that have a specific pattern to them. The process goes like this:

Factor: x3 + 3x2 + 2x + 6
  1. Rearrange the terms so that the exponents are in decreasing order, if they aren’t already.
  1. Group the first two and the last two terms together.
(x3 + 3x2) + (2x + 6)


  1. Factor each of the two groups separately. In our example, you can factor an x2 out of the first group and a 2 out of the second.
x2(x + 3) + 2(x + 3)


  1. Factor the common factor out of the two groups. In our example, both of the groups have an x + 3 in common. That’s what we’ll factor out.
(x + 3)(x2 + 2)


  1. Check the two factors to see if they can be factored further. Neither x2 + 2 or x + 3 can be factored further so (x + 3)(x2 + 2) is our final answer.
(x + 3)(x2 + 2)

Factoring by grouping requires the original polynomial to have a specific pattern that not all four term polynomials will have. If you do the factorization in step three and the two groups don’t have a common factor then you need to go back to square one and try a different approach

6) Use quadratic formula to first find the roots


The factors will be of the form (x-root).

For example to factorise  3x2-5x-2,  substitute a=3, b=-5 and c=-2  into the quadratic formula to get the roots which work out as 2 and -1/3.

The factors will then be (x-2) and (x + 1/3).

However we do not like fractions here, we want only whole numbers appearing in our factors. But we solve the quadratic equation when it is equal to zero. So the product of the factors is equal to zero. So the factors themselves are zero. We can multiply zero by anything we like and it will still give zero. To get rid of the fractions we simply multiply the factor by the denominator appearing in the fraction.

3(x+1/3) = 3x+1.

Our final factors are (x-2) and (3x+1). 



7)   Find the roots by trial and error and then use long division.

The roots of a function f(x) are the values of x that make the function value equal zero. For higher order functions we do not have convenient formulae like the quadratic one to solve them. Sometimes we can just guess the values of the roots and by substituting values by trial and error  hit upon the roots. Once we know a root we can form the factor. If we know one factor we can divide it into the function to be factorised by the method of long division.  



You might find these short videos on Factorising useful:

  1. Introduction to factorising
  2. Highest common factor (HCF)
  3. Factorising by grouping
  4. Factorising quadratic expressions