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
 Given a polynomial ax^{2}+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).
 a^{2} – b^{2} = (a+b)(ab) Difference of perfect squares.
 a^{3} – b^{3} = (ab)(a^{2}+ab+b^{2}) Difference of perfect cubes.
 a^{3} + b^{3} = (a+b)(a^{2} – ab+b^{2}) Sum of perfect cubes.
 Factoring by grouping, this is useful when a≠1 from method 1
 Using “b” formula to first get roots of the quadratic equation and then form factors.
Examples of Factoring
Factor polynomial of form ax^{2}+bx+c using method 1
1) x^{2}+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:
x^{2}+5x+4 = (x+1)(x+4)
Factor the following polynomial using method 5
2) 3x^{2}+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
3x^{2}+16x+5 = 3x^{2}+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
3x^{2}+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)
whereby
3x^{2}+16x+5 = (x+5)(3x+1)
Factor the difference of perfect squares using method 2
3) 9x^{2}81
first we need to write this in the form a^{2}b^{2}
9x^{2}81 = (3x)^{2} – 9^{2}
where a = 3x, and b=9. Since
a^{2} – b^{2} = (a+b)(ab)
then
(3x)^{2} – 9^{2} = (3x+9)(3x9)
and we can see that
9x^{2}81 = (3x+3)(3x3)
Factoring the difference of perfect squares can be regarded as a special case of method 1 above where the coefficient of the xterm is zero.
When given something like x^{2 }– 9, we can rewrite it as x^{2}+0x9.
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)(x3).
Generally, whenever we have x^{2}b^{2} we will always be able to factor it as (x+b)(xb).
Or (x^{2 }– number) will always be factored as (x+√number)(x√number).
For expressions more complex, such as 4x^{2}12y^{4};
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 (2x11y^{2})(2x11y^{2}).
Factor the difference of perfect cubes using method 3
4) 27x^{3}– 64y^{3}
first write this in the form a^{3}b^{3}
27x^{3}– 64y^{3} = (3x)^{3} – (4y)^{3}
where a = 3x, and b = 4y. Since
a^{3} – b^{3} = (ab)(a^{2}+ab+b^{2})
then
(3x)^{3} – (4y)^{3} = (3x4y)(9x^{2}+12xy+16y^{2})
and we can see that
27x^{3}– 64y^{3} = (3x4y)(9x^{2}+12xy+16y^{2})
5) Factoring by grouping
Factoring by grouping is a method of factoring that works on fourterm polynomials that have a specific pattern to them. The process goes like this:
Factor:  x^{3} + 3x^{2} + 2x + 6 



(x^{3} + 3x^{2}) + (2x + 6) 

x^{2}(x + 3) + 2(x + 3) 

(x + 3)(x^{2} + 2) 

(x + 3)(x^{2} + 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 roots:
If in an exam the question reads “solve for x to 2 decimal places” This means it’s time to use the equation. But if we solve for x then we can work back and get the factors.
You might find these short videos on Factorising useful: