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). - a
^{2}– b^{2}= (a+b)(a-b) - a
^{3}– b^{3}= (a-b)(a^{2}+ab+b^{2}) - a
^{3}+ b^{3}= (a+b)(a^{2}– ab+b^{2}) - Factoring by grouping, this is useful when a≠1 from method 1

**Examples of Factoring**

Factor the following polynomial 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 polynomial 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)(a-b)

then

(3x)^{2} – 9^{2} = (3x+9)(3x-9)

and we can see that

9x^{2}-81 = (3x+3)(3x-3)

Factor the polynomial 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} = (a-b)(a^{2}+ab+b^{2})

then

(3x)^{3} – (4y)^{3} = (3x-4y)(9x^{2}+12xy+16y^{2})

and we can see that

27x^{3}– 64y^{3} = (3x-4y)(9x^{2}+12xy+16y^{2})

**5)** Use quadratic formula:

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: