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Factoring Trinomials: Part 3

You can use the distributive law to see that

3 ( 4 n + 5 ) = 12 n + 15 ,

and you can use FOIL to see that

( n + 2 ) ( n + 3 ) = n ⋅ n + n ⋅ 3 + 2 ⋅ n + 2 ⋅ 3                                                           = n 2 + 3 n + 2 n + 6                                                           = n 2 + 5 n + 6

But how can you start with the answer and find the factors?

Factoring x 2 + b x + c when c is negative

In this case, you need two numbers with opposite signs (so that their product is negative).

Example 1:

Factor x 2 + 6 x − 16 .

Here we need to find two numbers with opposite signs which have − 16 as a product and 6 as a sum.

The factor pairs for − 16 are:

− 16 = ( − 16 ) ( 1 ) ; − 16 + 1 = − 15 − 16 = ( − 8 ) ( 2 ) ; − 8 + 2 = − 6 − 16 = ( − 4 ) ( 4 ) ; − 4 + 4 = 0 − 16 = ( − 2 ) ( 8 ) ; − 2 + 8 = 6 − 16 = ( − 1 ) ( 16 ) ; − 1 + 16 = 15

− 2 and 8 work. So we can factor the polynomial as

x 2 + 6 x − 16 = ( x − 2 ) ( x + 8 ) .

Example 2:

Factor x 2 − x − 20 .

Here we need to find two numbers with opposite signs which have − 20 as a product and − 1 as a sum.

The factor pairs for − 20 are:

− 20 = ( − 20 ) ( 1 ) ; − 20 + 1 = − 19 − 20 = ( − 10 ) ( 2 ) ; − 10 + 2 = − 8 − 20 = ( − 5 ) ( 4 ) ; − 5 + 4 = − 1 − 20 = ( − 4 ) ( 5 ) ; − 4 + 5 = 1 − 20 = ( − 2 ) ( 10 ) ; − 2 + 10 = 8 − 20 = ( − 1 ) ( 20 ) ; − 1 + 20 = 19

− 5 and 4 work. So we can factor the polynomial as

x 2 − x − 20 = ( x − 5 ) ( x + 4 ) .

See also Factoring: Parts 1 , 2 , and 4 ; factoring by grouping ; and irreducible polynomials .