# Factoring Trinomials: Part 3

You can use the distributive law to see that

$3\left(4n+5\right)=12n+15$ ,

and you can use FOIL to see that

$\begin{array}{l}\left(n+2\right)\left(n+3\right)=n\cdot n+n\cdot 3+2\cdot n+2\cdot 3\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={n}^{2}+3n+2n+6\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={n}^{2}+5n+6\end{array}$

## Factoring ${x}^{2}+bx+c$ when $c$ is negative

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

Example 1:

Factor ${x}^{2}+6x-16$ .

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:

$\begin{array}{l}-16=\left(-16\right)\left(1\right);-16+1=-15\\ -16=\left(-8\right)\left(2\right);-8+2=-6\\ -16=\left(-4\right)\left(4\right);-4+4=0\\ -16=\left(-2\right)\left(8\right);-2+8=6\\ -16=\left(-1\right)\left(16\right);-1+16=15\end{array}$

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

${x}^{2}+6x-16=\left(x-2\right)\left(x+8\right)$ .

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:

$\begin{array}{l}-20=\left(-20\right)\left(1\right);-20+1=-19\\ -20=\left(-10\right)\left(2\right);-10+2=-8\\ -20=\left(-5\right)\left(4\right);-5+4=-1\\ -20=\left(-4\right)\left(5\right);-4+5=1\\ -20=\left(-2\right)\left(10\right);-2+10=8\\ -20=\left(-1\right)\left(20\right);-1+20=19\end{array}$

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

${x}^{2}-x-20=\left(x-5\right)\left(x+4\right)$ .

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