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Factoring Trinomials: Part 2

You can use the distributive law to see that

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

and you can use FOIL to see that

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

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

Factoring $x^2+bx+c$ when $b$ is negative, $c$ is positive

In this case, you need two negative numbers, so that their product is positive but their sum is negative.

The negative factor pairs for $10$ are:

$\begin{array}{l}10=\left(-10\right)\left(-1\right);-10-1=-11\\ 10=\left(-5\right)\left(-2\right);-5-2=-7\end{array}$

So the polynomial can be factored as

$x^2-7x+10=\left(x-2\right)\left(x-5\right)$ .