# Question #42384

Mar 24, 2017

${a}^{2} + 11 a + 30$

#### Explanation:

When multiplying together two binomials, a useful strategy is to "FOIL" it -- first, outside, inside, last. That is to say, multiply together the two first terms, the two outside terms, the two inside terms, and the two last terms. This is just an easy way to remember to multiply both things on the left by both things on the right, so however you want to remember that works.

FOIL might be easier to understand with an example. Say you have to find the product of $\left(a + b\right) \left(c - d\right)$. Remember that you need to FOIL it (deeper explanation below), giving you:
$a c + a \left(\text{-"d)+bc+b("-} d\right)$ (don't forget that $d$ is negative in this problem), which is the same as $a c - a d + b c - b d$.

For your problem, FOILed it is $a a + 6 a + 5 a + 5 \left(6\right)$, which when you multiply and combine like terms is ${a}^{2} + 11 a + 30$.

----FOIL explanation----

FOIL comes from the distributive property, which states that $x \left(y + z\right) = x y + x z$

Going back to $\left(a + b\right) \left(c - d\right)$, first let $\left(a + b\right) = n$ so the problem is now $n \left(c - d\right)$. Look familiar? Distribute, so now you have $n c + n \left(\text{-d}\right)$ (don't forget the negative!)

Plug $\left(a + b\right)$ back in for $n$ to give $\left(a + b\right) c + \left(a + b\right) \left(\text{-} d\right)$. Both terms of this can be distributed again, giving $a c + b c + a \left(\text{-"d)+b("-} d\right)$, which multiplied out is $a c + b c - a d - b d$, which using the commutative property is the same $a c - a d + b c - b d$ we got from FOIL.