Question

How do you FOIL $\left( x-10 \right)\left( x+10 \right)$?

Answer 1

Hint: We first explain the meaning of the word FOIL. We multiply the terms according to their positions. At the end we add all the terms. We also verify the final result using the identity of $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$.

Complete step by step answer:
We have been given multiplication of two linear equations. We have to do the breakings of the polynomials in order of FOIL. The word FOIL stands for First-Outside-Inside-Last. It is a technique to distribute the binomials.
There are two terms in each polynomial.
We start by multiplying the first terms of $\left( x-10 \right)$ and $\left( x+10 \right)$. The terms are x and x.
The multiplication gives the result of $x\times x={{x}^{2}}$.
We now multiply the outside terms of $\left( x-10 \right)$ and $\left( x+10 \right)$. The terms are x and 10.
The multiplication gives a result of $x\times 10=10x$.
Then we multiply the inside terms of $\left( x-10 \right)$ and $\left( x+10 \right)$. The terms are -10 and x.
The multiplication gives the result of $\left( -10 \right)\times x=-10x$.
We end by multiplying the last terms of $\left( x-10 \right)$ and $\left( x+10 \right)$. The terms are -10 and 10.
The multiplication gives the result of $\left( -10 \right)\times 10=-100$.
Now we add all the terms and get the final solution as
$\left( x-10 \right)\left( x+10 \right)={{x}^{2}}+10x-10x-100={{x}^{2}}-100$.

Note: Although we have used the FOIL technique to find the multiplied form of $\left( x-10 \right)\left( x+10 \right)$. We can also verify the result using the identity of $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$. We convert to the equation by taking values as $a=x;b=10$.
Putting the values, we get $\left( x-10 \right)\left( x+10 \right)={{x}^{2}}-{{10}^{2}}={{x}^{2}}-100$.
Thus, verified the result of the FOIL technique.