# How do you simplify the product (a - 6)(a + 8) and write it in standard form?

Mar 3, 2016

${a}^{2} + 2 a - 48$

#### Explanation:

For simplifying a quadratic equation into standard form, the F.O.I.L. (first, outside, inside, last) method is often used to expand the brackets. Here is what you will need to know before we start:

$1$. Use the F.O.I.L. (first, outside, inside, last) method to expand the brackets.

(color(red)a color(blue)(-6))(color(orange)a color(green)(+8))

$2$. For the "F" (first) in F.O.I.L., multiply the two $a$'s together.

$\textcolor{red}{a} \left(\textcolor{\mathmr{and} a n \ge}{a}\right)$

$= \textcolor{p u r p \le}{{a}^{2}}$

$3$. For the "O" (outside) in F.O.I.L., multiply the first $\textcolor{red}{a}$ and $\textcolor{g r e e n}{8}$ together.

$\textcolor{p u r p \le}{{a}^{2}}$ $\textcolor{red}{+ a} \left(\textcolor{g r e e n}{8}\right)$

$= \textcolor{p u r p \le}{{a}^{2}}$ $\textcolor{p u r p \le}{+ 8 a}$

$4$. For the "I" (inside) in F.O.I.L., multiply $\textcolor{b l u e}{- 6}$ and the second $\textcolor{\mathmr{and} a n \ge}{a}$ together.

$\textcolor{p u r p \le}{{a}^{2}}$ $\textcolor{p u r p \le}{+ 8 a}$ $\textcolor{b l u e}{- 6} \left(\textcolor{\mathmr{and} a n \ge}{a}\right)$

$= \textcolor{p u r p \le}{{a}^{2}}$ $\textcolor{p u r p \le}{+ 8 a}$ $\textcolor{p u r p \le}{- 6 a}$

$5$. For the "L" (last) in F.O.I.L., multiply $\textcolor{b l u e}{- 6}$ and $\textcolor{g r e e n}{8}$ together.

$\textcolor{p u r p \le}{{a}^{2}}$ $\textcolor{p u r p \le}{+ 8 a}$ $\textcolor{p u r p \le}{- 6 a}$ $\textcolor{b l u e}{- 6} \textcolor{g r e e n}{\left(8\right)}$

$= \textcolor{p u r p \le}{{a}^{2}}$ $\textcolor{p u r p \le}{+ 8 a}$ $\textcolor{p u r p \le}{- 6 a}$ $\textcolor{p u r p \le}{- 48}$

$6$. Recall that the general quadratic equation in standard form is: $a {x}^{2} + b x + c = 0$. Thus, simplify the equation.

$= \textcolor{p u r p \le}{{a}^{2} + 2 a - 48}$

$\therefore$, the equation in standard form is ${a}^{2} + 2 a - 48$.