# What is the standard form of a polynomial (5k + 2)(3k + 1)?

Mar 18, 2016

#### Answer:

$15 {k}^{2} + 11 k + 2 = 0$

#### Explanation:

Recall that the standard form of a polynomial is written in the form:

$\textcolor{t e a l}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} a {x}^{2} + b x + c = 0 \textcolor{w h i t e}{\frac{a}{a}} |}}} \textcolor{w h i t e}{X} , \textcolor{w h i t e}{X}$where $a \ne 0$

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:

$1$. Assuming the given equation is equal to $0$, locate the terms, as well as their appropriate positive or negative signs.

(color(red)(5k) color(blue)(+2))(color(orange)(3k) color(green)(+1))=0

$2$. For the "F" (first) in F.O.I.L., multiply $\textcolor{red}{5 k}$ and $\textcolor{\mathmr{and} a n \ge}{3 k}$ together.

$\textcolor{red}{5 k} \left(\textcolor{\mathmr{and} a n \ge}{3 k}\right)$

$= \textcolor{p u r p \le}{15 {k}^{2}}$

$3$. For the "O" (outside) in F.O.I.L., multiply $\textcolor{red}{+ 5 k}$ and $\textcolor{g r e e n}{1}$ together.

$\textcolor{p u r p \le}{15 {k}^{2}}$ $\textcolor{red}{+ 5 k} \left(\textcolor{g r e e n}{1}\right)$

$= \textcolor{p u r p \le}{15 {k}^{2}}$ $\textcolor{p u r p \le}{+ 5 k}$

$4$. For the "I" (inside) in F.O.I.L., multiply $\textcolor{b l u e}{+ 2}$ and $\textcolor{\mathmr{and} a n \ge}{3 k}$ together.

$\textcolor{p u r p \le}{15 {k}^{2}}$ $\textcolor{p u r p \le}{+ 5 k}$ $\textcolor{b l u e}{+ 2} \left(\textcolor{\mathmr{and} a n \ge}{3 k}\right)$

$= \textcolor{p u r p \le}{15 {k}^{2}}$ $\textcolor{p u r p \le}{+ 5 k}$ $\textcolor{p u r p \le}{+ 6 k}$

$5$. For the "L" (last) in F.O.I.L., multiply $\textcolor{b l u e}{+ 2}$ and $\textcolor{g r e e n}{1}$ together.

$\textcolor{p u r p \le}{15 {k}^{2}}$ $\textcolor{p u r p \le}{+ 5 k}$ $\textcolor{p u r p \le}{+ 6 k}$ $\textcolor{b l u e}{+ 2} \textcolor{g r e e n}{\left(1\right)}$

$= \textcolor{p u r p \le}{15 {k}^{2}}$ $\textcolor{p u r p \le}{+ 5 k}$ $\textcolor{p u r p \le}{+ 6 k}$ $\textcolor{p u r p \le}{+ 2}$

$6$. Simplify the equation.

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 15 {k}^{2} + 11 k + 2 = 0 \textcolor{w h i t e}{\frac{a}{a}} |}}}$