# How do you combine (3k + 4) ( 3k - 5)?

Feb 20, 2017

$9 {k}^{2} - 3 k - 20$

#### Explanation:

Each term in the second bracket must be multiplied by each term in the first bracket as shown below.

$\left(\textcolor{red}{3 k + 4}\right) \left(3 k - 5\right)$

$= \textcolor{red}{3 k} \left(3 k - 5\right) \textcolor{red}{+ 4} \left(3 k - 5\right)$

distributing brackets gives.

$= 9 {k}^{2} - 15 k + 12 k - 20$

collect like terms.

$= 9 {k}^{2} - 3 k - 20$

Feb 20, 2017

$9 {k}^{2} - 3 k - 20$

#### Explanation:

$\left(3 k + 4\right) \left(3 k - 5\right)$

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} 3 k + 4$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} \underline{3 k - 5}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} 9 {k}^{2} + 12 k$
color(white)(..............................)ul(-15k-20
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \underline{9 {k}^{2} - 3 k - 20}$

or

$\left(3 k + 4\right) \left(3 k - 5\right)$

$3 k \times 3 k = \textcolor{w h i t e}{\ldots . .} \textcolor{red}{9 {k}^{2}}$

4 xx 3k=color(white)(.......)color(red)(12k

-5 xx 3k=color(red)(-15k

4 xx -5=color(white)(.)-ulcolor(red)(20

:.=color(red)(9k^2+12k-15k-20
$\therefore = 9 {k}^{2} - 3 k - 20$