# How do you find the product of (k+4)(k^2+3k-6)?

Apr 12, 2017

See the entire solution process below:

#### Explanation:

To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

$\left(\textcolor{red}{k} + \textcolor{red}{4}\right) \left(\textcolor{b l u e}{{k}^{2}} + \textcolor{b l u e}{3 k} - \textcolor{b l u e}{3}\right)$ becomes:

$\left(\textcolor{red}{k} \times \textcolor{b l u e}{{k}^{2}}\right) + \left(\textcolor{red}{k} \times \textcolor{b l u e}{3 k}\right) - \left(\textcolor{red}{k} \times \textcolor{b l u e}{6}\right) + \left(\textcolor{red}{4} \times \textcolor{b l u e}{{k}^{2}}\right) + \left(\textcolor{red}{4} \times \textcolor{b l u e}{3 k}\right) - \left(\textcolor{red}{4} \times \textcolor{b l u e}{6}\right)$

${k}^{3} + 3 {k}^{2} - 6 k + 4 {k}^{2} + 12 k - 24$

We can now group and combine like terms:

${k}^{3} + 3 {k}^{2} + 4 {k}^{2} + 12 k - 6 k - 24$

${k}^{3} + \left(3 + 4\right) {k}^{2} + \left(12 - 6\right) k - 24$

${k}^{3} + 7 {k}^{2} + 6 k - 24$

Apr 12, 2017

$= {k}^{3} + 7 {k}^{2} + 6 k - 24$

#### Explanation:

Each term in the first bracket must be multiplied by each term in the second bracket. This means there will be 6 terms at first.

$\left(\textcolor{red}{k} + \textcolor{b l u e}{4}\right) \left({k}^{2} + 3 k - 6\right)$

$= \textcolor{red}{k} \left({k}^{2} + 3 k - 6\right) + \textcolor{b l u e}{4} \left({k}^{2} + 3 k - 6\right)$

=color(red)(k^3+3k^2-6k +color(blue)(4k^2+12k-24)

$= {k}^{3} + 7 {k}^{2} + 6 k - 24$