# How do you multiply (x+4)(3x^2-8x-7)?

Mar 18, 2018

$3 {x}^{3} + 4 {x}^{2} - 39 x - 28$

#### Explanation:

To multiply two polynomials requires distribution. Each term of one polynomial must multiple each of the terms in the other polynomial.

$\left(x + 4\right) \left(3 {x}^{2} - 8 x - 7\right)$

$= x \left(3 {x}^{2} - 8 x - 7\right) + 4 \left(3 {x}^{2} - 8 x - 7\right)$

First we multiply $x$ from the first polynomial by each term of the second polynomial:

$= 3 {x}^{3} - 8 {x}^{2} - 7 x + 4 \left(3 {x}^{2} - 8 x - 7\right)$

Now we multiply $4$ from the first polynomial by each term of the second polynomial:

$= 3 {x}^{3} - 8 {x}^{2} - 7 x + 12 {x}^{2} - 32 x - 28$

Now we simplify:

$= \textcolor{b l u e}{3 {x}^{3}} - \textcolor{\mathmr{and} a n \ge}{8 {x}^{2}} - \textcolor{red}{7 x} + \textcolor{\mathmr{and} a n \ge}{12 {x}^{2}} - \textcolor{red}{32 x} - 28$

$= \textcolor{b l u e}{3 {x}^{3}} + \textcolor{\mathmr{and} a n \ge}{4 {x}^{2}} - \textcolor{red}{39 x} - 28$

$\implies \textcolor{g r e e n}{3 {x}^{3} + 4 {x}^{2} - 39 x - 28}$