# How do I find f f*g and f/g if f(x)=(x^2+3x-4) and g(x)=(x+4)?

## How do I find f $f \cdot g$ and $\frac{f}{g}$ if $f \left(x\right) = \left({x}^{2} + 3 x - 4\right)$ and $g \left(x\right) = \left(x + 4\right)$?

Jan 4, 2018

See a solution process below:

#### Explanation:

First, we can factor $\left({x}^{2} + 3 x - 4\right)$ as: $\left(x + 4\right) \left(x - 1\right)$

Therefore: $f \cdot g$ is:

$f \cdot g = f \left(x\right) \times g \left(x\right) = \left({x}^{2} + 3 x - 4\right) \left(x + 4\right) =$

$\left(x + 4\right) \left(x - 1\right) \left(x + 4\right) = {\left(x + 4\right)}^{2} \left(x - 1\right)$

Or, we can expand $\left({x}^{2} + 3 x - 4\right) \left(x + 4\right)$ as:

$f \cdot g = \left({x}^{2} + 3 x - 4\right) \left(x + 4\right)$

$f \cdot g = \left({x}^{2} \times x\right) + \left(3 x \times x\right) - \left(4 \times x\right) + \left({x}^{2} \times 4\right) + \left(3 x \times 4\right) - \left(4 \times + 4\right)$

$f \cdot g = {x}^{3} + 3 {x}^{2} - 4 x + 4 {x}^{2} + 12 x - 16$

$f \cdot g = {x}^{3} + 3 {x}^{2} + 4 {x}^{2} - 4 x + 12 x - 16$

$f \cdot g = {x}^{3} + 7 {x}^{2} + 8 x - 16$ or $f \cdot g = {\left(x + 4\right)}^{2} \left(x - 1\right)$

Then, $\frac{f}{g}$ is:

$\frac{f}{g} = f \frac{x}{g} \left(x\right) = \frac{{x}^{2} + 3 x - 4}{x + 4} =$

$\frac{\left(x + 4\right) \left(x - 1\right)}{x + 4} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + 4\right)}}} \left(x - 1\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x + 4}}}}$

$\frac{f}{g} = x - 1$

However, because we cannot divide by zero we must find the exclusions:

$x + 4 \ne 0$ therefore $x \ne - 4$

$\frac{f}{g} = x - 1$ where $x \ne - 4$