# How do you find g(x+1) given g(x)=x^3-2x^2?

Feb 17, 2018

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

#### Explanation:

A function takes in an input and returns an output.

In this function $g \left(x\right)$, you input $x$ and it returns ${x}^{3} - 2 {x}^{2}$. We can also look at this as this: you input $x$ (what's enclosed by the parentheses) and this function $g \left(x\right)$ takes the input (x), cubes it (${x}^{3}$), subtracts two times the input squared ($- 2 {x}^{2}$). Putting this all together, this particular function $g \left(x\right)$ takes in an input $x$ and returns the expression ${x}^{3} - 2 {x}^{2}$, with $x$ representing the input (which could be anything).

Cool! So what exactly does the question mean by $g \left(x + 1\right)$? What this means is that the entire expression $x + 1$ is the input! So, because

$g \left(x\right) = {x}^{3} - 2 {x}^{2}$

so,

$g \left(x + 1\right) = {\left(x + 1\right)}^{3} - 2 {\left(x + 1\right)}^{2}$

So now, all we have to do is use algebra to simplify this expression! We have

$g \left(x + 1\right) = {\left(x + 1\right)}^{3} - 2 {\left(x + 1\right)}^{2}$
$g \left(x + 1\right) = \left(x + 1\right) \left(x + 1\right) \left(x + 1\right) - 2 \left({x}^{2} + 2 x + 1\right)$
$g \left(x + 1\right) = \left(x + 1\right) \left({x}^{2} + 2 x + 1\right) - 2 \left({x}^{2} + 2 x + 1\right)$

Here, we see that we have a $\left({x}^{2} + 2 x + 1\right)$ term in common on both sides of the minus sign. We invoke the distributive property:

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

And we're done!

SIDENOTE:

Alternatively, you could also write the answer like this:

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

Either way is fine.