# Problem of the Week

## Updated at Oct 11, 2021 5:18 PM

This week we have another equation problem:

How can we solve the equation $$4(3-n)(3-{n}^{2})=52$$?

Let's start!

$4(3-n)(3-{n}^{2})=52$

1
Expand.
$36-12{n}^{2}-12n+4{n}^{3}=52$

2
Move all terms to one side.
$36-12{n}^{2}-12n+4{n}^{3}-52=0$

3
Simplify  $$36-12{n}^{2}-12n+4{n}^{3}-52$$  to  $$-16-12{n}^{2}-12n+4{n}^{3}$$.
$-16-12{n}^{2}-12n+4{n}^{3}=0$

4
Factor out the common term $$4$$.
$-4(4+3{n}^{2}+3n-{n}^{3})=0$

5
Factor $$4+3{n}^{2}+3n-{n}^{3}$$ using Polynomial Division.
$-4(-{n}^{2}-n-1)(n-4)=0$

6
Divide both sides by $$-4$$.
$(-{n}^{2}-n-1)(n-4)=0$

7
Solve for $$n$$.
$n=4$

8
$n=\frac{1+\sqrt{3}\imath }{-2},\frac{1-\sqrt{3}\imath }{-2}$
$n=4,\frac{1+\sqrt{3}\imath }{-2},\frac{1-\sqrt{3}\imath }{-2}$
$n=4,-\frac{1+\sqrt{3}\imath }{2},-\frac{1-\sqrt{3}\imath }{2}$