# Problem of the Week

Updated at Jul 1, 2019 1:10 PM

This week's problem comes from the equation category.

How would you solve the equation $${(v-3)}^{2}(4v+2)=24$$?

Let's begin!

${(v-3)}^{2}(4v+2)=24$

1
Expand.
$4{v}^{3}+2{v}^{2}-24{v}^{2}-12v+36v+18=24$

2
Simplify $$4{v}^{3}+2{v}^{2}-24{v}^{2}-12v+36v+18$$ to $$4{v}^{3}-22{v}^{2}+24v+18$$.
$4{v}^{3}-22{v}^{2}+24v+18=24$

3
Move all terms to one side.
$4{v}^{3}-22{v}^{2}+24v+18-24=0$

4
Simplify $$4{v}^{3}-22{v}^{2}+24v+18-24$$ to $$4{v}^{3}-22{v}^{2}+24v-6$$.
$4{v}^{3}-22{v}^{2}+24v-6=0$

5
Factor out the common term $$2$$.
$2(2{v}^{3}-11{v}^{2}+12v-3)=0$

6
Factor $$2{v}^{3}-11{v}^{2}+12v-3$$ using Polynomial Division.
$2(2{v}^{2}-9v+3)(v-1)=0$

7
Solve for $$v$$.
$v=1$

8
$v=\frac{9+\sqrt{57}}{4},\frac{9-\sqrt{57}}{4}$
$v=1,\frac{9+\sqrt{57}}{4},\frac{9-\sqrt{57}}{4}$