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
Use the Quadratic Formula.
\[v=\frac{9+\sqrt{57}}{4},\frac{9-\sqrt{57}}{4}\]

9
Collect all solutions from the previous steps.
\[v=1,\frac{9+\sqrt{57}}{4},\frac{9-\sqrt{57}}{4}\]

Done

Decimal Form: 1, 4.137459, 0.362541
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