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
Use the Quadratic Formula.
\[n=\frac{1+\sqrt{3}\imath }{-2},\frac{1-\sqrt{3}\imath }{-2}\]

9
Collect all solutions from the previous steps.
\[n=4,\frac{1+\sqrt{3}\imath }{-2},\frac{1-\sqrt{3}\imath }{-2}\]

10
Simplify solutions.
\[n=4,-\frac{1+\sqrt{3}\imath }{2},-\frac{1-\sqrt{3}\imath }{2}\]

Done
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