Problem of the Week

Updated at Jul 1, 2019 1:10 PM

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Mar 24, 2025

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

Expand.

4{v}^{3}+2{v}^{2}-24{v}^{2}-12v+36v+18=24

Simplify

4{v}^{3}+2{v}^{2}-24{v}^{2}-12v+36v+18

4{v}^{3}-22{v}^{2}+24v+18

4{v}^{3}-22{v}^{2}+24v+18=24

Move all terms to one side.

4{v}^{3}-22{v}^{2}+24v+18-24=0

Simplify

4{v}^{3}-22{v}^{2}+24v+18-24

4{v}^{3}-22{v}^{2}+24v-6

4{v}^{3}-22{v}^{2}+24v-6=0

Factor out the common term

2

2(2{v}^{3}-11{v}^{2}+12v-3)=0

Factor

2{v}^{3}-11{v}^{2}+12v-3

using Polynomial Division.

Factor the following.

2{v}^{3}-11{v}^{2}+12v-3

First, find all factors of the constant term 3.

1, 3

Try each factor above using the Remainder Theorem.

Substitute 1 into v. Since the result is 0, v-1 is a factor..

2\times {1}^{3}-11\times {1}^{2}+12\times 1-3 = 0

v-1

Polynomial Division: Divide

2{v}^{3}-11{v}^{2}+12v-3

v-1

Rewrite the expression using the above.

2{v}^{2}-9v+3

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2(2{v}^{2}-9v+3)(v-1)=0

Solve for

v

v=1

Use the Quadratic Formula.

In general, given

a{x}^{2}+bx+c=0

, there exists two solutions where:

x=\frac{-b+\sqrt{{b}^{2}-4ac}}{2a},\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}

In this case,

a=2

b=-9

and

c=3

{v}^{}=\frac{9+\sqrt{{(-9)}^{2}-4\times 2\times 3}}{2\times 2},\frac{9-\sqrt{{(-9)}^{2}-4\times 2\times 3}}{2\times 2}

Simplify.

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

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

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