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Chapter 6: Q. 312 (page 627)

Solve quadratic equations by factoring.

2y3+2y2=12y

Short Answer

Expert verified

The roots are0,2,-3.

Step by step solution

01

Step 1. Rearrange the terms

Rearranging the terms to bring them on a single side,

2y3+2y2-12y=0

02

Step 2. Factor the greatest common factor

Factoring out the greatest common factor first,

2y(y2+y-6)=0

03

Step 3. Simplify the quadratic equation

We factor the trinomial first,

2y(y2+y-6)=02y(y2+3y-2y-6)=02y(y(y+3)-2(y+3))=02y(y+3)(y-2)=0

Use the Zero Product Property to set each factor to 0, we get,

when 2y=0,

y=0.

When y+3=0,

y=-3.

When y-2=0,

y=2.

04

Step 4. Check

Resubstitute each of the roots separately into the original equation.

When y=0,

2y3+2y2=12y2(0)3+2(0)2=12(0)0+0=00=0

This is true.

When y=-3,

role="math" localid="1644682832455" 2y3+2y2=12y2(-3)3+2(-3)2=12(-3)-54+18=-36-36=-36

This is true.

When y=2,

2y3+2y2=12y2(2)3+2(2)2=12(2)16+8=2424=24

This is true as well.

Thus, all the roots satisfy the original equation.

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