Problem of the Week

Updated at Jul 31, 2023 10:17 AM

Recent Posts

Mar 31, 2025

How would you solve ${(3-y)}^{2}(4+4y)=32$ ?

Below is the solution.

{(3-y)}^{2}(4+4y)=32

Expand.

36+36y-24y-24{y}^{2}+4{y}^{2}+4{y}^{3}=32

Simplify

36+36y-24y-24{y}^{2}+4{y}^{2}+4{y}^{3}

36+12y-20{y}^{2}+4{y}^{3}

36+12y-20{y}^{2}+4{y}^{3}=32

Move all terms to one side.

36+12y-20{y}^{2}+4{y}^{3}-32=0

Simplify

36+12y-20{y}^{2}+4{y}^{3}-32

4+12y-20{y}^{2}+4{y}^{3}

4+12y-20{y}^{2}+4{y}^{3}=0

Factor out the common term

4

4(1+3y-5{y}^{2}+{y}^{3})=0

Factor

1+3y-5{y}^{2}+{y}^{3}

using Polynomial Division.

Factor the following.

1+3y-5{y}^{2}+{y}^{3}

First, find all factors of the constant term 1.

1

Try each factor above using the Remainder Theorem.

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

1+3\times 1-5\times {1}^{2}+{1}^{3} = 0

y-1

Polynomial Division: Divide

1+3y-5{y}^{2}+{y}^{3}

y-1

Rewrite the expression using the above.

{y}^{2}-4y-1

To get access to all 'How?' and 'Why?' steps, join Cymath Plus!

4({y}^{2}-4y-1)(y-1)=0

Solve for

y

y=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=1

b=-4

and

c=-1

{y}^{}=\frac{4+\sqrt{{(-4)}^{2}+4}}{2},\frac{4-\sqrt{{(-4)}^{2}+4}}{2}

Simplify.

y=\frac{4+2\sqrt{5}}{2},\frac{4-2\sqrt{5}}{2}

To get access to all 'How?' and 'Why?' steps, join Cymath Plus!

y=\frac{4+2\sqrt{5}}{2},\frac{4-2\sqrt{5}}{2}

Collect all solutions from the previous steps.

y=1,\frac{4+2\sqrt{5}}{2},\frac{4-2\sqrt{5}}{2}

Simplify solutions.

y=1,2+\sqrt{5},2-\sqrt{5}

Done

Decimal Form: 1, 4.236068, -0.236068