Problem of the Week

Updated at Apr 7, 2025 11:17 AM

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Apr 14, 2025

How would you solve $\frac{4p{(3-p)}^{2}}{5}=\frac{16}{5}$ ?

Below is the solution.

\frac{4p{(3-p)}^{2}}{5}=\frac{16}{5}

Multiply both sides by

5

4p{(3-p)}^{2}=16

Expand.

36p-24{p}^{2}+4{p}^{3}=16

Move all terms to one side.

36p-24{p}^{2}+4{p}^{3}-16=0

Factor out the common term

4

4(9p-6{p}^{2}+{p}^{3}-4)=0

Factor

9p-6{p}^{2}+{p}^{3}-4

using Polynomial Division.

Factor the following.

9p-6{p}^{2}+{p}^{3}-4

First, find all factors of the constant term 4.

1, 2, 4

Try each factor above using the Remainder Theorem.

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

9\times 1-6\times {1}^{2}+{1}^{3}-4 = 0

p-1

Polynomial Division: Divide

9p-6{p}^{2}+{p}^{3}-4

p-1

Rewrite the expression using the above.

{p}^{2}-5p+4

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

Factor

{p}^{2}-5p+4

Ask: Which two numbers add up to

-5

and multiply to

4

-4

and

-1

Rewrite the expression using the above.

(p-4)(p-1)

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4(p-4)(p-1)(p-1)=0

Solve for

p

Ask: When will

(p-4)(p-1)(p-1)

equal zero?

When

p-4=0

p-1=0

, or

p-1=0

Solve each of the 3 equations above.

p=4,1

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p=4,1

Done