# Problem of the Week

## Updated at Mar 25, 2024 5:43 PM

This week we have another equation problem:

How would you solve the equation $${(4p)}^{2}+\frac{3}{2+p}=17$$?

Let's start!

${(4p)}^{2}+\frac{3}{2+p}=17$

1
Use Multiplication Distributive Property: $${(xy)}^{a}={x}^{a}{y}^{a}$$.
${4}^{2}{p}^{2}+\frac{3}{2+p}=17$

2
Simplify  $${4}^{2}$$  to  $$16$$.
$16{p}^{2}+\frac{3}{2+p}=17$

3
Multiply both sides by $$2+p$$.
$16{p}^{2}(2+p)+3=17(2+p)$

4
Simplify.
$32{p}^{2}+16{p}^{3}+3=34+17p$

5
Move all terms to one side.
$32{p}^{2}+16{p}^{3}+3-34-17p=0$

6
Simplify  $$32{p}^{2}+16{p}^{3}+3-34-17p$$  to  $$32{p}^{2}+16{p}^{3}-31-17p$$.
$32{p}^{2}+16{p}^{3}-31-17p=0$

7
Factor $$32{p}^{2}+16{p}^{3}-31-17p$$ using Polynomial Division.
$(16{p}^{2}+48p+31)(p-1)=0$

8
Solve for $$p$$.
$p=1$

9
Use the Quadratic Formula.
$p=\frac{-48+8\sqrt{5}}{32},\frac{-48-8\sqrt{5}}{32}$

10
Collect all solutions from the previous steps.
$p=1,\frac{-48+8\sqrt{5}}{32},\frac{-48-8\sqrt{5}}{32}$

11
Simplify solutions.
$p=1,-\frac{6-\sqrt{5}}{4},-\frac{6+\sqrt{5}}{4}$

Done

Decimal Form: 1, -0.940983, -2.059017