# Problem of the Week

## Updated at Jan 8, 2024 8:46 AM

For this week we've brought you this equation problem.

How would you solve the equation $$\frac{3}{(2+w){(w-3)}^{2}}=\frac{1}{4}$$?

Here are the steps:

$\frac{3}{(2+w){(w-3)}^{2}}=\frac{1}{4}$

1
Multiply both sides by $$(2+w){(w-3)}^{2}$$.
$3=\frac{1}{4}(2+w){(w-3)}^{2}$

2
Simplify  $$\frac{1}{4}(2+w){(w-3)}^{2}$$  to  $$\frac{(2+w){(w-3)}^{2}}{4}$$.
$3=\frac{(2+w){(w-3)}^{2}}{4}$

3
Multiply both sides by $$4$$.
$12=(2+w){(w-3)}^{2}$

4
Expand.
$12=2{w}^{2}-12w+18+{w}^{3}-6{w}^{2}+9w$

5
Simplify  $$2{w}^{2}-12w+18+{w}^{3}-6{w}^{2}+9w$$  to  $$-4{w}^{2}-3w+18+{w}^{3}$$.
$12=-4{w}^{2}-3w+18+{w}^{3}$

6
Move all terms to one side.
$12+4{w}^{2}+3w-18-{w}^{3}=0$

7
Simplify  $$12+4{w}^{2}+3w-18-{w}^{3}$$  to  $$-6+4{w}^{2}+3w-{w}^{3}$$.
$-6+4{w}^{2}+3w-{w}^{3}=0$

8
Factor $$-6+4{w}^{2}+3w-{w}^{3}$$ using Polynomial Division.
$(-{w}^{2}+3w+6)(w-1)=0$

9
Solve for $$w$$.
$w=1$

10
$w=\frac{-3+\sqrt{33}}{-2},\frac{-3-\sqrt{33}}{-2}$
$w=1,\frac{-3+\sqrt{33}}{-2},\frac{-3-\sqrt{33}}{-2}$
$w=1,-\frac{-3+\sqrt{33}}{2},-\frac{-3-\sqrt{33}}{2}$