Question

1. the length of a rectangle is $\frac{w + 3}{w^{2}+w - 12}$ and the width is $\frac{w^{2}+7w + 12}{w^{2}-9}$. what is the area of the rectangle in simplest form?
2. explain how you can simplify the expression $\frac{6x^{3}-6x^{2}}{x^{4}+5x^{3}}div\frac{3x^{2}-15x + 12}{2x^{2}+2x - 40}\times\frac{2x^{2}-10x + 12}{4x^{3}+16x^{2}-20x}$

a) simplify the expression. show your work.
b) state any restrictions on the variable.

Explanation:

3.

Step1: Recall area formula

The area $A$ of a rectangle is $A = l\times w$, where $l$ is the length and $w$ is the width. Here, $l=\frac{w + 3}{w^{2}+w - 12}$ and $w=\frac{w^{2}+7w + 12}{w^{2}-9}$. So $A=\frac{w + 3}{w^{2}+w - 12}\times\frac{w^{2}+7w + 12}{w^{2}-9}$.

Step2: Factor the polynomials

Factor $w^{2}+w - 12=(w + 4)(w-3)$, $w^{2}+7w + 12=(w + 3)(w + 4)$ and $w^{2}-9=(w + 3)(w - 3)$. Then $A=\frac{w + 3}{(w + 4)(w-3)}\times\frac{(w + 3)(w + 4)}{(w + 3)(w - 3)}$.

Step3: Simplify the expression

Cancel out the common factors $(w + 3)$ and $(w + 4)$ in the numerator and denominator. We get $A=\frac{w + 3}{(w-3)^{2}}$.

Step1: Rewrite division as multiplication

Recall that $\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\times\frac{d}{c}$. So $\frac{6x^{3}-6x^{2}}{x^{4}+5x^{3}}\div\frac{3x^{2}-15x + 12}{2x^{2}+2x - 40}\times\frac{2x^{2}-10x + 12}{4x^{3}+16x^{2}-20x}=\frac{6x^{3}-6x^{2}}{x^{4}+5x^{3}}\times\frac{2x^{2}+2x - 40}{3x^{2}-15x + 12}\times\frac{2x^{2}-10x + 12}{4x^{3}+16x^{2}-20x}$.

Step2: Factor the polynomials

• $6x^{3}-6x^{2}=6x^{2}(x - 1)$
• $x^{4}+5x^{3}=x^{3}(x + 5)$
• $3x^{2}-15x + 12=3(x^{2}-5x + 4)=3(x - 1)(x - 4)$
• $2x^{2}+2x - 40=2(x^{2}+x - 20)=2(x + 5)(x - 4)$
• $2x^{2}-10x + 12=2(x^{2}-5x + 6)=2(x - 2)(x - 3)$
• $4x^{3}+16x^{2}-20x=4x(x^{2}+4x - 5)=4x(x + 5)(x - 1)$

Then the expression becomes $\frac{6x^{2}(x - 1)}{x^{3}(x + 5)}\times\frac{2(x + 5)(x - 4)}{3(x - 1)(x - 4)}\times\frac{2(x - 2)(x - 3)}{4x(x + 5)(x - 1)}$.

Step3: Simplify the expression

Cancel out the common factors:
$\frac{6x^{2}(x - 1)\times2(x + 5)(x - 4)\times2(x - 2)(x - 3)}{x^{3}(x + 5)\times3(x - 1)(x - 4)\times4x(x + 5)(x - 1)}=\frac{2(x - 2)(x - 3)}{x^{2}(x + 5)(x - 1)}$

The original expression has denominators $x^{4}+5x^{3}=x^{3}(x + 5)$, $3x^{2}-15x + 12=3(x - 1)(x - 4)$, $2x^{2}+2x - 40=2(x + 5)(x - 4)$ and $4x^{3}+16x^{2}-20x=4x(x + 5)(x - 1)$.
Set each denominator equal to zero and solve for $x$.

• $x^{3}(x + 5)=0$ gives $x=0$ or $x=-5$
• $(x - 1)(x - 4)=0$ gives $x = 1$ or $x=4$

Answer:

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

4.

a)