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:

Step1: Recall area formula for rectangle

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

Step2: Factor the polynomials

Factor the denominators and numerators: $x^{2}+x - 12=(x + 4)(x - 3)$, $x^{2}+7x + 12=(x + 3)(x+4)$, $x^{2}-9=(x + 3)(x - 3)$. So $A=\frac{x + 3}{(x + 4)(x - 3)}\times\frac{(x + 3)(x + 4)}{(x + 3)(x - 3)}$.

Step3: Simplify the product

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

for question 4:

a)

Step1: Factor the polynomials in the expression

For $6x^{3}-6x=6x(x^{2}-1)=6x(x + 1)(x - 1)$; $x^{4}+5x^{3}=x^{3}(x + 5)$; $3x^{2}-15x + 12=3(x^{2}-5x + 4)=3(x - 1)(x - 4)$; $2x^{2}+2x - 40=2(x^{2}+x - 20)=2(x + 5)(x - 4)$; $2x^{2}-10x + 12=2(x^{2}-5x + 6)=2(x - 2)(x - 3)$; $4x^{3}+16x^{2}-20x=4x(x^{2}+4x - 5)=4x(x + 5)(x - 1)$.
The original expression $\frac{6x^{3}-6x}{x^{4}+5x^{3}}\div\frac{3x^{2}-15x + 12}{2x^{2}+2x - 40}\times\frac{2x^{2}-10x + 12}{4x^{3}+16x^{2}-20x}$ becomes $\frac{6x(x + 1)(x - 1)}{x^{3}(x + 5)}\times\frac{2(x + 5)(x - 4)}{3(x - 1)(x - 4)}\times\frac{2(x - 2)(x - 3)}{4x(x + 5)(x - 1)}$.

Step2: Simplify the product of fractions

Cancel out the common factors:
Cancel out $x$, $(x + 1)$, $(x - 1)$, $(x + 5)$ and $(x - 4)$ terms where possible.
$\frac{6}{x^{2}}\times\frac{2}{3}\times\frac{2(x - 2)(x - 3)}{4(x - 1)}=\frac{24(x - 2)(x - 3)}{12x^{2}(x - 1)}=\frac{2(x - 2)(x - 3)}{x^{2}(x - 1)}=\frac{2(x^{2}-5x + 6)}{x^{2}(x - 1)}=\frac{2x^{2}-10x + 12}{x^{3}-x^{2}}$

b)

The restrictions on the variable come from the denominators of the original rational - expressions.
Set $x^{4}+5x^{3}=x^{3}(x + 5)
eq0$, $3x^{2}-15x + 12 = 3(x - 1)(x - 4)
eq0$, $2x^{2}+2x - 40=2(x + 5)(x - 4)
eq0$, $4x^{3}+16x^{2}-20x=4x(x + 5)(x - 1)
eq0$.
The restrictions are $x
eq0,x
eq1,x
eq3,x
eq4,x
eq - 5$.

Answer:

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