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? 4. 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 for rectangle

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

Step2: Factor the quadratic expressions

Factor \(w^{2}+w - 12=(w + 4)(w-3)\), \(w^{2}+7w + 12=(w + 3)(w + 4)\) and \(w^{2}-9=(w + 3)(w - 3)\).
So \(l=\frac{w + 3}{(w + 4)(w-3)}\) and \(w=\frac{(w + 3)(w + 4)}{(w + 3)(w - 3)}\).

Step3: Calculate the area

\[

$$\begin{align*} A&=l\times w\\ &=\frac{w + 3}{(w + 4)(w-3)}\times\frac{(w + 3)(w + 4)}{(w + 3)(w - 3)}\\ &=\frac{w + 3}{(w-3)^2} \end{align*}$$

\]

Step1: Rewrite division as multiplication

Recall that dividing by a rational - expression is the same as multiplying by its reciprocal. So \(\frac{6x^{3}-6x^{2}}{x^{4}+5x^{3}}\div\frac{3x^{2}-15x + 12}{2x^{2}+2x - 40}\times\frac{2x^{2}-10x + 12}{4x^{3}+16x^{2}-20x}\) becomes \(\frac{6x^{3}-6x^{2}}{x^{4}+5x^{3}}\times\frac{2x^{2}+2x - 40}{3x^{2}-15x + 12}\times\frac{2x^{2}-10x + 12}{4x^{3}+16x^{2}-20x}\).

Step2: Factor each polynomial

• Factor \(6x^{3}-6x^{2}=6x^{2}(x - 1)\), \(x^{4}+5x^{3}=x^{3}(x + 5)\), \(2x^{2}+2x - 40=2(x^{2}+x - 20)=2(x + 5)(x - 4)\), \(3x^{2}-15x + 12=3(x^{2}-5x + 4)=3(x - 1)(x - 4)\), \(2x^{2}-10x + 12=2(x^{2}-5x + 6)=2(x - 2)(x - 3)\) and \(4x^{3}+16x^{2}-20x=4x(x^{2}+4x - 5)=4x(x + 5)(x - 1)\).

Step3: Substitute the factored forms

\[

$$\begin{align*} &\frac{6x^{2}(x - 1)}{x^{3}(x + 5)}\times\frac{2(x + 5)(x - 4)}{3(x - 1)(x - 4)}\times\frac{2(x - 2)(x - 3)}{4x(x + 5)(x - 1)}\\ =&\frac{6\times2\times2x^{2}(x - 1)(x + 5)(x - 4)(x - 2)(x - 3)}{3\times4x^{3}(x + 5)(x - 1)(x - 4)(x - 1)}\\ \end{align*}$$

\]

Step4: Simplify the expression

Cancel out the common factors:
\[

$$\begin{align*} &\frac{6\times2\times2x^{2}(x - 1)(x + 5)(x - 4)(x - 2)(x - 3)}{3\times4x^{3}(x + 5)(x - 1)(x - 4)(x - 1)}\\ =&\frac{2(x - 2)(x - 3)}{x(x - 1)} \end{align*}$$

\]

Step5: Find the restrictions

The original rational - expressions are undefined when the denominators are equal to zero.
For \(x^{4}+5x^{3}=x^{3}(x + 5)=0\), \(x = 0\) or \(x=-5\); for \(3x^{2}-15x + 12 = 3(x - 1)(x - 4)=0\), \(x = 1\) or \(x = 4\); for \(4x^{3}+16x^{2}-20x=4x(x + 5)(x - 1)=0\), \(x = 0\), \(x=-5\) or \(x = 1\).
So the restrictions are \(x
eq0\), \(x
eq1\), \(x
eq4\), \(x
eq - 5\).

Answer:

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

4.