Question

1. what is the simplified form of the rational expression $\frac{x^{2}-36}{x^{2}+3x - 18}$? what is the domain?
2. find the product and give the domain of $\frac{y + 3}{y+2}cdot\frac{y^{2}+4y + 4}{y^{2}-9}$

directions: write an equivalent expression. state the domain.

1. $\frac{x^{3}+4x^{2}-12x}{x^{2}+x - 30}$
2. $\frac{3x^{2}+15x}{x^{2}+3x - 10}$

directions: what is the simplified form of each rational expression? what is the domain?
$\frac{y^{2}-5y - 24}{y^{2}+3y}$

1. $\frac{x^{2}+8x + 15}{x^{2}-x - 12}$
2. $\frac{x^{3}+9x^{2}-10x}{x^{3}-9x^{2}-10x}$

Explanation:

1.

Step1: Factor the numerator and denominator

The numerator $x^{2}-36=(x + 6)(x - 6)$ using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$. The denominator $x^{2}+3x - 18=(x+6)(x - 3)$ by factoring the quadratic $ax^{2}+bx + c$ where $a = 1$, $b = 3$, $c=-18$ and finding two numbers that multiply to $ac=-18$ and add up to $b = 3$ (the numbers are 6 and - 3).

Step2: Simplify the rational expression

$\frac{x^{2}-36}{x^{2}+3x - 18}=\frac{(x + 6)(x - 6)}{(x + 6)(x - 3)}=\frac{x - 6}{x - 3}$, for $x
eq - 6$ and $x
eq3$.

Step3: Find the domain

The domain is all real numbers except the values that make the denominator equal to zero. Set $x^{2}+3x - 18=0$, factored as $(x + 6)(x - 3)=0$. The solutions are $x=-6$ and $x = 3$. So the domain is $\{x\in\mathbb{R}|x
eq-6,x
eq3\}$.

Step1: Factor the expressions

$y^{2}+4y + 4=(y + 2)^{2}$ using the perfect - square formula $(a + b)^{2}=a^{2}+2ab + b^{2}$ with $a=y$ and $b = 2$, and $y^{2}-9=(y + 3)(y - 3)$ using the difference - of - squares formula.

Step2: Multiply the rational expressions

$\frac{y + 3}{y+2}\cdot\frac{y^{2}+4y + 4}{y^{2}-9}=\frac{y + 3}{y+2}\cdot\frac{(y + 2)^{2}}{(y + 3)(y - 3)}=\frac{y + 2}{y - 3}$ for $y
eq-3$, $y
eq-2$, $y
eq3$.

Step3: Find the domain

Set the denominators of the original fractions equal to zero. For $y + 2=0$, $y=-2$; for $y^{2}-9=0$, $(y + 3)(y - 3)=0$, so $y=-3$ or $y = 3$. The domain is $\{y\in\mathbb{R}|y
eq-3,y
eq-2,y
eq3\}$.

Step1: Factor the numerator and denominator

The numerator $x^{3}+4x^{2}-12x=x(x^{2}+4x - 12)=x(x + 6)(x - 2)$. The denominator $x^{2}+x - 30=(x + 6)(x - 5)$.

Step2: Simplify the rational expression

$\frac{x^{3}+4x^{2}-12x}{x^{2}+x - 30}=\frac{x(x + 6)(x - 2)}{(x + 6)(x - 5)}=\frac{x(x - 2)}{x - 5}$, for $x
eq-6$ and $x
eq5$.

Step3: Find the domain

Set $x^{2}+x - 30=0$, factored as $(x + 6)(x - 5)=0$. The solutions are $x=-6$ and $x = 5$. So the domain is $\{x\in\mathbb{R}|x
eq-6,x
eq5\}$.

Answer:

Simplified form: $\frac{x - 6}{x - 3}$, Domain: $\{x\in\mathbb{R}|x
eq-6,x
eq3\}$

2.