Question

multiple choice
1 determine the vertical asymptote(s) of:
$f(x)=\frac{1}{(x+6)(x+1)}$
$x=7, x=6$
$x=-7, x=-6$
$x=6, x=1$
$x=-6, x=-1$

2 simplify completely: $\frac{b^{2}+5b}{b^{2}+10b+25}$
$\frac{b}{b+5}$
$\frac{b^{2}}{b+5}$
$b(b+5)$
$\frac{5b}{b+10}$

3 divide: $\frac{p-6}{p+2}\div\frac{p^{2}-2p-24}{p^{2}+4p}$
$\frac{p}{p+2}$
$\frac{p+4}{p^{2}+2p}$
$\frac{p+4}{p+2}$
$\frac{p+2}{p}$

4 divide: $\frac{x+3}{x-3}\cdot\frac{3-x}{x^{2}-9}$
$\frac{x+3}{x-3}$
$-\frac{1}{x-3}$
$-\frac{1}{x-1}$
$\frac{1}{x-3}$

5 add: $\frac{r}{4r-9}+\frac{-9}{16r-36}$
$\frac{r-9}{4r-9}$
$\frac{r}{4}$
$\frac{r-9}{20r-45}$
$\frac{1}{4}$

6 subtract: $\frac{d+7x}{xe}-\frac{d-2x}{xe}$
$\frac{5}{e}$
$\frac{(2d+5x)}{xe}$
$\frac{9}{e}$
$\frac{(2d+9x)}{xe}$

Explanation:

Problem 1

Step1: Find denominator roots

Set $(x+6)(x+1)=0$

Step2: Solve for x

$x+6=0 \implies x=-6$; $x+1=0 \implies x=-1$

Step1: Factor numerator/denominator

Numerator: $b^2+5b = b(b+5)$
Denominator: $b^2+10b+25=(b+5)^2$

Step2: Cancel common factors

$\frac{b(b+5)}{(b+5)^2} = \frac{b}{b+5}$

Step1: Rewrite division as multiplication

$\frac{p-6}{p+2} \times \frac{p^2+4p}{p^2-2p-24}$

Step2: Factor all terms

$p^2+4p=p(p+4)$; $p^2-2p-24=(p-6)(p+4)$

Step3: Substitute and cancel

$\frac{p-6}{p+2} \times \frac{p(p+4)}{(p-6)(p+4)} = \frac{p}{p+2}$

Answer:

$x=-6, x=-1$

---

Problem 2