Question

1. completely factor the expression below.

$3a^3 + 2a^2 - 48a - 32$

1. simplify the expression below.

$\frac{2v^4 - 128v}{6v^3 - 24v^2}$

a. $\frac{v - 16}{3v}$ c. $\frac{v^2 + 4v + 16}{3v}$
b. $\frac{3v}{v^2 - 4v - 16}$ d. $\frac{v^2 - 4v + 16}{3v}$

1. simplify the expression below.

$\frac{p^2 - 3p - 28}{5p + 20} cdot \frac{10p^2}{49 - p^2}$

a. $-2p(p + 7)$ c. $-\frac{1}{2p(p + 7)}$
b. $-\frac{2p^2}{p + 7}$ d. $\frac{2p^2}{p + 7}$

1. simplify the expression below.

$\frac{3r - 5}{4r^2 - 4r + 1} div \frac{6r^2 - 7r - 5}{4r^2 - 1}$

a. $2r - 1$ c. $1$
b. $\frac{1}{2r - 1}$ d. $\frac{2r + 1}{2r - 1}$

1. simplify the expression below.

$\frac{2n}{n + 1} + \frac{n - 3}{n^2 - 1} - \frac{7}{n - 1}$

a. $\frac{2(n - 5)}{n - 1}$ c. $\frac{2(n + 5)}{n + 1}$
b. $2(n - 5)(n - 1)$ d. $\frac{2}{(n - 5)(n + 1)}$

1. simplify the expression below.

$\frac{\frac{1}{21k^3} - \frac{3}{7k}}{1 - \frac{1}{3k}}$

a. $\frac{7k^2}{3k + 1}$ c. $\frac{3k - 1}{7k^2}$
b. $\frac{-7k^2}{3k + 1}$ d. $\frac{-(3k + 1)}{7k^2}$

Explanation:

Problem 7:

Step1: Group terms for factoring

$3a^3 + 2a^2 - 48a - 32 = (3a^3 + 2a^2) + (-48a - 32)$

Step2: Factor out GCF from groups

$= a^2(3a + 2) - 16(3a + 2)$

Step3: Factor out common binomial

$= (a^2 - 16)(3a + 2)$

Step4: Factor difference of squares

$= (a - 4)(a + 4)(3a + 2)$

Problem 8:

Step1: Factor numerator and denominator

$\frac{2v(v^3 - 64)}{6v^2(v - 4)}$

Step2: Factor sum of cubes

$\frac{2v(v - 4)(v^2 + 4v + 16)}{6v^2(v - 4)}$

Step3: Cancel common factors

$\frac{v^2 + 4v + 16}{3v}$

Problem 9:

Step1: Factor all polynomials

$\frac{(p-7)(p+4)}{5(p+4)} \cdot \frac{10p^2}{(7-p)(7+p)}$

Step2: Rewrite $7-p$ as $-(p-7)$

$\frac{(p-7)(p+4)}{5(p+4)} \cdot \frac{10p^2}{-(p-7)(7+p)}$

Step3: Cancel common factors

$-\frac{2p^2}{p+7}$

Problem 10:

Step1: Rewrite division as multiplication

$\frac{3r-5}{(2r-1)^2} \cdot \frac{(2r-1)(2r+1)}{(3r-5)(2r+1)}$

Step2: Cancel common factors

$\frac{1}{2r-1}$

Problem 11:

Step1: Get common denominator $(n^2-1)$

$\frac{2n(n-1)}{n^2-1} + \frac{n-3}{n^2-1} - \frac{7(n+1)}{n^2-1}$

Step2: Combine numerators

$\frac{2n^2-2n + n - 3 -7n -7}{n^2-1}$

Step3: Simplify numerator

$\frac{2n^2-8n-10}{n^2-1} = \frac{2(n^2-4n-5)}{(n-1)(n+1)}$

Step4: Factor numerator and cancel

$\frac{2(n-5)(n+1)}{(n-1)(n+1)} = \frac{2(n-5)}{n-1}$

Problem 12:

Step1: Simplify numerator and denominator

Numerator: $\frac{1 - 9k^2}{21k^3}$, Denominator: $\frac{3k-1}{3k}$

Step2: Rewrite division as multiplication

$\frac{-(9k^2-1)}{21k^3} \cdot \frac{3k}{3k-1}$

Step3: Factor difference of squares

$\frac{-(3k-1)(3k+1)}{21k^3} \cdot \frac{3k}{3k-1}$

Step4: Cancel common factors

$\frac{-(3k+1)}{7k^2}$

Answer:

1. $(a - 4)(a + 4)(3a + 2)$
2. C. $\frac{v^2 + 4v + 16}{3v}$
3. B. $-\frac{2p^2}{p+7}$
4. B. $\frac{1}{2r-1}$
5. A. $\frac{2(n-5)}{n-1}$
6. D. $\frac{-(3k+1)}{7k^2}$