Question

1. $\frac{6x^{2}+17x + 5}{2x^{2}-2x - 24}cdot\frac{4x^{2}-13x - 12}{12x^{2}+13x + 3}$
2. $\frac{x^{2}-4}{x^{2}+3x - 10}div\frac{x^{2}+5x + 6}{x^{2}+8x + 15}$
3. $\frac{x^{2}+x - 12}{x^{2}+x - 30}cdot\frac{x^{2}+5x + 6}{x^{2}-2x - 3}div\frac{x + 3}{x^{2}+7x + 6}$
4. $\frac{\frac{a^{2}-b^{2}}{a^{2}+b^{2}}}{\frac{4a - 4b}{2a^{2}+2b^{2}}}$
5. $\frac{\frac{p^{3}-8q^{3}}{p^{2}-4q^{2}}}{\frac{p^{2}+4pq + 4q^{2}}{(p + 2q)^{2}}}$
6. $\frac{3x^{2}y^{4}}{4}cdot\frac{2}{9x^{3}y}$
7. $\frac{5ab^{3}}{3b^{4}}div\frac{10a^{2}b^{8}}{6b^{2}}$
8. $\frac{x + 1}{3}div\frac{3x + 3}{7}$

Explanation:

Step1: Factorize polynomials

For example, in problem 13, factorize $6x^{2}+17x + 5=(2x + 5)(3x+1)$, $2x^{2}-2x - 24 = 2(x^{2}-x - 12)=2(x - 4)(x+3)$, $4x^{2}-13x - 12=(4x + 3)(x - 4)$, $12x^{2}+13x + 3=(3x + 1)(4x+3)$.

Step2: Simplify the rational - function product

After factorizing, the product of rational functions $\frac{6x^{2}+17x + 5}{2x^{2}-2x - 24}\cdot\frac{4x^{2}-13x - 12}{12x^{2}+13x + 3}=\frac{(2x + 5)(3x + 1)}{2(x - 4)(x + 3)}\cdot\frac{(4x + 3)(x - 4)}{(3x + 1)(4x+3)}=\frac{2x + 5}{2(x + 3)}$.

We will solve problem 13 as a sample.
13.

Step1: Factorize numerators and denominators

• Factorize $6x^{2}+17x + 5$:

Using the AC - method, for $ax^{2}+bx + c$ ($a = 6$, $b = 17$, $c = 5$), $ac=6\times5 = 30$. We find two numbers that multiply to 30 and add up to 17, which are 15 and 2. So $6x^{2}+17x + 5=6x^{2}+15x+2x + 5 = 3x(2x + 5)+1(2x + 5)=(2x + 5)(3x + 1)$.

• Factorize $2x^{2}-2x - 24$: First, factor out the common factor 2, getting $2(x^{2}-x - 12)$. Then, for $x^{2}-x - 12$, we find two numbers that multiply to - 12 and add up to - 1, which are - 4 and 3. So $2x^{2}-2x - 24=2(x - 4)(x + 3)$.
• Factorize $4x^{2}-13x - 12$: Using the AC - method, $ac=4\times(-12)=-48$. The numbers are - 16 and 3. So $4x^{2}-13x - 12=4x^{2}-16x+3x - 12 = 4x(x - 4)+3(x - 4)=(4x + 3)(x - 4)$.
• Factorize $12x^{2}+13x + 3$: $ac = 12\times3=36$, the numbers are 9 and 4. So $12x^{2}+13x + 3=12x^{2}+9x+4x + 3 = 3x(4x + 3)+1(4x + 3)=(3x + 1)(4x + 3)$.

Step2: Multiply and simplify

\[

$$\begin{align*} &\frac{6x^{2}+17x + 5}{2x^{2}-2x - 24}\cdot\frac{4x^{2}-13x - 12}{12x^{2}+13x + 3}\\ =&\frac{(2x + 5)(3x + 1)}{2(x - 4)(x + 3)}\cdot\frac{(4x + 3)(x - 4)}{(3x + 1)(4x+3)}\\ =&\frac{2x + 5}{2(x + 3)} \end{align*}$$

\]

Answer:

$\frac{2x + 5}{2(x + 3)}$