Question

fully simplify the expression below and write your answer as a single fraction.
$\frac{x^{4}+x^{3}}{x^{2}+3x+2} cdot \frac{15x^{2}-60}{15x^{5}-75x^{4}+90x^{3}}$

Explanation:

Step1: Factor all numerators/denominators

1. $x^4+x^3 = x^3(x+1)$
2. $x^2+3x+2 = (x+1)(x+2)$
3. $15x^2-60 = 15(x^2-4)=15(x+2)(x-2)$
4. $15x^5-75x^4+90x^3 = 15x^3(x^2-5x+6)=15x^3(x-2)(x-3)$

Step2: Rewrite product with factored forms

$$\frac{x^3(x+1)}{(x+1)(x+2)} \cdot \frac{15(x+2)(x-2)}{15x^3(x-2)(x-3)}$$

Step3: Cancel common factors

Cancel $x^3$, $(x+1)$, $(x+2)$, $15$, $(x-2)$ from numerator/denominator:
$$\frac{\cancel{x^3}\cancel{(x+1)}}{\cancel{(x+1)}\cancel{(x+2)}} \cdot \frac{\cancel{15}\cancel{(x+2)}\cancel{(x-2)}}{\cancel{15}\cancel{x^3}\cancel{(x-2)}(x-3)}$$

Step4: Simplify remaining terms

$$\frac{1}{1} \cdot \frac{1}{(x-3)} \quad \text{correction: correction to step 2 denominator factorization error: } x^2-5x+6=(x-2)(x-3) \text{ is wrong, } x^2-5x+6=(x-2)(x-3) \text{ correct, but recheck full cancellation:}$$
Re-express correct cancellation:
$$\frac{x^3(x+1) \cdot 15(x+2)(x-2)}{(x+1)(x+2) \cdot 15x^3(x-2)(x-3)} = \frac{1}{x-3}$$
Correction of initial factoring oversight:
Re-express the original denominator $15x^5-75x^4+90x^3$ correctly:
$15x^3(x^2-5x+6)=15x^3(x-2)(x-3)$
After canceling all common factors: $x^3, (x+1), 15, (x+2), (x-2)$
Final simplified form: $\frac{1}{x-3}$

Corrected Final Answer:

$\frac{1}{x-3}$

Corrected Explanation:

Step1: Factor all components

$x^4+x^3=x^3(x+1)$; $x^2+3x+2=(x+1)(x+2)$; $15x^2-60=15(x+2)(x-2)$; $15x^5-75x^4+90x^3=15x^3(x-2)(x-3)$

Step2: Substitute factored forms

$$\frac{x^3(x+1)}{(x+1)(x+2)} \cdot \frac{15(x+2)(x-2)}{15x^3(x-2)(x-3)}$$

Step3: Cancel common factors

Eliminate matching terms top/bottom: $x^3, (x+1), 15, (x+2), (x-2)$

Step4: Write simplified fraction

$$\frac{1}{x-3}$$

Answer:

$\frac{1}{x(x-2)}$