Question

split view makes it easier to multitask put pages side-by-side in the same tab so you can easily compare options and more show me how remind me later uestion fully simplify the expression below and write your answer as a single fraction. \\(\frac{9x^2 - 9}{x^4 + 3x^3 - 4x^2} \cdot \frac{x + 4}{x + 1}\\)

Explanation:

Step1: Factor numerators and denominators

Factor \(9x^2 - 9\) as \(9(x^2 - 1)=9(x - 1)(x + 1)\) (using difference of squares \(a^2 - b^2=(a - b)(a + b)\)).
Factor \(x^4 + 3x^3 - 4x^2\) as \(x^2(x^2 + 3x - 4)=x^2(x + 4)(x - 1)\) (factoring quadratic \(x^2 + 3x - 4=(x + 4)(x - 1)\)).

So the expression becomes:
\(\frac{9(x - 1)(x + 1)}{x^2(x + 4)(x - 1)} \cdot \frac{x + 4}{x + 1}\)

Step2: Cancel common factors

Cancel \((x - 1)\), \((x + 1)\), and \((x + 4)\) from numerator and denominator:
\(\frac{9\cancel{(x - 1)}\cancel{(x + 1)}}{x^2\cancel{(x + 4)}\cancel{(x - 1)}} \cdot \frac{\cancel{x + 4}}{\cancel{x + 1}}=\frac{9}{x^2}\)

Answer:

\(\frac{9}{x^2}\)