Question

2x² - 7x - 15
x² - 49
15 - 3x
x² + 2x - 63

Explanation:

Step1: Rewrite as fraction multiplication

$\frac{2x^2 - 7x - 15}{x^2 - 49} \div \frac{15 - 3x}{x^2 + 2x - 63} = \frac{2x^2 - 7x - 15}{x^2 - 49} \times \frac{x^2 + 2x - 63}{15 - 3x}$

Step2: Factor all polynomials

• $2x^2 -7x -15=(2x+3)(x-5)$
• $x^2 -49=(x+7)(x-7)$
• $x^2 +2x -63=(x+9)(x-7)$
• $15-3x=-3(x-5)$

Substitute into the expression:
$\frac{(2x+3)(x-5)}{(x+7)(x-7)} \times \frac{(x+9)(x-7)}{-3(x-5)}$

Step3: Cancel common factors

Cancel $(x-5)$ and $(x-7)$:
$\frac{(2x+3)}{(x+7)} \times \frac{(x+9)}{-3}$

Step4: Multiply remaining terms

$\frac{(2x+3)(x+9)}{-3(x+7)} = -\frac{(2x+3)(x+9)}{3(x+7)}$

Answer:

$-\frac{(2x+3)(x+9)}{3(x+7)}$ (or expanded form $-\frac{2x^2+21x+27}{3x+21}$, where $x
eq -7,5,7$)