Question

I need help with this math problem

Answer 1

Answer:

1). $\frac{2}{x^{2}-x-12 }=\frac{2}{(x+3)(x-4)}$

2). $\frac{1}{x^{2}-16 }=\frac{1}{(x-4)(x+4)}$

Step-by-step explanation:

In this question we have to write the fractions in the factored form.

Rational expressions are $\frac{2}{x^{2}-x-12 }$ and $\frac{1}{x^{2}-16 }$.

1). $\frac{2}{x^{2}-x-12 }$

Factored form of the denominator (x² - x - 12) = x² - 4x + 3x - 12

                                                                           = x(x - 4) + 3(x - 4)

                                                                           = (x + 3)(x - 4)

Therefore. $\frac{2}{x^{2}-x-12 }=\frac{2}{(x+3)(x-4)}$

2). $\frac{1}{x^{2}-16 }$

Factored form of the denominator (x² - 16) = (x - 4)(x + 4)

[Since (a²- b²) = (a - b)(a + b)]

Therefore, $\frac{1}{x^{2}-16 }=\frac{1}{(x-4)(x+4)}$