Question

Please help me I am confused on how to do these.

Answer 1

1. There's so much wrong with the suggested solution that it's hard to pinpoint the exact error. The "logic" is inconsistent - why do the $x$ and constant term stay in the denominator but the $x^2$ term does not?

What should be done is factorization in the numerator and denominator:

$x^2 + 2x - 8 = (x + 4) (x - 2)$

$x^2 + 6x - 16 = (x + 8) (x - 2)$

Then in the quotient, the factors of $x-2$ cancel so that

$\dfrac{x^2+2x-8}{x^2+6x-16} = \dfrac{(x+4)(x-2)}{(x+8)(x-2)} = \boxed{\dfrac{x+4}{x+8}}$

2. Use the same strategy: factorize everything everything you can, then cancel anything you can.

$x^2 - 9 = (x-3) (x+3)$

$x^2+x-2 = (x+2)(x-1)$

$x^2+x-12 = (x+4)(x-3)$

$x^2+6x+8 = (x+4)(x+2)$

Then using the algebraic properties of multiplication/division, we have

$\dfrac{x^2-9}{x^2+x-2} \div \dfrac{x^2+x-12}{x^2+6x+8} = \dfrac{x^2-9}{x^2+x-2} \times \dfrac{x^2+6x+8}{x^2+x-12} \\\\ ~~~~~~~~= \dfrac{(x-3)(x+3)(x+4)(x+2)}{(x+2)(x-1)(x+4)(x-3)} \\\\ ~~~~~~~~= \boxed{\dfrac{x+3}{x-1}}$