Question

What is the simplified form of x minus 4 over x squared minus x minus 12 ⋅ x minus 4 over x squared minus 8x plus 16 ?

Answer 1

Answer:

$\frac{1}{(x+3)(x-4)}$

Step-by-step explanation:

The expression we are given is

$\frac{x-4}{x^2-x-12}\times \frac{x-4}{x^2-8x+16}$

We factor the denominator of each expression.

To factor

x²-x-12, we find factors of -12 that sum to -1.

-4(3) = -12 and -4+3 = -1; this gives us

$\frac{x-4}{(x-4)(x+3)}$

Cancelling the common factors in the numerator and denominator, we have

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

To factor

x²-8x+16, we find factors of 16 that sum to -8;

-4(-4) = 16 and -4+-4 = -8.  This gives us

$\frac{x-4}{(x-4)(x-4)}$

Cancelling the common factors in the numerator and denominator, we have

$\frac{1}{x-4}$

This gives us

$\frac{1}{x+3}\times \frac{1}{x-4}\\\\=\frac{1}{(x+3)(x-4)}$

Answer 2

Answer

1 / (x + 3)(x + 4)

That is, 1 over the quantity x plus 3 times the quantity x minus 4.

Step by step explanation

The given expression = $\frac{x-4}{x^2 - x - 12} . \frac{x-4}{x^2 - 8x + 16}$

Let's factor the denominators.

= $\frac{x-4}{(x - 4)(x + 3)} .\frac{x-4}{(x - 4)(x - 4)}$

Cancelling (x- 4) both in the numerator and the denominator, we get

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

= 1/(x + 3)(x - 4)

The answer is 1 over the quantity x plus 3 times the quantity x minus 4.

Thank you.