Question

What is the simplified form of x plus 3 over x squared minus x minus 12 ⋅ x minus 4 over x squared minus 8x plus 16 ?

Answer 1

Answer:

$\dfrac{1}{x^2-8x+16}$

Step-by-step explanation:

The expression, written in full, looks like this:

$\dfrac{x+3}{x^2-x-12}\cdot\dfrac{x-4}{x^2-8x+16}$

To simplify this expression, it would help us out a lot if we could factor the expressions in the denominators. Let's handle $x^2-x-12$ first:

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

Next, we can factor $x^2-8x+16$:

$x^2-8x+16=\\=x^2-4x-4x+16\\=x(x-4)-4(x-4)\\=(x-4)(x-4)\\=(x-4)^2$

Substituting these back into our original expression, we get

$\dfrac{x+3}{(x-4)(x+3)}\cdot\dfrac{x-4}{(x-4)^2}$

On the left, we can cancel an (x+3) in the numerator and denominator, and on the right, we can cancel an (x-4), simplifying the expression to

$\dfrac{1}{x-4}\cdot\dfrac{1}{x-4}$

Multiplying the two together gives us the fraction

$\dfrac{1\cdot1}{(x-4)\cdot(x-4)}=\dfrac{1}{(x-4)^2}$

Since $(x-4)^2=x^2-8x+16$, we can rewrite this fraction in simplified form as

$\dfrac{1}{x^2-8x+16}$