Question

X^2 - 2x + 1 x^2 - 2x - 1 x^2 + 2x + 1 In the denominator

Answer 1

Answer:

$x^{2} -4x+4$ in the numerator

$x^{2} -2x+1$ in the denominator

Step-by-step explanation:

First fraction simplifies to:

$\frac{x^{2}+x-6}{x^{2}-6x+5}\\\frac{x^{2}-2x+3x-6}{x^{2}-x-5x+5}\\\frac{x(x-2)+3(x-2)}{x(x-1)+5(x-1)}\\\frac{(x+3)(x-2)}{(x-1)(x+5)}$

Dividing my a fraction is the same as multiplying by the reciprocal, so the second fraction simplifies to:

$\frac{x^{2}-7x+10}{x^{2}+2x-3}\\\frac{x^{2}-2x+5x-10}{x^{2}-x+3x-3}\\\frac{x(x-2)+5(x-2)}{x(x-1)+3(x-1)}\\\frac{(x+5)(x-2)}{(x-1)(x+3)}$

So the equation becomes:

$\frac{(x+3)(x-2)}{(x+5)(x-1)} * \frac{(x+5)(x-2)}{(x-1)(x+3)}$

The (x+3) and (x+5) terms cancel out, so what your left with is

$\frac{(x-2)^{2}}{(x-1)^{2}}$

Expanding this term you get the answer:

$x^{2} -4x+4$ in the numerator

$x^{2} -2x+1$ in the denominator