Question

The quotient of x^2+x-6/x^2-6x+5*x^2+2x-3/x^2-7x+10 has ___ in the numerator and ______ in the denominator.

Answer 1

Answer:

So the quotient of $\frac{x^{2} + x - 6}{x^{2} -6x + 5} X \frac{x^{2} + 2x - 3}{x^{2} -7x + 10}$ has (x + 3)² in the numerator and (x + 5)² in the denominator.

Step-by-step explanation:

$\frac{x^{2} + x - 6}{x^{2} -6x + 5} X \frac{x^{2} + 2x - 3}{x^{2} -7x + 10}$

Factorizing the expressions we have

$\frac{x^{2} + 3x -2x - 6}{x^{2} -x - 5x + 5} X \frac{x^{2} + 3x - x - 3}{x^{2} -2x -5x + 10}$

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

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

Cancelling out the like factors, (x -1) and (x - 2), we have

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

= $\frac{(x + 3)^{2} }{(x + 5)^{2} }$

So the quotient of $\frac{x^{2} + x - 6}{x^{2} -6x + 5} X \frac{x^{2} + 2x - 3}{x^{2} -7x + 10}$ has (x + 3)² in the numerator and (x + 5)² in the denominator.