Question

solve this :-pleswehsk

Answer 1

Solution:-

$\sf \to \dfrac{(x-3)(x-5)(x^{2}-14x+24)}{(x-7)(x^{2}-8x+15)}$

• Do the middle term splitting in the denominator.

$\sf \to \dfrac{(x-3)(x-5)(x^{2}-14x+24)}{(x-7)(x^{2}-3x-5x+15)}$

• Take out the common terms.

$\sf \to \dfrac{(x-3)(x-5)(x^{2}-14x+24)}{(x-7)(x(x-3)-5(x-3))}$

• Factor out x - 3 from the expression.

$\sf \to \dfrac{(x-3)(x-5)(x^{2}-14x+24)}{(x-7)(x-5)(x-3)}$

• Cancel out the terms.

$\sf \to \dfrac{\cancel{(x-3)}\cancel{(x-5)}(x^{2}-14x+24)}{(x-7)\cancel{(x-5)}\cancel{(x-3)}}$

$\sf \to \dfrac{(x^{2}-14x+24)}{(x-7)}$

• Do the middle term splitting in the numerator.

$\sf \to \dfrac{(x^{2}-2x-12x+24)}{(x-7)}$

• Take out the common terms.

$\sf \to \dfrac{(x(x-2)-12(x-2))}{(x-7)}$

• Factor out x - 2 from the expression.

$\sf \to \dfrac{(x-2)(x-12)}{(x-7)}$