Please Help me Simplify?

Question

(2x^2-x-6)/(x^2-3x+2) divided by (2x^2+5x+3)/(3x^2-3)

Answer 1

First factor out the expressions........

(2x+3)(x-2)/(x-1)(x-2) divided by (2x+3)(x+1)/3(x+1)(x-1) becomes

(2x+3)/(x-1) divided by (2x+3)/3(x-1) which of course is 3

Answer 2

(2x^2-x-6)/(x^2-3x+2) = (2x + 3)(x - 2)/(x - 2)(x - 1)--------> remove x - 2

(2x + 3)/(x - 1)

(2x^2+5x+3)/(3x^2-3) = (2x + 3)(x + 1)/(3(x + 1)(x - 1)----> remove x + 1

(2x + 3)/3(x - 1)

{(2x + 3)/(x - 1)} / {2x + 3)/3(x - 1)} -------> when dividing by faction, just inverse multiply

{(2x + 3)/(x - 1)} * {3(x - 1)/(2x + 3)}----> remove x - 1 and 2x + 3

1/1 * 3/1 = 3

Answer 3

[(2x^2-x-6)/(x^2-3x+2)]/[(2x^2+5x+3)/(3x...

To divide fractions, multiply the big numerator by the reciprocal of the big denominator

[(2x^2-x-6)/(x^2-3x+2)] [(3x^2 - 3)/(2x^2 + 5x + 3)]

Now factor where possible to see what cancels

{[(x - 2)(2x + 3)]/[(x - 1)(x - 2)]} {[3(x + 1)(x - 1)]/[(2x + 3)(x + 1)}

3

If you have trouble with the harder factoring

2x^2 - x - 6

ac = 2(-6) = -12

Factors of -12 that add to -1 are +3 & -4

2x^2 + 3x - 4x - 12 {rewrite for grouping}

x(2x + 3) - 2(3x + 3)

(x - 2)(2x + 3)

Answer 4

(2x+3)(x-2)/(x-2)(x-1) * 3(x+1)(x-1)/(2x+3)(x+1)

= 3