Question

What is the simplified form of the complex fraction?

Answer 1

`((2x^2+x-6)/(x^2-x-6))/((4x^2-9)/(x^2+5x+6))=(*)\\---------------------\\2x^2+x-6=2x^2+4x-3x-6=2x(x+2)-3(x+2)=(x+2)(2x-3)\\-----------\\x^2-x-6=x^2+2x-3x-6=x(x+2)-3(x+2)=(x+2)(x-3)\\-----------\\x^2+5x+6=x^2+2x+3x+6=x(x+2)+3(x+2)=(x+2)(x+3)\\-----------\\4x^2-9=(2x)^2-3^2=(2x-3)(2x+3)\\----------------------`

`(*)=((x+2)(2x-3))/((x+2)(x-3))\cdot((x+2)(x+3))/((2x-3)(2x+3))=(**)\\----------------------\\(x+2)-canceled\\(2x-3)-canceled\\----------------------\\(**)=((x+2)(x+3))/((x-3)(2x+3))\\\\Answer:\ \boxed{\boxed{((x+2)(x+3))/((2x+3)(x-3))}}`

Answer 2

[(x+3)(x+2)] /[(2x+3)(x-3)]

Explanation:

[(2x²+x-6)/(x²-x-6)]/[(4x²-9)/(x²+5x+6)]

= (2x²+x-6)/(x²-x-6) × (x²+5x+6) /(4x²-9) Factor

= [(2x-3)(x+2)]/[(x-3)(x+2] × [(x+3)(x+2)] /[(2x+3)(2x-3)] Cancel terms

= (2x-3)/(x-3) × [(x+3)(x+2)] /[(2x+3)(2x-3)] Cancel terms

= [(x+3)(x+2)] /[(2x+3)(x-3)]