Question 195791

{{{((2x^2+5x-12)/(9x^2-16))/((2x^2-7x+6)/(3x^2-x-4))}}} Start with the given expression.



{{{((2x^2+5x-12)/(9x^2-16))((3x^2-x-4)/(2x^2-7x+6))}}} Multiply the first fraction {{{(2x^2+5x-12)/(9x^2-16)}}} by the reciprocal of the second fraction {{{(2x^2-7x+6)/(3x^2-x-4)}}}.



{{{(((x+4)(2x-3))/(9x^2-16))((3x^2-x-4)/(2x^2-7x+6))}}} Factor {{{2x^2+5x-12}}} to get {{{(x+4)(2x-3)}}}.



{{{(((x+4)(2x-3))/((3x-4)(3x+4)))((3x^2-x-4)/(2x^2-7x+6))}}} Factor {{{9x^2-16}}} to get {{{(3x-4)(3x+4)}}}.



{{{(((x+4)(2x-3))/((3x-4)(3x+4)))(((x+1)(3x-4))/(2x^2-7x+6))}}} Factor {{{3x^2-x-4}}} to get {{{(x+1)(3x-4)}}}.



{{{(((x+4)(2x-3))/((3x-4)(3x+4)))(((x+1)(3x-4))/((2x-3)(x-2)))}}} Factor {{{2x^2-7x+6}}} to get {{{(2x-3)(x-2)}}}.



{{{((x+4)(2x-3)(x+1)(3x-4))/((3x-4)(3x+4)(2x-3)(x-2))}}} Combine the fractions. 



{{{((x+4)highlight((2x-3))(x+1)highlight((3x-4)))/(highlight((3x-4))(3x+4)highlight((2x-3))(x-2))}}} Highlight the common terms. 



{{{((x+4)cross((2x-3))(x+1)cross((3x-4)))/(cross((3x-4))(3x+4)cross((2x-3))(x-2))}}} Cancel out the common terms. 



{{{((x+4)(x+1))/((3x+4)(x-2))}}} Simplify. 



{{{(x^2+5x+4)/(3x^2-2x-8)}}} FOIL



So {{{((2x^2+5x-12)/(9x^2-16))/((2x^2-7x+6)/(3x^2-x-4))}}} simplifies to {{{(x^2+5x+4)/(3x^2-2x-8)}}}.



In other words, {{{((2x^2+5x-12)/(9x^2-16))/((2x^2-7x+6)/(3x^2-x-4))=(x^2+5x+4)/(3x^2-2x-8)}}} where {{{x<>-4/3}}}, {{{x<>-1}}}, {{{x<>4/3}}}, {{{x<>3/2}}}, or {{{x<>2}}}