You can put this solution on YOUR website! Divide:
[ 2x^(2)+5x–12 / 9x^(2)–16 ] ÷ [ 2x^(2)–7x+6 / 3x^(2)–x–4 ]
Factor where you can to get:
[ 2x^2 + 8x - 3x -12 / (3x-4)(3x+4) ] / [ 2x^2 -3x -4x + 6 / (3x^2-4x + 3x - 4 ]
= [ 2x(x+4) - 3(x+4) / (3x-4)(3x+4) ] / [ x(2x-3) -2(2x-3) / x(3x-4) + (3x-4) ]
= [ (x+4)(2x-3) / (3x-4)(3x+4) ] / [ (2x-3)(x-2) / (3x-4)(x+1) ]
Invert the denominator and multiply to get:
= [ (x+4)(2x-3) / (3x-4)(3x+4) ] * [(3x-4)(x+1) / (2x-3)(x-2) ]
Cancel any factors that are common to a numerator and denominator
such as (2x-3), (3x-4)
= [ (x+4) / (3x+4) ] * [(x+1) / (x-2) ]
= [(x+4)(x+1)] / [(3x+4)(x-2)]
=========================================
Cheers,
Stan H.
(