Question 189464
{{{((x^2+6x-7)/(x^2+5x))((3x^2+16x+5)/(2x^2+7x-9))}}} Start with the given expression



{{{(((x+7)(x-1))/(x^2+5x))((3x^2+16x+5)/(2x^2+7x-9))}}} Factor the first numerator



{{{(((x+7)(x-1))/(x(x+5)))((3x^2+16x+5)/(2x^2+7x-9))}}} Factor the first denominator



{{{(((x+7)(x-1))/(x(x+5)))(((x+5)(3x+1))/(2x^2+7x-9))}}} Factor the second numerator



{{{(((x+7)(x-1))/(x(x+5)))(((x+5)(3x+1))/((2x+9)(x-1)))}}} Factor the second denominator



{{{((x+7)(x-1)(x+5)(3x+1))/(x(x+5)(2x+9)(x-1))}}} Combine the fractions.



{{{((x+7)highlight((x-1))highlight((x+5))(3x+1))/(x*highlight((x+5))(2x+9)highlight((x-1)))}}} Highlight the common terms



{{{((x+7)cross((x-1))cross((x+5))(3x+1))/(x*cross((x+5))(2x+9)cross((x-1)))}}} Cancel out the common terms.



{{{((x+7)(3x+1))/(x(2x+9))}}} Simplify



{{{(3x^2+22x+7)/(x(2x+9))}}} FOIL



{{{(3x^2+22x+7)/(2x^2+9x)}}} Distribute



So {{{((x^2+6x-7)/(x^2+5x))((3x^2+16x+5)/(2x^2+7x-9))}}} simplifies to {{{(3x^2+22x+7)/(2x^2+9x)}}}



In other words, {{{((x^2+6x-7)/(x^2+5x))((3x^2+16x+5)/(2x^2+7x-9))=(3x^2+22x+7)/(2x^2+9x)}}} where {{{x<>-5}}}, {{{x<>-9/2}}}, {{{x<>0}}}, or {{{x<>1}}} (these are the restrictions)