Question 119550
{{{((16x)/(25x^2-5))((2x^2-x-1)/(8x))}}} Start with the given expression


{{{((16(x))/(25x^2-5))((2x^2-x-1)/(8x))}}}   Factor {{{16x}}} to get {{{16(x)}}} 


{{{((16(x))/(5(5x^2-1)))((2x^2-x-1)/(8x))}}}   Factor {{{25x^2-5}}} to get {{{5(5x^2-1)}}}


{{{((16(x))/(5(5x^2-1)))(((x-1)(2x+1))/(8x))}}}   Factor {{{2x^2-x-1}}} to get {{{(x-1)(2x+1)}}} 


{{{((16(x))/(5(5x^2-1)))(((x-1)(2x+1))/(8(x)))}}}   Factor {{{8x}}} to get {{{8(x)}}} 



{{{16(x)(x-1)(2x+1)/5(5x^2-1)8(x)}}} Combine the fractions



{{{16cross(x)(x-1)(2x+1)/5(5x^2-1)8cross(x)}}} Cancel like terms




{{{2(x-1)(2x+1)/5(5x^2-1)}}} Divide and simplify




So {{{((16x)/(25x^2-5))((2x^2-x-1)/(8x))}}} simplifies to {{{2(x-1)(2x+1)/5(5x^2-1)}}}




In other words, {{{((16x)/(25x^2-5))((2x^2-x-1)/(8x))=2(x-1)(2x+1)/5(5x^2-1)}}}