How do you evaluate $\frac { x ^ { 3} y } { x ^ { 2} - 3x - 10} \cdot \frac { 5x ^ { 2} - 26x + 5} { x y ^ { 6} }$ ?

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Answer 1 · 2018-05-21T21:50:44

$(x^3-x^2)/(xy^5+2y^5)$

$(x^3y)/(x^2-3x-10)xx(5x^2-26x+5)/(x y^6)$

$(x^3y)/((x +2)(x-5))xx((x-5)(5x-1))/(x y^6)$

$(x^3y)/(x +2)xx(5x-1)/(xy^6)$

$x^2/(x +2)xx(5x-1)/y^5$

$(x^3-x^2)/(xy^5+2y^5)$

How do you evaluate \frac { x ^ { 3} y } { x ^ { 2} - 3x - 10} \cdot \frac { 5x ^ { 2} - 26x + 5} { x y ^ { 6} }\frac { x ^ { 3} y } { x ^ { 2} - 3x - 10} \cdot \frac { 5x ^ { 2} - 26x + 5} { x y ^ { 6} }?