How do you divide $\frac{1 - 9 y^{2}}{y^{2} - 25} \div \frac{y^{2} - 3 y - 10}{3 y^{2} + 5 y - 2}$ $\frac { 1- 9y ^ { 2} } { y ^ { 2} - 25} \div \frac { y ^ { 2} - 3y - 10} { 3y ^ { 2} + 5y - 2}$ ?

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Answer 1 · 2018-03-03T06:53:33

$\frac{- ((3 y + 1) (3 y - 1) (3 y - 1))}{(y + 5) (y - 5) (y - 5)}$ $(-((3y+1)(3y-1)(3y-1)))/((y+5)(y-5)(y-5))$

$(\frac{1 - 9 y^{2}}{y^{2} - 25}) . \frac{3 y^{2} + 5 y - 2}{y^{2} - 3 y - 10}$ $((1-9y^2)/(y^2-25) ). (3y^2+5y-2)/(y^2-3y-10)$

$\frac{- (9 y^{2} + 1)}{y^{2} - 25} . \frac{3 y^{2} + 5 y - 2}{y^{2} - 3 y - 10}$ $(-(9y^2+1))/(y^2-25) . (3y^2+5y-2)/(y^2-3y-10)$

$- \frac{(3 y + 1) (3 y - 1)}{(y + 5) (y - 5)} . \frac{3 y^{2} + 5 y - 2}{(y + 2) (y - 5)}$ $-((3y+1)(3y-1))/((y+5)(y-5)).(3y^2+5y-2)/((y+2)(y-5))$

$- \frac{(3 y + 1) (3 y - 1)}{(y + 5) (y - 5)} . \frac{3 y - 1}{y - 5}$ $-((3y+1)(3y-1))/((y+5)(y-5)).(3y-1)/(y-5)$

$\frac{- ((3 y + 1) (3 y - 1) (3 y - 1))}{(y + 5) (y - 5) (y - 5)}$ $(-((3y+1)(3y-1)(3y-1)))/((y+5)(y-5)(y-5))$

How do you divide 1−9y2y2−25÷y2−3y−103y2+5y−2\frac { 1- 9y ^ { 2} } { y ^ { 2} - 25} \div \frac { y ^ { 2} - 3y - 10} { 3y ^ { 2} + 5y - 2}?