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

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Answer 1 · 2018-03-03T06:53:33

$(-((3y+1)(3y-1)(3y-1)))/((y+5)(y-5)(y-5))$

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

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

$-((3y+1)(3y-1))/((y+5)(y-5)).(3y^2+5y-2)/((y+2)(y-5))$

$-((3y+1)(3y-1))/((y+5)(y-5)).(3y-1)/(y-5)$

$(-((3y+1)(3y-1)(3y-1)))/((y+5)(y-5)(y-5))$

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