Question

39y^3(50x^2-98)÷26y^2(5y+7)

Answer 1

The question should be this:

39y^3(50y^2-98)÷26y^2(5y+7)

It was tough but I solved it:

Here its:

dividing 39y^3 by 26y^y2  ....(1)

then taking 2 as common from (50x^2-98)

then it would be 2(25x^2-49) ....(2)

by (1) and (2)

=>3y * 2(25y^2 - 49) / 2 * (5y + 7)

=>3y * 2({5y}^2 - (7)^2) / 2 * (5y + 7)

=>3y * 2( {5y-7} * {5y+7}) / 2 * (5y + 7)           by a^2 - b^2 identity....

2 and 5y +7  will be cancel

=> 3y (5y-7)

15y^2 - 21y

Recheck ur question though...