Solve this question and answer please quick.
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siddhartharao77:
I got 3y(5y - 7) (or) 15y^2 - 21y. is it correct?
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I think this one is....
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Answered by
1
Given 39y^3(50y^2 - 98)/ 26y^2(5y + 7) can be written as..
= 13 * 3y^3(50y^2 - 98)/13 * 2y^2(5y + 7)
= 3y^3(50y^2 - 98)/2y^2(5y + 7)
= 3y(50y^2 - 98)/2(5y + 7)
= 2 * 3y(25y^2 - 49)/2(5y + 7)
= 6y(25y^2 - 49)/2(5y + 7)
= 3y(25y^2 - 49)/(5y + 7)
We know that (a^2 - b^2) = (a - b)(a+b)
= 3y(5y - 7)(5y + 7)/(5y + 7)
= 3y(5y - 7) (or) 15y^2 - 21y.
Hope this helps!
= 13 * 3y^3(50y^2 - 98)/13 * 2y^2(5y + 7)
= 3y^3(50y^2 - 98)/2y^2(5y + 7)
= 3y(50y^2 - 98)/2(5y + 7)
= 2 * 3y(25y^2 - 49)/2(5y + 7)
= 6y(25y^2 - 49)/2(5y + 7)
= 3y(25y^2 - 49)/(5y + 7)
We know that (a^2 - b^2) = (a - b)(a+b)
= 3y(5y - 7)(5y + 7)/(5y + 7)
= 3y(5y - 7) (or) 15y^2 - 21y.
Hope this helps!
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