Question

39y^3(50^2-98)÷26y^2(5y-7)

Answer 1

this if the answer........

Answer 2

Heya ! Here is u r Ans >
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$= 39y {}^{3} (50 {}^{2} - 98) \div 26y {}^{2} (5y - 7)$

$= \frac{39y {}^{2} \times 50 {}^{2} - 98 }{26} \times y {}^{2} \times (5y - 7)$

$= \frac{3y {}^{3} \times (50 {}^{2} - 98) }{2} \times y {}^{2} \times (5y - 7)$

$= \frac{3y {}^{5} \times (50 {}^{2} - 98)}{2} \times (5y - 7)$

$= \frac{3y {}^{5} \times (50 {}^{2} - 98) \times 5y - 3y {}^{5} \times (50 {}^{2} - 98) \times 7 }{2}$

$= \frac{15 \times 50 {}^{2}y {}^{6} - 1470y {}^{6} - 21 \times 50 {}^{2}y {}^{5} + 2058y {}^{5} }{2}$

$= \frac{15 \times 2500y {}^{6} - 1470y {}^{6} - 21 \times 50 {}^{2}y {}^{5} + 2058y {}^{5} }{2}$

$= \frac{37500y {}^{6} - 1470y {}^{6} - 21 \times 2500y {}^{5} + 2058y {}^{5} }{2}$

$= \frac{36030y {}^{6} - 21 \times 2500y {}^{5} + 2058y {}^{5} }{2}$

$= \frac{36030y {}^{6} - 52500y {}^{5} + 2058y {}^{5} }{2}$

$= \frac{36030y {}^{6} - 50442y {}^{5} }{2}$

$= \frac{2(18015y {}^{6} - 25221y {}^{5}) }{2}$

$= 18015y {}^{6} - 25221y {}^{5}$
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