Divide the polynomial $39 y^{3} (50 y^{2} - 98)$ by $36 y^{2} (5 y + 7)$

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Solution

Factorising the polynomial $39 y^{3} (50 y^{2} - 98)$

we get $39 y^{3} (50 y^{2} - 98) = 39 \times 2 y^{3} (5 y - 7) (5 y + 7)$

Thus dividing the above polynomial by $36 y^{2} (5 y + 7)$ , we get

$\frac{39 \times 2 y^{3} (5 y - 7) (5 y + 7)}{36 y^{2} (5 y + 7)} = \frac{13}{6} y (5 y - 7)$

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