Find the remainder after dividing $(6 y^{5} + 15 y^{4} + 16 y^{3} + 4 y^{2} + 10 y - 35)$ by $(3 y^{2} + 5)$

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Solution

We divide $6 y^{5} + 15 y^{4} + 16 y^{3} + 4 y^{2} + 10 y - 35$ by $3 y^{2} + 5$ as shown in the above image:

From the above division, we get that the quotient is $2 y^{3} + 5 y^{2} + 2 y - 7$ and the remainder is $0$

Hence, the remainder is $0$ .

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(15 y^{4} - 16 y^{3} + 9 y^{2} - \frac{1}{3} y - \frac{50}{9})

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The remainder obtained when $6 y^{5} + 30 y^{4} + 18 y^{3} + 6 y^{2} + 15 y + 3$ is divided by $2 y^{3} + 1$ is $3 y^{2} - 6$ .

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