Find the remainder after diving $9 x^{4} + 3 x^{3} y + 16 x^{2} y^{2} + 24 x y^{3} + 32 y^{4}$ by $(3 x^{2} + 5 x y + 4 y^{2})$ .

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Solution

We divide $9 x^{4} + 3 x^{3} y + 16 x^{2} y^{2} + 24 x y^{3} + 32 y^{4}$ by $3 x^{2} + 5 x y + 4 y^{2}$ as shown in the above image:

From the above division, we get that the quotient is $3 x^{2} - 4 x y + 8 y^{2}$ and the remainder is $0$

Hence, the remainder is $0$ .

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(9 x^{4} + 3 x^{3} y + 16 x^{2} y^{2} + 24 x y^{3} + 32 y^{4})

(3 x^{2} + 5 x y + 4 y^{2})

(3 x^{2} + 5 x y + 4 y^{2})

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