Find the remainder when we divide by $(6 a^{5} + 3 a^{4} + 8 a^{3} + 2 a^{2} + 26 a + 35)$ by $(2 a^{2} + 3 a + 5)$

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Solution

We divide $6 a^{5} + 3 a^{4} + 8 a^{3} + 2 a^{2} + 26 a + 35$ by $2 a^{2} + 2 a + 5$ as shown in the above image:

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

Hence, the remainder is $0$ .

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Similar questions

(6 a^{5} + 8 a^{4} + 8 a^{3} + 2 a^{2} + 26 a + 35)

(2 a^{2} + 3 a + 5)

3 a^{3} - 3 a^{2} + a + 7

a^{3} + 2 a^{2} - 5 a x - 7

a + 1

R_{1}

R_{1} = 14

x

\sqrt{3} a^{4} + 2 \sqrt{3} a^{3} + 3 a^{2} - 6 a by 3 a

6 a^{5} + 5 a^{4} b - 8 a^{3} b^{2} - 6 a^{2} b^{3} - 6 a b^{4}

2 a^{3} + 3 a^{2} b - b^{3}

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