What will be the quotient and remainder if $6 x^{4} + 3 x^{2} - 9 + 5 x + 5 x^{3} \div (x^{2} - 1)$ ?

Quotient =

(6 x^{2} + 5 x + 9)

Remainder =

(10 x)

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Quotient =

(6 + 5 x + 9 x)

Remainder =

(10 x^{3})

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Remainder=

(6 x^{2} + 5 x + 9)

Quotient =

(10 x)

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Remainder=

(6 x^{2} + 5 x + 9)

Quotient =

(10 x^{2})

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Solution

The correct option is A Quotient = $(6 x^{2} + 5 x + 9)$
Remainder = $(10 x)$
Given: $(6 x^{4} + 3 x^{2} - 9 + 5 x + 5 x^{3}) \div (x^{2} - 1)$
$\begin{matrix} 6 x^{2} + 5 x + 9 (x^{2} - 1) 6 x^{4} + 5 x^{3} + 3 x^{2} + 5 x - 9 6 x^{4} - 0 x^{3} - 6 x^{2} - 0 x - + + + - --------------------------- - 5 x^{3} + 9 x^{2} + 5 x - 9 5 x^{3} + 0 - 5 x - 0 - - + + - ------------------- - 9 x^{2} + 10 x - 9 9 x^{2} + 0 - 9 - - + - ------------- - 10 x - ------------------ - \end{matrix}$
Quotient = $6 x^{2} + 5 x + 9$
Remainder = $10 x$

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