What must be added to $6 x^{4} - x^{3} + 3 x^{2} + 4 x - 2$ , so that the result is exactly divisible by $3 x^{2} - 2 x + 4$ ? Also find the degree of the quotient.

2 x - 2, 2

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2 x + 2, 2

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2 x - 3, 2

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None of these

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Solution

The correct option is A $2 x - 2, 2$
$a^{2} + b^{2} + a c^{2} + 2 (a b + 2 b c + 2 c a) = (a + b + 2 c)^{2}$
when we divide $(a + b + 2 c)^{2}$ by $(a + b + 2 c)$ we get $(a + b + 2 c)$
$\Rightarrow$ Remainder is zero
$∴ R = - 2 x + 2$
$\Rightarrow$ Adding $(2 x - 2)$ in given polynomial will made R = 0 & hence exactly divisible by $(3 x^{2} - 2 x + 4)$
having a degree of $2$

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