The quotient and the remainder when $x^{4} - 11 x^{3} + 44 x^{2} - 76 x + 48$ is divided by $x^{2} - 7 x + 12$ , are respectively

x^{2} - 4 x - 2; 0

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x^{2} - 4 x + 4; 0

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x^{2} - 4 x; x + 12

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x^{2} + 4 x; x + 12

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Solution

The correct option is D $x^{2} - 4 x + 4; 0$
Let $x^{4} - 11 x^{3} + 44 x^{2} - 76 x + 48 = (x^{2} - 7 x + 12) (a x^{2} + b x + c) + d x + e$ , where $d x + e$ is the remainder.
We have,
$(x^{2} - 7 x + 12) (a x^{2} + b x + c) + d x + e$
$= (a) x^{4} + (- 7 a + b) x^{3} + (c - 7 b + 12 a) x^{2} + (- 7 c + 12 b + d) x + 12 c + e$
Comparing the coefficients we get $a = 1$
$∴ (- 7 a + b) = - 11$
$\Rightarrow b = - 4$
and $c - 7 b + 12 a = 44$
$\Rightarrow c = 4$
$- 7 c + 12 b + d = - 76$
$\Rightarrow d = 0$
$12 c + e = 48$
$\Rightarrow e = 0$
Hence, the remainder is $0$ and the quotient is $x^{2} - 4 x + 4$ .

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