On dividing $f (x) = 2 x^{5} + 3 x^{4} + 4 x^{3} + 4 x^{2} + 3 x + 2$ by a polynomial $g (x)$ , where $g (x) = x^{3} + x^{2} + x + 1$ , the quotient obtained as $2 x^{2} + x + 1$ . Find the remainder $r (x)$ .

r (x) = x^{2} - 1

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r (x) = 2 x + 1

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r (x) = x - 2

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r (x) = x + 1

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Solution

The correct option is D $r (x) = x + 1$
By remainder theorem,
$f (x) = q (x) g (x) + r (x)$
$∴ 2 x^{5} + 3 x^{4} + 4 x^{3} + 4 x^{2} + 3 x + 2 = (2 x^{2} + x + 1) (x^{3} + x^{2} + x + 1) + r (x)$
$= 2 x^{2} (x^{3} + x^{2} + x + 1) + x (x^{3} + x^{2} + x + 1) + 1 (x^{3} + x^{2} + x + 1) + r (x)$
$= 2 x^{5} + 2 x^{4} + 2 x^{3} + 2 x^{2} + x^{4} + x^{3} + x^{2} + x + x^{3} + x^{2} + x + 1 + r (x)$
$= 2 x^{5} + 3 x^{4} + 4 x^{3} + 4 x^{2} + 2 x + 1 + r (x)$
$r (x) = x + 1$

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On dividing f(x) = x³ - 3x² + x + 2 by a polynomial g(x) the quotient and remainder q(x) and r(x) are (x – 2) and (-2x + 4) respectively, then g(x) is

Q. A polynomial

p (x)

is divided by

g (x)

, the obtained quotient

q (x)

and the remainder

r (x)

are given in the table. Find

p (x)

in each case.

Sl.	$p (x)$	$g (x)$	$q (x)$	$r (x)$
i	?	$x - 2$	$x^{2} - x + 1$	$4$
ii	?	$x + 3$	$2 x^{2} + x + 5$	$3 x + 1$
iii	?	$2 x + 1$	$x^{3} + 3 x^{2} - x + 1$	$0$
iv	?	$x - 1$	$x^{3} - x^{2} - x - 1$	$2 x - 4$
v	?	$x^{2} + 2 x + 1$	$x^{4} - 2 x^{2} + 5 x - 7$	$4 x + 12$