On dividing $f (x) = 3 x^{3} + x^{2} + 2 x + 5$ by a polynomial $g (x) = x^{2} + 2 x + 1$ , the remainder $r (x) = 9 x + 10$ . Find the quotient polynomial $q (x)$

q (x) = x - 15

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q (x) = 3 x - 5

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q (x) = 4 x - 18

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None of these

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Solution

The correct option is B $q (x) = 3 x - 5$
By remainder theorem,
$f (x) = q (x) g (x) + r (x)$
$∴ 3 x^{3} + x^{2} + 2 x + 5 = q (x) (x^{2} + 2 x + 1) + 9 x + 10$
$\Rightarrow q (x) (x^{2} + 2 x + 1) = 3 x^{3} + x^{2} - 7 x - 5$
Now,
$x^{2} + 2 x + 1) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 3 x^{3} + x^{2} - 7 x - 5$ ( $3 x - 5 = q (x)$
$3 - x^{3} + - 6 x^{2} + - 3 x - ------------- -$
$- 5 x^{2} - 10 x - 5$
$- + 5 x^{2} - + 10 x - + 5 - -------------- -$
$0$
Hence, $q (x) = 3 x - 5$

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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	?	$x^{2} + 2 x + 1$	$x^{4} - 2 x^{2} + 5 x - 7$	$4 x + 12$