Match the two columns. Column-II gives remainder when $x^{3} + 3 x^{2} + 3 x + 1$ is divided by expression given in Column-I.

Column 1 Column 2

$x + 1$

$\frac{27}{8}$

$x$

$- \frac{27}{8}$

$x - \frac{1}{2}$

$1$

$5 + 2 x$

$0$

Column 1	Column 2
$x + 1$	$\frac{27}{8}$
$x$	$- \frac{27}{8}$
$x - \frac{1}{2}$	$1$
$5 + 2 x$	$0$

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Solution

Finding the remainder of a polynomial
The remainder theorem will be applied i.e. when a polynomial $a (x)$ is divided by a linear polynomial $b (x)$ where $x = k$ then the remainder $r = a (k)$
When $f (x) = x^{3} + 3 x^{2} + 3 x + 1$ is divided by
Explanation for (A)
(A) $x + 1$ we get
$\Rightarrow x + 1 = 0 \Rightarrow x = - 1$
$f (- 1) = {(- 1)}^{3} + 3 {(- 1)}^{2} + 3 (- 1) + 1 = - 1 + 3 - 3 + 1 = 0$
Therefore, the remainder is $0$
Therefore, (A) matches to (IV)
Explanation for (B)
(B) $x$ we get
$\Rightarrow x = 0$
$f (0) = {(0)}^{3} + 3 {(0)}^{2} + 3 (0) + 1 = 0 + 0 + 0 + 1 = 1$
Therefore, the remainder is $1$
Therefore, (B) matches to (III)
Explanation for (C)
(C) $x - \frac{1}{2}$ , we get
$\Rightarrow x - \frac{1}{2} = 0 \Rightarrow x = \frac{1}{2}$
$\begin{array}{rcl} f (\frac{1}{2}) & = & {(\frac{1}{2})}^{3} + 3 {(\frac{1}{2})}^{2} + 3 (\frac{1}{2}) + 1 \\ = & \frac{1}{8} + \frac{3}{4} + \frac{3}{2} + 1 \\ = & \frac{1 + 6 + 12 + 8}{8} \\ = & \frac{27}{8} \end{array}$
Therefore, the remainder is $\frac{27}{8}$
Therefore, (C) matches to (II)
Explanation for (D)
(D) $5 + 2 x$ we get
$\Rightarrow 5 + 2 x = 0 \Rightarrow x = \frac{- 5}{2}$
$\begin{array}{rcl} f (\frac{- 5}{2}) & = & {(\frac{- 5}{2})}^{3} + 3 {(\frac{- 5}{2})}^{2} + 3 (\frac{- 5}{2}) + 1 \\ = & \frac{- 125}{8} + \frac{75}{4} + \frac{- 15}{2} + 1 \\ = & \frac{- 125 + 150 + - 60 + 8}{8} \\ = & \frac{- 27}{8} \end{array}$
Therefore, the remainder is $\frac{- 27}{8}$ .
Therefore, (D) matches to (II)
Column-I Column-II
(A) $x + 1$ (IV) $0$
(B) $x$ (III) $1$
(C) $x - \frac{1}{2}$ (I) $\frac{27}{8}$
(D) $5 + 2 x$ (II) $- \frac{27}{8}$

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Q. The polynomial which when divided by −x² + x − 1 gives a quotient x − 2 and remainder 3, is

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(b) −x³ − 3x² − 3x − 5
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Question 151
Given below are two coloumns - Column I and Column II. Match each item of Column I with the corresponding item of Column II.
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Match the two columns. Column- II gives value of $k$ for polynomials given in Column- I when it is divided by $x - 1$ .

Column 1	Column 2
$k x^{2} - 3 x + k$	$- 2$
$x^{2} + x + k$	$\frac{3}{2}$
$2 x^{2} + k x + \sqrt{2}$	$\sqrt{2} - 1$
$k x^{2} - \sqrt{2} x + 1$	$- (2 + \sqrt{2})$

Match the two columns. Column-II gives factors for expression given in Column-I.

Column 1	Column 2
$9 x^{2} + 6 x y + y^{2}$	$(2 x + 3 y - 4 z) (2 x + 3 y - 4 z)$
$4 x^{2} + 9 y^{2} + 16 z^{2} + 12 x y - 24 y z - 16 x z$	$(x + \frac{y}{10}) (x - \frac{y}{10})$
$27 x^{3} + y^{3} + z^{3} - 9 x y z$	$(3 x + y + z) (9 x^{2} + y^{2} + z^{2} - 3 x y - y z - 3 z x)$
$x^{2} - \frac{y^{2}}{100}$	$(3 x + y) (3 x + y)$

Match the two columns. Column II gives the degree of polynomials given in column I match them correctly –

Column 1	Column 2
$2 - y^{2} - y^{3} + 2 y^{8}$	$2$
$2$	$1$
$5 x - \sqrt{7}$	$0$
$4 - x^{2}$	$8$

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Column-I	Column-II
(A) $x + 1$	(IV) $0$
(B) $x$	(III) $1$
(C) $x - \frac{1}{2}$	(I) $\frac{27}{8}$
(D) $5 + 2 x$	(II) $- \frac{27}{8}$

Match the two columns. Column-II gives remainder when x3+3x2+3x+1 is divided by expression given in Column-I.Column 1Column 2 x+1 278 x -278 x-12 1 5+2x 0

Match the two columns. Column-II gives remainder when $x^{3} + 3 x^{2} + 3 x + 1$ is divided by expression given in Column-I.

Column 1 Column 2

$x + 1$

$\frac{27}{8}$

$x$

$- \frac{27}{8}$

$x - \frac{1}{2}$

$1$

$5 + 2 x$

$0$