# Polynomials : Exercise 2.3 (Mathematics NCERT Class 9th) Q.1В В В В В  Find the remainder when ${x^3} + 3{x^2} + 3x + 1$ is divided by
В В В В В В В В В В В В В В В  (i) x + 1 В  В  В  В  В  В  В  В  В  В  В  В В  (ii) $x - {1 \over 2}$
В В В В В В В В В В В В В В В  (iii) x В  В  В  В  В  В  В  В  В  В  В  В  В  В В  (iv) $x + \pi$В В В В В В В В В В В  (v) 5 + 2x
Sol.

(i) By remainder theorem, the required remainder is equal to p (вЂ“1).
Now, $p\left( x \right) = {x^3} + 3{x^2} + 3x + 1$
Therefore В $p\left( { - 1} \right) = {\left( { - 1} \right)^3} + 3{\left( { - 1} \right)^2} + 3\left( { - 1} \right) + 1$
$= - 1 + 3 - 3 + 1 = 0$
Hence, required remainder = p(вЂ“1) = 0

(ii) By remainder theorem, the required remainder is equal to $p\left( {{1 \over 2}} \right)$.
Now, $p\left( {{1 \over 2}} \right) = {x^3} + 3{x^2} + 3x + 1$
$= {\left( {{1 \over 2}} \right)^3} + 3{\left( {{1 \over 2}} \right)^2} + 3\left( {{1 \over 2}} \right) + 1$
$= {1 \over 8} + {3 \over 4} + {3 \over 2} + 1 = {{1 + 6 + 12 + 8} \over 8}$
$= {{27} \over 8}$
Hence, required remainder =$p\left( {{1 \over 2}} \right) = {{27} \over 8}$

(iii) By remainder В theorem, the required remainder is equal to p (0).
Now, $p\left( x \right) = {x^3} + 3{x^2} + 3x + 1$
Therefore В $p\left( 0 \right) = 0 + 0 + 0 + 1 = 1$
Hence, the required remainder = p(0) = 1

(iv) By remainder theorem the required remainder is $p\left( { - \pi } \right)$
Now , $p\left( x \right) = {x^3} + 3{x^2} + 3x + 1$
Therefore $p\left( { - \pi } \right) = {\left( { - \pi } \right)^3} + 3{\left( { - \pi } \right)^2} + 3\left( { - \pi } \right) + 1$
$= - {\pi ^3} + 3{\pi ^2} - 3\pi + 1$
Hence, the required remainder = p(x) $= - {\pi ^3} + 3{\pi ^2} - 3\pi + 1$

(v) By remainder theorem, the required remainder is $p\left( { - {5 \over 2}} \right)$
Now, $p\left( x \right) = {x^3} + 3{x^2} + 3x + 1$
Therefore $p\left( { - {5 \over 2}} \right)$
$= {\left( { - {5 \over 2}} \right)^3} + 3{\left( {{{ - 5} \over 2}} \right)^2} + 3\left( {{{ - 5} \over 2}} \right) + 1$
$= {{- 125} \over 8} + {{75} \over 4} - {{15} \over 2} + 1$
$= {{ - 125 + 150 -60 + 8} \over 8} = {{ - 27} \over 8}$
Hence, required remainder =$p\left( {{-5\over 2}} \right) = {{-27} \over 8}$

Q.2В В В В В  Find the remainder when В ${x^3} - a{x^2} + 6x - a$ В is divided by x вЂ“ a.
Sol.

Let $p\left( x \right) = {x^3} - a{x^2} + 6x - a$
By remainder theorem, when p(x) is divided by x - a.
Then remainder = p(a)
Therefore $p\left( a \right) = {a^3} - a.\,{a^2} + 6a - a$
$= {a^3} - \,{a^3} + 6a - a = 5a$
Hence, required remainder = p(x) = 5a

Q.3В В В В В  Check whether 7 + 3x is a factor of В $3{x^3} + 7x$
Sol.

7 + 3x will be a factor of В $p\left( x \right) = 3{x^3} + 7x\, if \, p\left( { - {7 \over 3}} \right) = 0$
Now, $p\left( { - {7 \over 3}} \right) = 3{\left( { - {7 \over 3}} \right)^3} + 7\left( { - {7 \over 3}} \right)$
$= 3 \times \left( {{{ - 343} \over {27}}} \right) - {{49} \over 3} = - {{343} \over 9} - {{49} \over 3} \ne 0$
Therefore 7 + 3 x is not a factor В of В $3{x^3} + 7 x$

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