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Polynomials : Exercise 2.3 (Mathematics NCERT Class 9th)


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Q.1      Find the remainder when {x^3} + 3{x^2} + 3x + 1 is divided by
                (i) x + 1                          (ii) x - {1 \over 2}
                (iii) x                              (iv) x + \pi             (v) 5 + 2x
Sol.

(i) By remainder theorem, the required remainder is equal to p (–1).
Now, p\left( x \right) = {x^3} + 3{x^2} + 3x + 1
Therefore   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(–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  theorem, the required remainder is equal to p (0).
Now, p\left( x \right) = {x^3} + 3{x^2} + 3x + 1
Therefore   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.2      Find the remainder when   {x^3} - a{x^2} + 6x - a  is divided by x – 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.3      Check whether 7 + 3x is a factor of   3{x^3} + 7x
Sol.

7 + 3x will be a factor of   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  of  3{x^3} + 7 x