NCERT Solutions for Class 9 Math Chapter 2.3 POLYNOMIALS ALL SOLUTION

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EXERCISE 2.1 EXERCISE 2.2 EXERCISE 2.4 EXERCISE 2.5

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1.    Find the remainder when `\x^3+3x^2+3x+1\ ` is divided by

(i)    `\x+1\ `

(ii)    `\x-\frac{1}{2}\ `

(iii)    `\x\ `

(iv)    `\x+π\ `

(v)    `\5+2x\ `

solution :

(i)    `\x+1\ `

    Here, `\p(x)=x^3+3x^2+3x+1\ `

    The zero of `\x+1\ ` is

    `\x+1=0\ `

    `\x=-1\ `

    So, `\P(−1)=(-1)^3+3(-1)^2+3(-1)+1\ `

    `\=-1+3+(-3)+1\ `

    `\=0\ `

    So, by the Remainder Theorem, 0 is the remainder when `\x^3+3x^2+3x+1\ ` is divided by `\x+1\ `.

(ii)    `\x-\frac{1}{2}\ `

    Here, `\p(x)=x^3+3x^2+3x+1\ `

    The zero of `\x-\frac{1}{2}\ ` is

     `\x-\frac{1}{2}=0\ `

    `\x=\frac{1}{2}\ `

    So, `\P(\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}\ `

    So, by the Remainder Theorem, `\frac{27}{8}\ ` is the remainder when `\x^3+3x^2+3x+1\ ` is divided by `\x-\frac{1}{2}\ `

(iii)    `\x\ `

    Here, `\p(x)=x^3+3x^2+3x+1\ `

    The zero of `\x\ ` is

    `\x=0\ `

    So, `\P(0)=0^3+3×0^2+3×0+1\ `

    `\=0+0+0+1\ `

    `\= 1\ `

    So, by the Remainder Theorem, 1 is the remainder when `\x^3+3x^2+3x+1\ ` is divided by `\x\ `.

(iv)    `\x+π\ `

    Here, `\p(x)=x^3+3x^2+3x+1\ `

    The zero of `\x+pi\ ` is

    `\x+pi=0\ `

    `\x=-pi\ `

    So, `\P(-pi)=-pi^3+3(-pi)^2+3(-pi)+1\ `

    `\=\-pi^3+3pi^2-3pi+1\ `

    So, by the Remainder Theorem, `\=\-pi^3+3pi^2-3pi+1\ ` is the remainder when `\x^3+3x^2+3x+1\ ` is divided by `\x+pi\ `.

(v)    `\5+2x\ `

    Here, `\p(x)=x^3+3x^2+3x+1\ `

    The zero of `\5+2x\ ` is

    `\5+2x=0\ `

    `\2x=-5\ `

    `\x=\frac{-5}{2}\ `

    So, `\P(\frac{-5}{2})=(\frac{-5}{2})^3+3(\frac{-5}{2})^2+3(\frac{-5}{2})+1\ `

    `\=\(\frac{-125}{8})+(\frac{75}{8})+(\frac{-15}{8})+1\ `

    `\=\frac{-27}{8}\ `

    So, by the Remainder Theorem, `\frac{-27}{8}\ ` is the remainder when `\x^3+3x^2+3x+1\ ` is divided by `\5 +2x\ `.

2.    Find the remainder when `\x^3-ax^2+6x-a\ ` is divided by `\x-a\ `.

solution :

    Let, `\p(x)=x^3-ax^2+6x-a\ `

    The zero of `\x-a\ ` is

    `\x-a=0\ `

    `\x=a\ `

    So, `\P(a)=a^3-a×a^2+6a-a\ `

    `\=a^3-a^3+5a\ `

    `\=5a\ `

    So, by the Remainder Theorem, `\5a\ ` is the remainder when `\x^3-ax^2+6x-a\ ` is divided by `\x-a\ `.

3.    Check whether `\7+3x\ ` is a factor of `\3x^3+7x\ `.

solution :

    The zero of `\7+3x\ ` is

    `\7+3x=0\ `

    `\3x=-7\ `

    `\x=\frac{-7}{3}\ `

    So,

    `\p(\frac{-7}{3})=3(\frac{-7}{3})^3+7(\frac{-7}{3})\ `

    `\=3(\frac{-343}{27})+(\frac{-49}{3})\ `

    `\=(\frac{-343}{9})+(\frac{-49}{3})\ `

    `\=(\frac{(-343)+(-147)}{9})\ `

    `\=(\frac{-490}{9})\ne0\ `

    `\=7+3x\ ` is not a factor of `\3x^3+7x\ `.

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EXERCISE 2.1 EXERCISE 2.2 EXERCISE 2.4 EXERCISE 2.5