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NCERT Exemplar · Class 9 Mathematics Polynomials

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EXERCISE 2.3 11–20 (part 5 of 8)

  1. Exercise 11

    Find the zeroes of the polynomial in each of the following: (i) p(x)=x−4\displaystyle p(x)=x-4 (ii) g(x)=3−6x\displaystyle g(x)=3-6 x (iii) q(x)=2x−7\displaystyle q(x)=2 x-7 (iv) h(y)=2y\displaystyle h(y)=2 y

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    NCERT’s answer
    (i)
    $\displaystyle 4$ (ii) $\displaystyle \frac{1}{2}$ (iii) $\displaystyle \frac{7}{2}$ (iv) $\displaystyle 0$
    (i)
    \[x-4=0 \implies x=4 \]
    (ii)
    \[3-6x=0 \implies x=\tfrac{1}{2} \]
    (iii)
    \[2x-7=0 \implies x=\tfrac{7}{2} \]
    (iv)
    \[2y=0 \implies y=0 \]
    Answer: \(\displaystyle x=4;\ x=\tfrac{1}{2};\ x=\tfrac{7}{2};\ y=0\)

  2. Exercise 12

    Find the zeroes of the polynomial : p(x)=(x−2)2−(x+2)2p(x)=(x-2)^{2}-(x+2)^{2}

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    NCERT’s answer
    $\displaystyle 0$
    \[p(x) = (x-2)^2-(x+2)^2 = \big[(x-2)-(x+2)\big]\big[(x-2)+(x+2)\big] \] \[p(x) = (-4)(2x) = -8x \] \[p(x)=0 \implies -8x=0 \implies x=0 \]Answer: \(\displaystyle x=0\)

  3. Exercise 13

    By actual division, find the quotient and the remainder when the first polynomial is divided by the second polynomial : x4+1;x−1\displaystyle x^{4}+1 ; x-1

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    NCERT’s answer
    $\displaystyle x^{3}+x^{2}+x+1,2$
    \[x^4+1 = x^3(x-1) + (x^3+1) \] \[x^3+1 = x^2(x-1) + (x^2+1) \] \[x^2+1 = x(x-1) + (x+1) \] \[x+1 = 1\cdot(x-1) + 2 \]Adding the quotient terms: \[x^4+1 = (x-1)(x^3+x^2+x+1) + 2 \]Answer: quotient \(\displaystyle x^3+x^2+x+1\), remainder \(\displaystyle 2\)

  4. Exercise 14

    By Remainder Theorem find the remainder, when p(x)\displaystyle p(x) is divided by g(x)\displaystyle g(x), where (i) p(x)=x3−2x2−4x−1,g(x)=x+1\displaystyle p(x)=x^{3}-2 x^{2}-4 x-1, \quad g(x)=x+1 (ii) p(x)=x3−3x2+4x+50,g(x)=x−3\displaystyle p(x)=x^{3}-3 x^{2}+4 x+50, \quad g(x)=x-3 (iii) p(x)=4x3−12x2+14x−3,g(x)=2x−1\displaystyle p(x)=4 x^{3}-12 x^{2}+14 x-3, g(x)=2 x-1 (iv) p(x)=x3−6x2+2x−4,g(x)=1−32x\displaystyle p(x)=x^{3}-6 x^{2}+2 x-4, \quad g(x)=1-\frac{3}{2} x

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    NCERT’s answer
    (i)
    $\displaystyle 0$ (ii) $\displaystyle 62$ (iii) $\displaystyle \frac{3}{2}$ (iv) $\displaystyle \frac{-136}{27}$
    (i)
    \(\displaystyle g(x)=x+1\), root \(\displaystyle x=-1\):
    \[p(-1) = (-1)^3-2(-1)^2-4(-1)-1 = -1-2+4-1 = 0 \]
    (ii)
    \(\displaystyle g(x)=x-3\), root \(\displaystyle x=3\):
    \[p(3) = 3^3-3(3)^2+4(3)+50 = 27-27+12+50 = 62 \]
    (iii)
    \(\displaystyle g(x)=2x-1\), root \(\displaystyle x=\tfrac{1}{2}\):
    \[p\left(\tfrac{1}{2}\right) = 4\left(\tfrac{1}{2}\right)^3-12\left(\tfrac{1}{2}\right)^2+14\left(\tfrac{1}{2}\right)-3 = \tfrac{1}{2}-3+7-3 = \tfrac{3}{2} \]
    (iv)
    \(\displaystyle g(x)=1-\tfrac{3}{2}x\), root \(\displaystyle x=\tfrac{2}{3}\):
    \[p\left(\tfrac{2}{3}\right) = \left(\tfrac{2}{3}\right)^3-6\left(\tfrac{2}{3}\right)^2+2\left(\tfrac{2}{3}\right)-4 = \tfrac{8}{27}-\tfrac{8}{3}+\tfrac{4}{3}-4 = -\tfrac{136}{27} \]
    Answer: remainders \(\displaystyle 0,\ 62,\ \tfrac{3}{2},\ -\tfrac{136}{27}\)

  5. Exercise 15

    Check whether p(x)\displaystyle p(x) is a multiple of g(x)\displaystyle g(x) or not : (i) p(x)=x3−5x2+4x−3,g(x)=x−2\displaystyle p(x)=x^{3}-5 x^{2}+4 x-3, \quad g(x)=x-2 (ii) p(x)=2x3−11x2−4x+5,g(x)=2x+1\displaystyle p(x)=2 x^{3}-11 x^{2}-4 x+5, \quad g(x)=2 x+1

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    NCERT’s answer
    (i)
    No (ii) No
    (i)
    \(\displaystyle g(x)=x-2\), root \(\displaystyle x=2\):
    \[p(2) = 2^3-5(2)^2+4(2)-3 = 8-20+8-3 = -7 \]
    \(\displaystyle p(2)\neq0\), so \(\displaystyle g(x)\) is not a factor of \(\displaystyle p(x)\).
    (ii)
    \(\displaystyle g(x)=2x+1\), root \(\displaystyle x=-\tfrac{1}{2}\):
    \[p\left(-\tfrac{1}{2}\right) = 2\left(-\tfrac{1}{2}\right)^3-11\left(-\tfrac{1}{2}\right)^2-4\left(-\tfrac{1}{2}\right)+5 = -\tfrac{1}{4}-\tfrac{11}{4}+2+5 = 4 \]
    \(\displaystyle p\left(-\tfrac{1}{2}\right)\neq0\), so \(\displaystyle g(x)\) is not a factor of \(\displaystyle p(x)\).
    Answer: \(\displaystyle p(x)\) is not a multiple of \(\displaystyle g(x)\) in either (i) or (ii)

  6. Exercise 16

    Show that: (i) x+3\displaystyle x+3 is a factor of 69+11x−x2+x3\displaystyle 69+11 x-x^{2}+x^{3}. (ii) 2x−3\displaystyle 2 x-3 is a factor of x+2x3−9x2+12\displaystyle x+2 x^{3}-9 x^{2}+12.

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    (i)
    \(\displaystyle p(x)=x^3-x^2+11x+69\); root of \(\displaystyle x+3\) is \(\displaystyle x=-3\):
    \[p(-3) = (-3)^3-(-3)^2+11(-3)+69 \]
    \[p(-3) = -27-9-33+69 = 0 \]
    \(\displaystyle p(-3)=0\), so by the Factor Theorem \(\displaystyle x+3\) is a factor.
    (ii)
    \(\displaystyle p(x)=2x^3-9x^2+x+12\); root of \(\displaystyle 2x-3\) is \(\displaystyle x=\tfrac{3}{2}\):
    \[p\left(\tfrac{3}{2}\right) = 2\left(\tfrac{3}{2}\right)^3-9\left(\tfrac{3}{2}\right)^2+\tfrac{3}{2}+12 \]
    \[p\left(\tfrac{3}{2}\right) = \tfrac{27}{4}-\tfrac{81}{4}+\tfrac{6}{4}+\tfrac{48}{4} = 0 \]
    \(\displaystyle p\left(\tfrac{3}{2}\right)=0\), so by the Factor Theorem \(\displaystyle 2x-3\) is a factor.
    Answer: (i) \(\displaystyle x+3\) is a factor of \(\displaystyle x^3-x^2+11x+69\); (ii) \(\displaystyle 2x-3\) is a factor of \(\displaystyle 2x^3-9x^2+x+12\).

  7. Exercise 17

    Determine which of the following polynomials has x−2\displaystyle x-2 a factor: (i) 3x2+6x−24\displaystyle 3 x^{2}+6 x-24 (ii) 4x2+x−2\displaystyle 4 x^{2}+x-2

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    NCERT’s answer
    (i)
    By the Factor Theorem, \(\displaystyle x-2\) divides \(\displaystyle p(x)\) exactly when \(\displaystyle p(2)=0\). \[p(x) = 3x^{2}+6x-24 \implies p(2) = 3(2)^{2}+6(2)-24 = 12+12-24 = 0 \] \[q(x) = 4x^{2}+x-2 \implies q(2) = 4(2)^{2}+2-2 = 16+2-2 = 16 \] \(\displaystyle p(2)=0\) but \(\displaystyle q(2)\neq 0\). Answer: \(\displaystyle x-2\) is a factor of (i) \(\displaystyle 3x^{2}+6x-24\) only.

  8. Exercise 18

    Show that p−1\displaystyle p-1 is a factor of p10−1\displaystyle p^{10}-1 and also of p11−1\displaystyle p^{11}-1.

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    By the Factor Theorem, \(\displaystyle p-1\) divides \(\displaystyle f(p)\) exactly when \(\displaystyle f(1)=0\). \[f(p) = p^{10}-1 \implies f(1) = 1^{10}-1 = 0 \] \[g(p) = p^{11}-1 \implies g(1) = 1^{11}-1 = 0 \] Answer: \(\displaystyle p-1\) is a factor of both \(\displaystyle p^{10}-1\) and \(\displaystyle p^{11}-1\), since \(\displaystyle f(1)=g(1)=0\).

  9. Exercise 19

    For what value of m\displaystyle m is x3−2mx2+16\displaystyle x^{3}-2 m x^{2}+16 divisible by x+2\displaystyle x+2 ?

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    NCERT’s answer
    $\displaystyle 1$
    \(\displaystyle x+2\) is a factor of \(\displaystyle f(x)=x^{3}-2mx^{2}+16\) exactly when \(\displaystyle f(-2)=0\) (Factor Theorem). \[f(-2) = (-2)^{3}-2m(-2)^{2}+16 = -8-8m+16 = 8-8m \] \[8-8m = 0 \implies m = 1 \] Answer: \(\displaystyle m = 1\).

  10. Exercise 20

    If x+2a\displaystyle x+2 a is a factor of x5−4a2x3+2x+2a+3\displaystyle x^{5}-4 a^{2} x^{3}+2 x+2 a+3, find a\displaystyle a.

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    NCERT’s answer
    $\displaystyle \frac{3}{2}$
    \(\displaystyle x+2a\) is a factor of \(\displaystyle f(x)=x^{5}-4a^{2}x^{3}+2x+2a+3\) exactly when \(\displaystyle f(-2a)=0\). \[f(-2a) = (-2a)^{5}-4a^{2}(-2a)^{3}+2(-2a)+2a+3 \] \[= -32a^{5}+32a^{5}-4a+2a+3 = 3-2a \] \[3-2a = 0 \implies a = \frac{3}{2} \] Answer: \(\displaystyle a = \dfrac{3}{2}\).