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NCERT Exemplar · Class 9 Mathematics Polynomials

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EXERCISE 2.3 1–10 (part 4 of 8)

  1. Exercise 1

    Classify the following polynomials as polynomials in one variable, two variables etc. (i) x2+x+1\displaystyle x^{2}+x+1 (ii) y3−5y\displaystyle y^{3}-5 y (iii) xy+yz+zx\displaystyle x y+y z+z x (iv) x2−2xy+y2+1\displaystyle x^{2}-2 x y+y^{2}+1

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    NCERT’s answer
    (i)
    One variable (ii) One variable (iii) Three variable (iv) Two variables
    \[\text{(i) } x^{2}+x+1 : \text{one variable, } x \]\[\text{(ii) } y^{3}-5y : \text{one variable, } y \]\[\text{(iii) } xy+yz+zx : \text{three variables, } x, y, z \]\[\text{(iv) } x^{2}-2xy+y^{2}+1 : \text{two variables, } x, y \]Answer: (i) one variable (ii) one variable (iii) three variables (iv) two variables

  2. Exercise 2

    Determine the degree of each of the following polynomials : (i) 2x−1\displaystyle 2 x-1 (ii) -10\displaystyle 10 (iii) x3−9x+3x5\displaystyle x^{3}-9 x+3 x^{5} (iv) y3(1−y4)\displaystyle y^{3}\left(1-y^{4}\right)

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    NCERT’s answer
    (i)
    $\displaystyle 1$ (ii) $\displaystyle 0$ (iii) $\displaystyle 5$ (iv) $\displaystyle 7$
    \[\text{(i) } 2x-1 : \deg = 1 \]\[\text{(ii) } -10 : \deg = 0 \]\[\text{(iii) } x^{3}-9x+3x^{5} : \deg = 5 \]\[\text{(iv) } y^{3}(1-y^{4}) = y^{3}-y^{7} : \deg = 7 \]Answer: (i) $\displaystyle 1$ (ii) $\displaystyle 0$ (iii) $\displaystyle 5$ (iv) $\displaystyle 7$

  3. Exercise 3

    For the polynomial x3+2x+15−72x2−x6\displaystyle \frac{x^{3}+2 x+1}{5}-\frac{7}{2} x^{2}-x^{6}, write (i) the degree of the polynomial (ii) the coefficient of x3\displaystyle x^{3} (iii) the coefficient of x6\displaystyle x^{6} (iv) the constant term

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    NCERT’s answer
    (i)
    $\displaystyle 6$ (ii) $\displaystyle \frac{1}{5}$ (iii) -$\displaystyle 1$ (iv) $\displaystyle \frac{1}{5}$
    \[\frac{x^{3}+2x+1}{5}-\frac{7}{2}x^{2}-x^{6} = -x^{6}-\frac{7}{2}x^{2}+\frac{1}{5}x^{3}+\frac{2}{5}x+\frac{1}{5} \]\[\text{(i) degree} = 6 \]\[\text{(ii) coefficient of } x^{3} = \frac{1}{5} \]\[\text{(iii) coefficient of } x^{6} = -1 \]\[\text{(iv) constant term} = \frac{1}{5} \]Answer: degree \(\displaystyle 6\); coefficient of \(\displaystyle x^{3}\) is \(\displaystyle \tfrac{1}{5}\); coefficient of \(\displaystyle x^{6}\) is \(\displaystyle -1\); constant term \(\displaystyle \tfrac{1}{5}\)

  4. Exercise 4

    Write the coefficient of x2\displaystyle x^{2} in each of the following : (i) π6x+x2−1\displaystyle \frac{\pi}{6} x+x^{2}-1 (ii) 3x−5\displaystyle 3 x-5 (iii) (x−1)(3x−4)\displaystyle (x-1)(3 x-4) (iv) (2x−5)(2x2−3x+1)\displaystyle (2 x-5)\left(2 x^{2}-3 x+1\right)

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    NCERT’s answer
    (i)
    $\displaystyle 1$ (ii) $\displaystyle 0$ (iii) $\displaystyle 3$ (iv) -$\displaystyle 16$
    \[\text{(i) } \frac{\pi}{6}x+x^{2}-1 : \text{coeff. of } x^{2} = 1 \]\[\text{(ii) } 3x-5 : \text{coeff. of } x^{2} = 0 \]\[(x-1)(3x-4) = 3x^{2}-7x+4 \quad\Rightarrow\quad \text{coeff. of } x^{2} = 3 \]\[(2x-5)(2x^{2}-3x+1) = 4x^{3}-16x^{2}+17x-5 \quad\Rightarrow\quad \text{coeff. of } x^{2} = -16 \]Answer: (i) $\displaystyle 1$ (ii) $\displaystyle 0$ (iii) $\displaystyle 3$ (iv) -$\displaystyle 16$

  5. Exercise 5

    Classify the following as a constant, linear, quadratic and cubic polynomials : (i) 2−x2+x3\displaystyle 2-x^{2}+x^{3} (ii) 3x3\displaystyle 3 x^{3} (iii) 5t−7\displaystyle 5 t-\sqrt{7} (iv) 4−5y2\displaystyle 4-5 y^{2} (v) 3\displaystyle 3 (vi) 2+x\displaystyle 2+x (vii) y3−y\displaystyle y^{3}-y (viii) 1+x+x2\displaystyle 1+x+x^{2} (ix) t2\displaystyle t^{2} (x) 2x−1\displaystyle \sqrt{2} x-1

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    NCERT’s answer
    Constant Polynomial : (v) Linear Polynomials : (iii), (vi), (x) Quadratic Polynomials : (iv), (viii), (ix) Cubic Polynomials : (i), (ii), (vii)
    \[\text{(i) } 2-x^{2}+x^{3}: \deg 3 \Rightarrow \text{cubic} \]\[\text{(ii) } 3x^{3}: \deg 3 \Rightarrow \text{cubic} \]\[\text{(iii) } 5t-\sqrt{7}: \deg 1 \Rightarrow \text{linear} \]\[\text{(iv) } 4-5y^{2}: \deg 2 \Rightarrow \text{quadratic} \]\[\text{(v) } 3: \deg 0 \Rightarrow \text{constant} \]\[\text{(vi) } 2+x: \deg 1 \Rightarrow \text{linear} \]\[\text{(vii) } y^{3}-y: \deg 3 \Rightarrow \text{cubic} \]\[\text{(viii) } 1+x+x^{2}: \deg 2 \Rightarrow \text{quadratic} \]\[\text{(ix) } t^{2}: \deg 2 \Rightarrow \text{quadratic} \]\[\text{(x) } \sqrt{2}x-1: \deg 1 \Rightarrow \text{linear} \]Answer: cubic (i), (ii), (vii); quadratic (iv), (viii), (ix); linear (iii), (vi), (x); constant (v)

  6. Exercise 6

    Give an example of a polynomial, which is : (i) monomial of degree 1\displaystyle 1 (ii) binomial of degree 20\displaystyle 20 (iii) trinomial of degree 2\displaystyle 2

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    NCERT’s answer
    (i)
    $\displaystyle 10 x$ (ii) $\displaystyle x^{20}+1$ (iii) $\displaystyle 2 x^{2}-x-1$
    \[\text{(i) monomial, deg 1:} \quad 2x \]\[\text{(ii) binomial, deg 20:} \quad x^{20}+1 \]\[\text{(iii) trinomial, deg 2:} \quad x^{2}+x+1 \]Other choices are equally valid; each meets the stated term count and degree.Answer: (i) \(\displaystyle 2x\) (ii) \(\displaystyle x^{20}+1\) (iii) \(\displaystyle x^{2}+x+1\)

  7. Exercise 7

    Find the value of the polynomial 3x3−4x2+7x−5\displaystyle 3 x^{3}-4 x^{2}+7 x-5, when x=3\displaystyle x=3 and also when x=−3\displaystyle x=-3.

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    NCERT’s answer
    $\displaystyle 61$, -$\displaystyle 143$
    \[p(x) = 3x^{3}-4x^{2}+7x-5 \]\[p(3) = 3(3)^{3}-4(3)^{2}+7(3)-5 = 81-36+21-5 = 61 \]\[p(-3) = 3(-3)^{3}-4(-3)^{2}+7(-3)-5 = -81-36-21-5 = -143 \]Answer: \(\displaystyle p(3)=61,\ p(-3)=-143\)

  8. Exercise 8

    If p(x)=x2−4x+3\displaystyle p(x)=x^{2}-4 x+3, evaluate : p(2)−p(−1)+p(12)\displaystyle p(2)-p(-1)+p\left(\frac{1}{2}\right)

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    NCERT’s answer
    $\displaystyle \frac{-31}{4}$
    \[p(x)=x^{2}-4x+3 \]\[p(2)=4-8+3=-1 \]\[p(-1)=1+4+3=8 \]\[p\left(\frac{1}{2}\right)=\frac{1}{4}-2+3=\frac{5}{4} \]\[p(2)-p(-1)+p\left(\frac{1}{2}\right) = -1-8+\frac{5}{4} = -\frac{31}{4} \]Answer: \(\displaystyle -\dfrac{31}{4}\)

  9. Exercise 9

    Find p(0),p(1),p(−2)\displaystyle p(0), p(1), p(-2) for the following polynomials : (i) p(x)=10x−4x2−3\displaystyle p(x)=10 x-4 x^{2}-3 (ii) p(y)=(y+2)(y−2)\displaystyle p(y)=(y+2)(y-2)

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    NCERT’s answer
    (i)
    -$\displaystyle 3$, $\displaystyle 3$, - $\displaystyle 39$ (ii) -$\displaystyle 4$, -$\displaystyle 3$, $\displaystyle 0$
    (i)
    \[p(0) = 10(0)-4(0)^2-3 = -3 \]
    \[p(1) = 10(1)-4(1)^2-3 = 3 \]
    \[p(-2) = 10(-2)-4(-2)^2-3 = -39 \]
    (ii)
    \(\displaystyle p(y)=(y+2)(y-2)\)
    \[p(0) = (0+2)(0-2) = -4 \]
    \[p(1) = (1+2)(1-2) = -3 \]
    \[p(-2) = (-2+2)(-2-2) = 0 \]
    Answer: (i) \(\displaystyle p(0)=-3,\ p(1)=3,\ p(-2)=-39\); (ii) \(\displaystyle p(0)=-4,\ p(1)=-3,\ p(-2)=0\)

  10. Exercise 10

    Verify whether the following are True or False : (i) -3\displaystyle 3 is a zero of x−3\displaystyle x-3 (ii) −13\displaystyle -\frac{1}{3} is a zero of 3x+1\displaystyle 3 x+1 (iii) −45\displaystyle \frac{-4}{5} is a zero of 4−5y\displaystyle 4-5 y (iv) 0\displaystyle 0 and 2\displaystyle 2 are the zeroes of t2−2t\displaystyle t^{2}-2 t (v) -3\displaystyle 3 is a z\displaystyle z ero of y2+y−6\displaystyle y^{2}+y-6

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    NCERT’s answer
    (i)
    False (ii) True (iii) False (iv) True (v) True
    (i)
    False. Zero of \(\displaystyle x-3\) is \(\displaystyle x=3\), not \(\displaystyle -3\):
    \[p(-3) = -3-3 = -6 \neq 0 \]
    (ii)
    True.
    \[3\left(-\tfrac{1}{3}\right)+1 = -1+1 = 0 \]
    (iii)
    False. Zero of \(\displaystyle 4-5y\) is \(\displaystyle y=\tfrac{4}{5}\), not \(\displaystyle -\tfrac{4}{5}\):
    \[4-5\left(-\tfrac{4}{5}\right) = 4+4 = 8 \neq 0 \]
    (iv)
    True.
    \[t^2-2t = t(t-2) = 0 \implies t=0,\ 2 \]
    (v)
    True.
    \[(-3)^2+(-3)-6 = 9-3-6 = 0 \]