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CBSE Class 9 Mathematics Question 2 of 16

Polynomials — Question 3

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Question 3

Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x) = 3x + 1, x = −13-\dfrac{1}{3}

(ii) p(x) = 5x - π, x = 45\dfrac{4}{5}

(iii) p(x) = x2 - 1, x = 1, -1

(iv) p(x) = (x + 1)(x - 2), x = -1, 2

(v) p(x) = x2, x = 0

(vi) p(x) = lx + m, x = −ml-\dfrac{m}{l}

(vii) p(x) = 3x2 - 1, x = −(13)-\Big(\dfrac{1}{\sqrt{3}}\Big), (23)\Big(\dfrac{2}{\sqrt{3}}\Big)

(viii) p(x) = 2x + 1, x = 12\dfrac{1}{2}

Answer

(i) p(x) = 3x + 1

Putting x = −13-\dfrac{1}{3} we get,

p(−13)\text{p}\Big(-\dfrac{1}{3}\Big) = 3(−13)3\Big(-\dfrac{1}{3}\Big) + 1

= -1 + 1

= 0

Hence, x = −13-\dfrac{1}{3} is zeroes of this polynomial.

(ii) p(x) = 5x - π

Putting x = 45\dfrac{4}{5} we get,

p(45)=5(45)−π\text{p}\Big(\dfrac{4}{5}\Big) = 5\Big(\dfrac{4}{5}\Big) - \pi

= 4 - π

Hence, x = 45\dfrac{4}{5} is not a zeroes of this polynomial.

(iii) p(x) = x2 - 1

Putting x = 1 we get,

p(1) = (1)2 - 1

= 0

Putting x = -1 we get

p(-1) = (-1)2 - 1

= 1 - 1

= 0

Hence, x = 1, -1 is zeroes of this polynomial.

(iv) p(x) = (x + 1)(x - 2)

Putting x = -1 we get,

p(1) = (-1 + 1)(-1 - 2)

= 0 x (-3)

= 0

Putting x = 2 we get,

p(2) = (2 + 1)(2 - 2)

= 3 x 0

= 0

Hence, x = -1, 2 are the zeroes of this polynomial.

(v) p(x) = x2

Putting x = 0 we get,

p(0) = (0)2

= 0

Hence, x = 0 is the zero of this polynomial.

(vi) p(x) = lx + m

Putting x = -ml\dfrac{m}{l} we get,

p - (ml)=lx−(ml)\Big(\dfrac{m}{l}\Big) = l x -\Big(\dfrac{m}{l}\Big) + m

= -m + m

= 0

Hence, x = −ml-\dfrac{m}{l} is the zero of this polynomial.

(vii) p(x) = 3x2 - 1

Putting x = −13-\dfrac{1}{\sqrt{3}} we get,

p-(13)\Big(\dfrac{1}{\sqrt{3}}\Big) = 3 x -(13)\Big(\dfrac{1}{\sqrt{3}}\Big)2 - 1

= 3 x 13\dfrac{1}{3} - 1

= 1 - 1

= 0

Putting x = 23\dfrac{2}{\sqrt{3}} we get,

p (23)\Big(\dfrac{2}{\sqrt{3}}\Big) = 3 x (23)\Big(\dfrac{2}{\sqrt{3}}\Big)2 - 1

= 3 x 43\dfrac{4}{3} - 1

= 4 - 1

= 3

Hence, x = −13-\dfrac{1}{\sqrt{3}} is zeroes of polynomial but x = 23\dfrac{2}{\sqrt{3}} is not a zeroes of this polynomial.

(viii) p(x) = 2x + 1

Putting x = 12\dfrac{1}{2} we get,

p(12)\Big(\dfrac{1}{2}\Big) = 2 x 12\dfrac{1}{2} + 1

= 1 + 1

= 2

No, x = 12\dfrac{1}{2} is not the zeroes of this polynomial.

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Polynomials — Interactive Study Guide

Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification

Not a polynomial: √x (fractional power), 1/x = x−1 (negative power), x + 1/x.

Polynomial: 5 (constant), 3x + 2 (linear), x² − 1 (quadratic), 2x³ + x − 1 (cubic).

Remainder and Factor Theorems — Quick Guide

Remainder Theorem: When p(x) is divided by (x − a), remainder = p(a).
Factor Theorem: (x − a) is a factor of p(x) ⇔ p(a) = 0.

Watch the sign! Dividing by (x + 3) means a = −3. So remainder = p(−3).

Identity Mastery Checklist

See This PatternUse This Identity
a² + 2ab + b²= (a + b)²
a² − 2ab + b²= (a − b)²
a² − b²= (a + b)(a − b)
a³ + b³= (a + b)(a² − ab + b²)
a³ − b³= (a − b)(a² + ab + b²)
a + b + c = 0⇒ a³ + b³ + c³ = 3abc

Quick Self-Check

  1. Degree of 5x³ − 2x + 1? (3)
  2. Remainder when x² + 3x + 2 is divided by (x + 1)? (p(−1) = 1 − 3 + 2 = 0)
  3. Expand: (2a + 3b)² (= 4a² + 12ab + 9b²)
  4. Factorise: 8x³ − 27 (= (2x − 3)(4x² + 6x + 9))

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