PREVIOUS

NEXT

#### NCERT Solution Chapter 2: Polynomials

TEXTBOOK’S EXERCISE – 2.1

• A polynomial can have many terms but not infinite terms.
• The exponent of a variable of a polynomial cannot be negative. This means a variable with power – 2, -3, -4, etc. are not allowed. If the power of a variable in an algebraic expression is negative, then that cannot be considered a polynomial.
• The exponent of a variable of a polynomial must be a whole number.
• The exponent of a variable of a polynomial cannot be a fraction. This means a variable with power 1/2, 3/2, etc. is not allowed. If the power of a variable in an algebraic expression is an infraction, then that cannot be considered a polynomial.
• A polynomial with the only constant term is called constant polynomial.
• The degree of a non-zero constant polynomial is zero.
• Degree of a zero polynomial is not defined.

Q 1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer

(i) 4x2−3x+7

Ans: Since the given expression has only one variable ‘x’, hence this is a polynomial in one variable.

(ii) y2+√2

Ans:  Since the given expression has only one variable ‘y’, hence this is a polynomial in one variable.

(iii) 3√t+t√2

Ans:  Since, variable, ‘t’ in this expression exponent of the variable is not a whole number. (Exponent of a variable in a polynomial must be a whole number. Expression with an exponent of a variable infraction is not considered as a polynomial.)

(iv) y+2/y

Ans:  Since the exponent of the variable is a negative integer, and not a whole number, hence it cannot be considered a polynomial.

(v) x10+y3+t50

Ans:  Since, given expression have three variables, i.e. x, y and t, so it is not a polynomial in one variable.

Q 2: Write the coefficients of x2 in each of the following:

(i) 2+x2+x

Ans: Given; 2+x2+x
Here, the coefficient of x2=1

(ii) 2–x2+x3

Ans:  Here, the coefficient of x2=1 (iii) π/2x2+x Answer: Here, the coefficient of x2=π/2 (iv) √2x–12x–1

Ans:  Here, the coefficient of x2=0

3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Sol. x35 + 5 is a binomial of degree 35.
2y100 is a monomial of degree 100.

4. Write the degree of each of the following polynomials :

(i) 5x3 + 4x2 + 7x (ii) 4 – y2 (iii) 5t –`√ 7 (iv) 3

Sol. (i) The degree is 3 as x3 is the highest power.
(ii) The degree is 2 as y2 is the highest power.
(iii) The degree is 1 as t is the highest power.
(iv) The degree is 0 as x0 is the highest power.

5. Classify the following as linear, quadratic and cubic polynomials:

(i) x2 + x (ii) x – x3 (iii) y + y2 + 4 (iv) 1 + x

(v) 3t (vi) r2 (vii) 7x3

Sol. (i) x2 + x is quadratic.

(ii) x x3 is cubic.

(iii) y + y2 + 4 is quadratic.

(iv) 1 + x is linear.

(v) 3t is linear.

(vii) 7x3 is cubic.

TEXTBOOK’S EXERCISE – 2.2

1. Find the value of the polynomial 5x – 4x2 + 3 at

(i) x = 0 (ii) x = –1 (iii) x = 2

Sol. p(x) = 5x – 4x2 + 3

(i) At x = 0, p(0) = 5 × 0 – 4 × 02 + 3 = 3

(ii) At x = –1,

p(–1) = 5 × (–1) – 4 × (–1)2 + 3

= –5 – 4 + 3 = – 6

(iii) At x = 2, p(2) = 5 × 2 – 4 × (2)2 + 3

= 10 – 16 + 3 = – 3

2. Find p(0), p(1) and p(2) for each of the following polynomials :

(i) p(y) = y2 – y + 1 (ii) p(t) = 2 + t + 2t2 – t3

(iii) p(x) = x3 (iv) p(x) = (x – 1) (x + 1)

Sol. (i) p(y) = y2 – y + 1

p(0) = 02 – 0 + 1 = 1

p(1) = 12 – 1 + 1 = 1

p(2) = 22 – 2 + 1 = 3.

(ii) p(t) = 2 + t + 2t2 – t3

p(0) = 2 + 0 + 2 × 02 – 03 = 2

p(1) = 2 + 1 + 2 × 12 – 13 = 4

p(2) = 2 + 2 + 2 × 22 – 23

= 4 + 8 – 8 = 4.

(iii) p(x) = x3

p(0) = 0

p(1) = 1

p(2) = 8.

(iv) p(x) = (x – 1) (x + 1)

p(0) = (–1) (1) = –1

p(1) = (1 – 1) (1 + 1) = 0

p(2) = (2 – 1) (2 + 1) = 3

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

(i) p(x) = 3x + 1, x = – 1/3 (ii) p(x) = 5x – π, x = 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 = – m/l

(vii) p(x) = 3x2 – 1, x = – 1/√3, 2/√3 ,

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

Sol. (i) Yes. 3x + 1 = 0, for x = – 1/3 .

(ii) No. 5x – π = 5 × 4/5 – π = 4 – π ≠ 0

(iii) Yes. x2 – 1 = 12 – 1 = 0 for x = 1 and x2 – 1 = (–1)2 –1 = 0 for x = 1

(iv) Yes. (x + 1) (x – 2) = 0 for x = –1, or, x = 2.

(v) Yes. x2 = 0 for x = 0

(vi) Yes. lx + m = 0 for x = – m/l

(vii) 3x2 – 1 = 3. 1/3 – 1 = 0 for x = −1/ √3 and 3x2 – 1 = 3· 4/3 –1 = 3 ≠ 0

Thus, for −1/ √3 is a zero but for – 2/ √3 is not a zero of the polynomial.

(viii) No. 2x + 1 ≠ 0 for x = 1/2.

4. Find the zero of the polynomial in each of the following cases :

(i) p(x) = x + 5

(ii) p(x) = x – 5

(iii) p(x) = 2x + 5

(iv) p(x) = 3x – 2

(v) p(x) = 3x

(vi) p(x) = ax, a ≠ 0

(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.

Sol. (i) x + 5 = 0, x = –5,

so, –5 is the zero of x + 5.

⇒ x = −5/2 ,

so – 5/2 is the zero of 2x + 5.

(iv) 3x – 2 = 0 ⇒ 3x = 2

⇒ x = 2/3 , so 2/3 is the zero of 3x – 2.

(v) 3x = 0 ⇒ x = 0, so 0 is the zero of 3x.

(vi) ax = 0 (a ≠ 0) ⇒ x = 0/a = 0,

so, 0 is the zero of ax.

(vii) cx + d = 0 (c ≠ 0)

⇒ cx = – d

⇒ x = −d/ c ,

so, −d /c is the 0 of cx + d.
PREVIOUS﻿