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NCERT Class 9 maths chapter 2 Polynomials
NCERT Solution Chapter 2: Polynomials
Important points about 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.
(vi) r2 is quadratic.
(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.
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