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NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials (Exercise 2.2)

Q1. Find the value of the polynomial 5x – 4x2 + 3 at 
(i) x = 0 
(ii) x = –1 
(iii) x = 2
Ans: 
Let f(x) = 5x−4x2+3
(i) When x = 0
f(0) = 5(0)-4(0)2+3
= 3


(ii) When x = -1
f(x) = 5x−4x2+3
f(−1) = 5(−1)−4(−1)2+3
= −5–4+3
= −6


(iii) When x = 2
f(x) = 5x−4x2+3
f(2) = 5(2)−4(2)2+3
= 10–16+3
= −3


Q2. Find p(0), p(1) and p(2) for each of the following polynomials: 
(i) p(y) = y− y + 1
Ans: p(y) = y2–y+1
∴ p(0) = (0)2− (0) + 1 = 1
p(1) = (1)– (1) + 1 =1
p(2) = (2)2–(2) + 1 = 3


(ii) p(t) = 2 + t + 2t− t3
Ans:  p(t) = 2 + t + 2t− t3
∴ p(0) = 2 + 0 + 2(0)– (0)= 2
p(1) = 2 + 1 + 2(1)– (1)= 2 + 1 + 2 – 1 = 4
p(2) = 2 + 2 + 2(2)– (2)= 2 + 2 + 8 – 8 = 4


(iii) p(x) = x3
Ans: ∴ p(0) = (0)3 = 0
p(1) = (1)3 = 1
p(2) = (2)3 = 8


(iv) P(x) = (x − 1) (x + 1)
Ans: ∴ p(0) = (0 – 1)(0 + 1) = (−1)(1) = –1
p(1) = (1 – 1)(1 + 1) = 0(2) = 0
p(2) = (2 – 1)(2 + 1) = 1(3) = 3


Q3. Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1, x = −1/3
Ans: For, x = -1/3, p(x) = 3x+1
∴ p(−1/3) = 3(-1/3)+1 = −1+1 = 0
∴ -1/3 is a zero of p(x).


(ii) p(x) = 5x – π, x = 4/5
Ans: For, x = 4/5, p(x) = 5x – π
∴ p(4/5) = 5(4/5) - π = 4  -π
∴ 4/5 is not a zero of p(x).


(iii) p(x) = x− 1, x = 1, −1
Ans: For, x = 1, −1;
p(x) = x2−1
∴ p(1) = 1− 1 = 1 − 1 = 0
p(−1) = (-1)− 1 = 1 − 1 = 0
∴ 1, −1 are zeros of p(x).


(iv) p(x) = (x+1)(x–2), x =−1, 2
Ans: For, x = −1,2;
p(x) = (x+1)(x–2)
∴ p(−1) = (−1+1)(−1–2)
= (0)(−3) = 0
p(2) = (2+1)(2–2) = (3)(0) = 0
∴ −1,2 are zeros of p(x).


(v) p(x) = x2, x = 0
Ans: For, x = 0 p(x) = x2
p(0) = 02 = 0
∴ 0 is a zero of p(x).


(vi) p(x) = lx + m, x = −m/l
Ans: For, x = -m/l ; p(x) = lx+m
∴ p(-m/l)= l(-m/l)+m = −m+m = 0
∴ -m/l is a zero of p(x).


(vii) p(x) = 3x2−1, x = -1/√3 , 2/√3
Ans: For, x = -1/√3 , 2/√3 ; p(x) = 3x2−1
∴ p(-1/√3) = 3(-1/√3)2-1 = 3(1/3)-1 = 1-1 = 0
∴ p(2/√3 ) = 3(2/√3)2-1 = 3(4/3)-1 = 4−1=3 ≠ 0
∴ -1/√3 is a zero of p(x) but 2/√3  is not a zero of p(x).


(viii) p(x) =2x + 1, x = 1/2
Ans: For, x = 1/2 p(x) = 2x + 1
∴ p(1/2) = 2(1/2) + 1 = 1 + 1 = 2 ≠ 0
∴ 1/2 is not a zero of p(x).


Q4. Find the zero of the polynomial in each of the following cases: 
(i) p(x) = x + 5 
Ans: p(x) = x + 5
⇒ x + 5 = 0
⇒ x = −5
∴ -5 is a zero polynomial of the polynomial p(x).


(ii) p(x) = x – 5
Ans: p(x) = x − 5
⇒ x − 5 = 0
⇒ x = 5
∴ 5 is a zero polynomial of the polynomial p(x).


(iii) p(x) = 2x + 5
Ans: p(x) = 2x + 5
⇒ 2x+5 = 0
⇒ 2x = −5
⇒ x = -5/2
∴ x = -5/2 is a zero polynomial of the polynomial p(x).


(iv) p(x) = 3x–2 
Ans: p(x) = 3x–2
⇒ 3x − 2 = 0
⇒ 3x = 2
⇒x = 2/3
∴ x = 2/3  is a zero polynomial of the polynomial p(x).


(v) p(x) = 3x 
Ans: p(x) = 3x
⇒ 3x = 0
⇒ x = 0
∴ 0 is a zero polynomial of the polynomial p(x).


(vi) p(x) = ax, a ≠ 0
Ans: p(x) = ax
⇒ ax = 0
⇒ x = 0
∴ x = 0 is a zero polynomial of the polynomial p(x).


(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
Ans: p(x) = cx + d
⇒ cx + d =0
⇒ x = -d/c
∴ x = -d/c is a zero polynomial of the polynomial p(x).

The document NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials (Exercise 2.2) is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials (Exercise 2.2)

1. What are polynomials?
Ans. Polynomials are mathematical expressions that consist of variables, coefficients, and exponents. They involve addition, subtraction, multiplication, and non-negative integer exponents. For example, 3x^2 + 2x - 5 is a polynomial.
2. How are polynomials classified?
Ans. Polynomials are classified based on the degree of the highest power of the variable. If the highest power is 1, it is a linear polynomial. If the highest power is 2, it is a quadratic polynomial. If the highest power is 3, it is a cubic polynomial. And so on.
3. What is the degree of a polynomial?
Ans. The degree of a polynomial is the highest power of the variable in the polynomial. For example, the degree of the polynomial 3x^2 + 2x - 5 is 2.
4. How can we add or subtract polynomials?
Ans. To add or subtract polynomials, we combine the like terms. Like terms have the same variables raised to the same powers. We add or subtract the coefficients of the like terms while keeping the variables and their exponents unchanged.
5. How can we multiply polynomials?
Ans. To multiply polynomials, we use the distributive property. We multiply each term of one polynomial by each term of the other polynomial and then combine like terms. The resulting expression is the product of the two polynomials.
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