CBSE Class 9  >  Class 9 Notes  >  Mathematics (Maths)   >  NCERT Solutions: Polynomials (Exercise 2.2)

NCERT Solutions: 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: Polynomials (Exercise 2.2) is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on NCERT Solutions: Polynomials (Exercise 2.2)

1. How do I find the zeros of a polynomial in Exercise 2.2?
Ans. Zeros of a polynomial are values of the variable that make the polynomial equal to zero. To find them, set the polynomial expression equal to zero and solve for the variable. For example, in p(x) = x - 2, the zero is x = 2. The Factor Theorem helps identify zeros by testing factors of the constant term, especially useful for cubic and higher-degree polynomials in CBSE Class 9 mathematics.
2. What's the difference between a root and a zero of a polynomial?
Ans. A root and a zero mean the same thing-both refer to values that make the polynomial equal to zero. The terms are used interchangeably in polynomial algebra. If p(a) = 0, then 'a' is both a root and a zero of p(x). Understanding this terminology helps avoid confusion when solving polynomial equations during Exercise 2.2 problem-solving.
3. How do I use the Factor Theorem to solve Exercise 2.2 problems?
Ans. The Factor Theorem states that (x - a) is a factor of p(x) if and only if p(a) = 0. To apply it, substitute potential values into the polynomial; if the result is zero, that value is a zero and corresponds to a linear factor. This shortcut avoids lengthy division and speeds up polynomial factorisation in Class 9 mathematics assessments.
4. Why do some polynomials have no real zeros?
Ans. A polynomial may have no real zeros when its values never actually equal zero for any real number. For instance, p(x) = x² + 1 has no real zeros because x² is always non-negative, making x² + 1 always positive. Such polynomials have complex zeros instead, a concept building towards higher mathematics beyond standard CBSE Class 9 curriculum scope.
5. How can I check if a number is a zero without fully solving the polynomial?
Ans. Substitute the number directly into the polynomial expression and evaluate. If the result equals zero, it's confirmed as a zero. This direct substitution method is faster than algebraic manipulation and helps verify answers quickly during Exercise 2.2 problem attempts and exam practice for Class 9 polynomials assessments.
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