# NCERT solution class 9 chapter 2 Polynomials exercise 2.3 mathematics

## EXERCISE 2.3

#### Question 1:

Find the value of the polynomial at

(i) x = 0 (ii) x = −1 (iii) x = 2

(i)  (ii)  (iii)  #### Question 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 + + 2t2 − t3

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

(i) p(y) = y2 − y + 1

p(0) = (0)2 − (0) + 1 = 1

p(1) = (1)2 − (1) + 1 = 1

p(2) = (2)2 − (2) + 1 = 3

(ii) p(t) = 2 + + 2t2 − t3

p(0) = 2 + 0 + 2 (0)2 − (0)= 2

p(1) = 2 + (1) + 2(1)2 − (1)3

= 2 + 1 + 2 − 1 = 4

p(2) = 2 + 2 + 2(2)2 − (2)3

= 2 + 2 + 8 − 8 = 4

(iii) p(x) = x3

p(0) = (0)3 = 0

p(1) = (1)3 = 1

p(2) = (2)3 = 8

(iv) p(x) = (x − 1) (x + 1)

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

#### Question 3:

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

(i) (ii) (iii) p(x) = x2 − 1, x = 1, − 1 (iv) p(x) = (x + 1) (x − 2), x = − 1, 2

(v) p(x) = xx = 0 (vi) p(x) = lx m

(vii) (viii) (i) If is a zero of given polynomial p(x) = 3x + 1, then should be 0. Therefore, is a zero of the given polynomial.

(ii) If is a zero of polynomial p(x) = 5x − π , then should be 0. Therefore, is not a zero of the given polynomial.

(iii) If x = 1 and x = −1 are zeroes of polynomial p(x) = x2 − 1, then p(1) and p(−1) should be 0.

Here, p(1) = (1)2 − 1 = 0, and

p(− 1) = (− 1)2 − 1 = 0

Hence, x = 1 and −1 are zeroes of the given polynomial.

(iv) If x = −1 and x = 2 are zeroes of polynomial p(x) = (x +1) (x − 2), then p(−1) and p(2)should be 0.

Here, p(−1) = (− 1 + 1) (− 1 − 2) = 0 (−3) = 0, and

p(2) = (2 + 1) (2 − 2 ) = 3 (0) = 0

Therefore, x = −1 and = 2 are zeroes of the given polynomial.

(v) If x = 0 is a zero of polynomial p(x) = x2, then p(0) should be zero.

Here, p(0) = (0)= 0

Hence, x = 0 is a zero of the given polynomial.

(vi) If is a zero of polynomial p(x) = lx + m, then should be 0.

Here, Therefore, is a zero of the given polynomial.

(vii) If and are zeroes of polynomial p(x) = 3x2 − 1, then Hence, is a zero of the given polynomial. However, is not a zero of the given polynomial.

(viii) If is a zero of polynomial p(x) = 2x + 1, then should be 0. Therefore, is not a zero of the given polynomial.

#### Question 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≠ 0

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

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

(i) p(x) = x + 5

p(x) = 0

x + 5 = 0

x = − 5

Therefore, for x = −5, the value of the polynomial is 0 and hence, x = −5 is a zero of the given polynomial.

(ii) p(x) = x − 5

p(x) = 0

x − 5 = 0

x = 5

Therefore, for x = 5, the value of the polynomial is0 and hence, x = 5 is a zero of the given polynomial.

(iii) p(x) = 2x + 5

p(x) = 0

2x + 5 = 0

2x = − 5 Therefore, for , the value of the polynomial is 0 and hence, is a zero of the given polynomial.

(iv) p(x) = 3x − 2

p(x) = 0

3x − 2 = 0 Therefore, for , the value of the polynomial is 0 and hence, is a zero of the given polynomial.

(v) p(x) = 3x

p(x) = 0

3x = 0

x = 0

Therefore, for x = 0, the value of the polynomial is 0 and hence, x = 0 is a zero of the given polynomial.

(vi) p(x) = ax

p(x) = 0

ax = 0

x = 0

Therefore, for = 0, the value of the polynomial is 0 and hence, x = 0 is a zero of the given polynomial.

(vii) p(x) = cx + d

p(x) = 0

cx+ d = 0 Therefore, for , the value of the polynomial is 0 and hence, is a zero of the given polynomial.

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