Class 9 Math Exercise 2.2 | Question 3

Question 3: Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1, x = -\\frac{1}{3})
(ii) p(x) = 5x – π, x = \\frac{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 = -\\frac{m}{l})
(vii) p(x) = 3x² - 1, x = -\\frac{1}{√3}), \\frac{2}{√3})
(viii) p(x) = 2x + 1, x = \\frac{1}{2})

Explanatory Answer | Exercise 2.2 Question 3

Verify : (i) p(x) = 3x + 1, x = -\\frac{1}{3})

Approach: Find the value of p (-\\frac{1}{3})). If it is zero, then the answer is Yes.
p (-\\frac{1}{3})) = 3(-\\frac{1}{3})) + 1 = - 1 + 1 = 0
YES, x = -\\frac{1}{3}) is a zero of the polynomial.

Verify : (ii) p(x) = 5x – π, x = \\frac{4}{5})

Approach: Find the value of p (\\frac{4}{5})). If it is zero, then the answer is Yes.
p (\\frac{4}{5})) = 5(\\frac{4}{5})) - π = 4 – π ≠ 0
NO, x = -\\frac{1}{3}) is not a zero of the polynomial.

Verify : (iii) p(x) = x² – 1, x = 1, -1

Approach: Find the value of p (1) and p (-1). If these values are zero, the answer is Yes.
p (1) = 1² – 1 = 0
YES, x = 1 is a zero of the polynomial.
p (-1) = (-1)² – 1 = 1 – 1 = 0
YES, x = - 1 is a zero of the polynomial.

Verify : (iv) p(x) = (x + 1)(x – 2), x = -1, 2

Approach: Find the value of p (-1) and p (2). If these values are zero, the answer is Yes.
p (-1) = (-1 + 1)(-1 – 2) = 0(- 3) = 0
YES, x = - 1 is a zero of the polynomial.
p (2) = (2 + 1)(2 – 2) = 3(0) = 0
YES, x = 2 is a zero of the polynomial.

Verify : (v) p(x) = x², x = 0

Approach: Find the value of p (0). If the value is zero, the answer is Yes.
p (0) = 0
YES, x = 0 is a zero of the polynomial.

Verify : (vi) p(x) = lx + m, x = -\\frac{m}{l})

Approach: Find the value of p (-\\frac{m}{l})). If the value is zero, the answer is Yes.
p (-\\frac{m}{l})) = (l(-\\frac{m}{l}))) + m = - m + m = 0
YES, x = -\\frac{m}{l}) is a zero of the polynomial.

Verify : (vii) p(x) = 3x² - 1, x = -\\frac{1}{√3}), \\frac{2}{√3})

Approach: Find the value of p (-\\frac{1}{√3})) and p (\\frac{2}{√3})). If the value is zero, the answer is Yes.
p (-\\frac{1}{√3})) = 3(-\\frac{1}{√3}))² – 1 = 3(\\frac{1}{3})) – 1 = 1 – 1 = 0
YES, x = -\\frac{1}{√3}) is a zero of the polynomial.
p(\\frac{2}{√3})) = 3(\\frac{2}{√3}))² – 1 = (3(\\frac{1}{3})) – 1 = 4 – 1 = 3 ≠ 0
NO, x = \\frac{2}{√3}) is not a zero of the polynomial.

Verify : (viii) p(x) = 2x + 1, x = \\frac{1}{2})

Approach: Find the value of p (\\frac{1}{2})). If the value is zero, the answer is Yes.
p (\\frac{1}{2})) = (2(\\frac{1}{2}))) + 1 = 1 + 1 = 2
NO, x = \\frac{1}{2}) is not a zero of the polynomial.