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 = π/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 = 1/2

In order to verify the values are zeros of polynomial p(x), we must replace the variable x with the given values. If p(x) = 0, then that given value is zero of polynomial p(x)
(i) p(x) = 3x + 1: Put x = -1/3, we get, p(x) = p(-1/3) = 3(-1/3) + 1 = -1 + 1 = 0. So, x = -1/3 is the zero of the given polynomial p(x)
(ii) p(x) = 5x - π: Put x = π/5, we get, p(x) = p(π/5) = 5(π/5) - π = π - π = 0. So, x = π/5 is the zero of the given polynomial p(x)
(iii) p(x) = x² - 1: Put x = 1, we get, p(x) = p(1) = (1)² - 1 = 1 - 1 = 0. So, x = 1 is the zero of the given polynomial p(x). Now put x = -1, we get, p(x) p(-1) = (-1)² - 1 = 1 - 1 = 0. So, x = -1 is the zero of the given polynomial p(x)
(iv) p(x) = (x + 1)(x - 2): Put x = -1, we get, p(x) = (-1 + 1)(-1 - 2) = 0(-3) = 0. So, x = -1 is the zero of the given polynomial p(x). Now put x = 2, we get, p(x) = (2 + 1)(2 - 2) = 3(0) = 0. So, x = 2 is the zero of the given polynomial p(x)
(v) p(x) = x²: Put x = 0, we get, p(x) = p(0) = (0)² = 0. So, x = 0 is the zero of the given polynomial p(x)
(vi) p(x) = lx + m: Put x = -m/l, we get, p(x) = p(-m/l) = l(-m/l) + m = -m + m = 0. So, x = -m/l is the zero of the given polynomial p(x)
(vii) p(x) = 3x² - 1: Put x = 1/√3, we get, p(x) = p(1/√3) = 3(1/√3)² - 1 = (3/3) - 1 = 1 - 1 = 0. So, x = 1/√3 is the zero of the given polynomial p(x). Now put x = 2√3, we get, p(x) = p(2√3) = 3(2√3)² - 1 = (3 × 12) - 1 = 36 - 1 = 35 ≠ 0. So, x = 2√3 is not the zero of the given polynomial p(x)
(viii) p(x) = 2x + 1: Put x = 1/2, we get, p(x) = p(1/2) = 2(1/2) + 1 = 1 + 1 = 2 ≠ 0. So, x = 1/2 is not the zero of the given polynomial p(x).