Question

Verify whether the following are zeroes of the polynomial, induced 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) = 1x + m; x = - \(\frac{m}{l}\)

(vii) p(x) = 3x² – 1; x = - \(\frac{1}{\sqrt 3}, \frac{2}{\sqrt3}\)

(viii) p(x) = 2x + 1; x = - \(\frac{1}{2}\)

Answer 1

(i) p(x) = 3x + 1; x = -\(\frac{1}{3}\)

Here value of polynomial is zero. 

 x = -\(\frac{1}{3}\) is not zero of the polynomial

  (ii) p(x) = 5x – π; x = \(\frac{4}{5}\)

Here value of polynomial is not zero. 

 x = \(\frac{4}{5}\) is not zero of the polynomial

(iii) p(x) = x² – 1; x = 1, -1 

p(1) = (1)² – 1 = 1 – 1 

p(1) = 0 Here value of 

p(x) is zero. 

hence its zero is 1. 

p(x) = x² – 1; x = -1 

p(-1) = (-1)² – 1 = 1 – 1 

p(-1) = 0 

Here value of p(x) is zero. 

∴ -1 is zero. 

(iv) p(x) = (x – 1)(x – 2); x = -1, 2 

p(x) = x² – 2x + x – 2 

p(x) = x² – x + 2 

x = -1 p(-1) = (-1)² – (-1)² + 2 = 1 + 1 + 2 

p(-1) = 4 

Here value of polynomila is not zero. 

∴ -1 is not zero. 

p(x) = x² – x + 2x = 2 

p(2) = (2)² – (2) + 2 = 4 – 2 + 2 

p(-1) = 4 

Here value of polynomial is not zero. 

∴ 2 is not zero. 

(v) p(x) = x² ; x = 0 

p(0) = (0)² p(0) = 0

Here value of p(x) is zero. 

∴ 0 is its zero.