Question

Find the zeroes of the polynomial f(x) = x² ˗ 2x ˗ 8 and verify the relation between its zeroes and coefficients.

Answer 1

x² ˗ 2x ˗ 8 = 0 

⇒ x² ˗ 4x + 2x ˗ 8 = 0 

⇒ x(x ˗ 4) + 2(x ˗ 4) = 0 

⇒ (x ˗ 4) (x + 2) = 0 

⇒ (x ˗ 4) = 0 or (x+2) = 0 

⇒ x = 4 or x = −2 

Sum of zeroes = 4 + (−2) = 2 = \(\frac{2}1\) = \(\frac{-(coefficient\,of\,x)}{(coefficient\,of\,x^2)}\)

Product of zeroes = (4) (−2) = \(\frac{-8}1\) = \(\frac{constant\,term}{(coefficient\,of\,x^2)}\)