Question

Find the zeroes of cubic polynomials.

i) - x³

ii) x² - x³

(iii) x³ – 5x² + 6x

Answer 1

i) Given polynomial is y = – x³ 

f(x) = -x³ ; f(x) = 0 ²

x³ = 0 

x = \(\sqrt[3]{0}\)= 0 

∴ Zero of the polynomial f(x) is only one i.e., 0.

ii) Given that y = x² – x³ 

f(x) = x² (1 – x) 

f(x) = 0 

⇒ x² (1 – x) = 0 

⇒ x² = 0 and 1 – x = 0 

⇒ x = 0 and x = 1 

∴ The zeroes of the polynomial f(x) are two i.e., 0 and 1.

iii) Given that x³ – 5x² + 6x Let f(x) = x³ – 5x² + 6x 

= x(x² – 5x + 6) 

= x(x² – 2x – 3x + 6) 

= x[x(x – 2) – 3(x – 2)] 

= x(x – 2) (x – 3) 

∴ The zeroes of the polynomial f(x) are x = 0 and x = 2 and x = 3