Question

Verify that 1, -1 and -3 are the zeroes of the cubic polynomial x³ + 3x² – x – 3 and check the relationship between zeroes and the coefficients.

Answer 1

Given cubic polynomial 

p(x) = x³ + 3x² – x – 3 

Comparing the given polynomial with ax³ + bx² + cx + d, we get a = 1, b = 3, c = -1, d = -3 

Futher given zeroes are 1,-1 and – 3 

p(1) = (1)³ + 3(1)² – 1 – 3 

= 1 + 3 – 1 – 3 = 0 

p(-1) = (-1)³ + 3(-1)² – 1 – 3 

= -1 + 3 + 1 – 3 = 0 

p(-3) = (-3)³ + 3(-3)² – (-3) – 3 

= -27 + 27 + 3 – 3 = 0

Therefore, 1, -1 and -3 are the zeroes of x³ + 3x² – x – 3. 

So, we take α = 1, β = -1 and γ = -3 

Now, α + β + γ = 1 + (-1) + (-3) = -3

αβ + βγ + γα = 1(-l) + (-1) (-3) + (-3)1 

= -1 + 3 – 3 

= -1 = c/a = -1/1 = -1 

αβγ = 1 (-1) (-3) = 3 = -d/a = -(-3)/1 = 3