Verify that 1, -1 and -3 are the zeroes of the cubic polynomial x3 + 3

1.	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» 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

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»

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