Question

If the zeroes of the polynomial f(x) = x3 12x2 + 39x + a are in ap, find the value of a.

Answer 1

Answer:

Please refer the image..........

Step-by-step explanation:

Answer 2

Answer:

a = 28

Step-by-step explanation:

Given

f(x) = x³ + 12x² + 39x + a

A = 1, b = 12, c = 39, d = a

( here a is constant term and A is the coefficient of x³, for convinence)

Roots are in A.P with a common difference  'd'(let's assume)

So the let the roots be x - d, x and x + d (the first term is x - d)

We  know that,

The sum of the roots = -b/A

⇒ x - r + x + x + r = -12

⇒ 3x = -12

∴ x = -4  

∴One root x = -4

⇒ f(-4) = 0

⇒ (-4)³ + 12(-4)² + 39(-4) + a = 0

⇒ -64 + 12(16) - 156 + a = 0

⇒ -220 + 192 + a = 0

∴ a = 28