Question

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Answer 1

Let a,b,c be the zeros of the polynomial.

Given f(x) = x^3 - 5x^2 - 2x + 24.

We know that sum of roots = -b/a

a + b + c = -(-5)/1

               = -(-5)

               = 5.   ------ (1)


We know that product of roots = -d/a

                              abc = -24/1

                                     = -24.  ----- (2)


Given that product of its two zeroes = 12.

ab = 12  ----- (3)


Substitute (3) in (2), we get

12 * c = -24

c = -2


Substitute c = -2 in (1),

a + b - 2 = 5

a + b = 7  ----- (4)


We know that (a + b)^2 = (a - b)^2 + 4ab

(7)^2 = (a - b)^2 + 4(12)

49 = (a - b)^2 + 48

a - b = 1.  ---- (5)


on solving 4 and 5, we get

a + b = 7

a - b = 1

--------------

2b = 8

b = 4


substitute b  4 in (5), we get

a  -b = 1

a = 3.


Therefore zeroes = 3,4,-2.

Answer 2

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