Question

Find the zeros of the polynomial f(x) = x^3 - 18x^2 + 99x - 162, if it is given that the zeros are in arithmetic progression.

Answer 1

Step-by-step explanation:

We have,

f(x)=x

3

−12x

2

+39x+k

Since, roots of this equation are in A.P.

Let a−d,a,a+d are roots.

Now, sum of roots =

a

−b

a−d+a+a+d=

1

12

3a=12

a=4

Sum of products of two consecutive roots =

a

c

(a−d)a+a(a+d)+(a−d)(a+d)=

1

39

a

2

−ad+a

2

+ad+a

2

−d

2

=39

3a

2

−d

2

=39

3×16−d

2

=39

48−d

2

=39

d

2

=

d=±3

Therefore, the roots are 1, 4, 7 or 7, 4, 1

Now, product of roots=(a−d)a(a+d)=

a

−d

1×4×7=−k

k=−28

Hence, the value of k is −28.

Answer 2

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