Question

find the value of k for which F(X) is equal to x cube + 3 X square + 3 X + k has all its 3 zeros equal.​

Answer 1

Step-by-step explanation:

Given :-

F(X) = X³+3X²+3X+K has all its three zeroes equal.

To find:-

Find the value of K ?

Solution :-

Given cubic polynomial is

F(X) = X³+3X²+3X+K

Given that

F(X) has 3 equal zeroes.

Let the zero be A

So the three zeroes are A, A and A

On comparing with the standard Cubic Polynomial ax³+bx²+cx+d

We have

a = 1

b = 3

c = 3

d = K

We know that

Sum of the zeroes = -b/a

=> A+A+A = -3/1

=> 3A = -3

=> A = -3/3

=> A = -1

Therefore , A = -1

The zeroes are -1 , -1 , -1

Product of the zeroes = -d/a

=> A×A×A = -K/1

=> -1×-1×-1 = -K

=> -1 = -K

=> 1 = K

=> K = 1

Therefore, K = 1

Answer:-

The value of K for the given problem is 1

Used formulae:-

• The standard Cubic Polynomial ax³+bx²+cx+d

• Sum of the zeroes = -b/a

• Product of the zeroes = -d/a

• Sum of the product of two zeroes taken at a time = c/a

Answer 2

Answer:

F(x)= x³+3x²+3x + K

F(3)= 3³+3(3)²+3(3)+K

0=27+27+9+K

K= -63

Step-by-step explanation:

IT MAY BE RIGHT.