Question

The roots of the polynomial \[p(x)={{x}^{3}}-3{{x}^{2}}+kx+4\] are in A.P., then \[\left| k \right|\] has the value equal to:
A. 0
B. 1
C. 2
D. 3

Answer 1

Hint: Given that \[p(x)\] is in A.P. Consider 3 terms in A.P. Find the sum of A.P. and sum of zeroes of the polynomial. Equate them to get the value of ‘a’ i.e. the first term. Then substitute in \[p(x)\] to get the value of k. Take \[\left| k \right|\] to get the final answer.

Complete step-by-step answer:
We have been given the polynomial \[p(x)={{x}^{3}}-3{{x}^{2}}+kx+4\] are in A.P. which is arithmetic progression.

It is a sequence of numbers in which each term after the first is obtained by adding a constant, d to the preceding term, where d is the common difference and a is the first term.
Let us assume that \[a-d,a,a+d\] are in A.P.

\[P(x)={{x}^{3}}-3{{x}^{2}}+kx+4=0\]

\[\therefore \]Sum of zeroes

\[=\dfrac{-Coefficient of{{x}^{2}}}{Coefficient of{{x}^{3}}}=\dfrac{-(-3)}{1}=3\]

Now let us take the sum of A.P.

\[\left( a+d \right)+a+\left( a-d \right)=\]Sum of zeroes \[=3\]

\[\begin{align}

  & \therefore a+d+a+a-d=3 \\

 & 3a=3 \\

\end{align}\] [Cancel out like terms]

Hence we got a = 1, which means that the first term of A.P. is 1.

Hence we can modify the A.P. we assumed by putting a = 1.

\[\therefore \]A.P. is \[(1-d),1,(1+d)\].

From this we can understand that 1 will satisfy p(x).

Put x = 1.

\[\begin{align}

  & P(x)={{x}^{3}}-3{{x}^{2}}+kx+4=0 \\

 & 1-3+k+4=0 \\

 & -2+k+4=0 \\

 & k+2=0 \\

 & \Rightarrow k=-2 \\

\end{align}\]

Thus, \[\left| k \right|\] = 2.

Hence we got the value of \[\left| k \right|\] = 2.

Option C is the correct answer.

Note: The sum of zeroes for a polynomial \[p(x):{{x}^{2}}-13x+42\] is taken as
Sum of zeroes \[=\dfrac{-Coefficientofx}{Coefficientof{{x}^{2}}}=\dfrac{-(-13)}{1}=13.\]
Similarly for the following:
for \[p(x)={{x}^{3}}-3{{x}^{2}}+2x+4=0\]
Sum of zeroes \[=\dfrac{-Coefficientof{{x}^{2}}}{Coefficientof{{x}^{3}}}=\dfrac{-(-3)}{1}=3.\]