# Find the condition that the zeros of the polynomial \[f(x)={{x}^{3}}+3p{{x}^{2}}+3qx+r\] are in an A.P.

Answer

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Hint: Consider \[a-d,a,a+d\] as a sum of polynomials. Find the expression for the sum of zeroes. Put \[f(a)=0\] and find the condition by substituting \[a=-p\] which is obtained from the sum of zeroes.

“Complete step-by-step answer:”

Given the polynomial,\[f(x)={{x}^{3}}+3p{{c}^{2}}+3qx+r\]

Let \[a-d,a,a+d\] be the zeroes of the polynomial f(x) which is in A.P. with common difference d.

The sum of zeroes\[=\dfrac{-Coefficient of{{x}^{2}}}{Coefficientof{{x}^{3}}}=(a-d)+a+(a+d)\]

\[\Rightarrow (a-d)+a+(a+d)=\dfrac{-3p}{1}\]

where coefficient of \[{{x}^{2}}=3p.\]

Coefficient of \[{{x}^{3}}=1\]

\[\begin{align}

& 3a=\dfrac{-3p}{1} \\

& \therefore a=-p \\

\end{align}\]

Since ‘a’ is a zero of the polynomial f(x),

\[\begin{align}

& \therefore f(x)={{x}^{3}}+3p{{x}^{2}}+3qx+r \\

& f(a)=0 \\

\end{align}\]

Put \[x=a.\]

\[\begin{align}

& f(a)={{a}^{3}}+3p{{a}^{2}}+3qa+r \\

& f(a)=0 \\

& \Rightarrow {{a}^{3}}+3p{{a}^{2}}+3qa+r=0 \\

\end{align}\]

Substitute \[a=-p.\]

\[\begin{align}

& \therefore {{(-p)}^{3}}+3p{{(-p)}^{2}}+3q(-p)+r=0 \\

& -{{p}^{3}}+3{{p}^{3}}+3pq+r=0 \\

& \Rightarrow 2{{p}^{3}}-3pq+r=0 \\

\end{align}\]

Hence the condition for the given polynomial is \[2{{p}^{3}}-3pq+r=0\].

Note:

We might think that for taking conditions, you might want to do differentiation, but that’s not right. Taking f’(x) won’t give us the required condition. As we have found \[a=-p\], find\[f(a)=0\], and then substitute \[a=-p\] to get the required condition.

“Complete step-by-step answer:”

Given the polynomial,\[f(x)={{x}^{3}}+3p{{c}^{2}}+3qx+r\]

Let \[a-d,a,a+d\] be the zeroes of the polynomial f(x) which is in A.P. with common difference d.

The sum of zeroes\[=\dfrac{-Coefficient of{{x}^{2}}}{Coefficientof{{x}^{3}}}=(a-d)+a+(a+d)\]

\[\Rightarrow (a-d)+a+(a+d)=\dfrac{-3p}{1}\]

where coefficient of \[{{x}^{2}}=3p.\]

Coefficient of \[{{x}^{3}}=1\]

\[\begin{align}

& 3a=\dfrac{-3p}{1} \\

& \therefore a=-p \\

\end{align}\]

Since ‘a’ is a zero of the polynomial f(x),

\[\begin{align}

& \therefore f(x)={{x}^{3}}+3p{{x}^{2}}+3qx+r \\

& f(a)=0 \\

\end{align}\]

Put \[x=a.\]

\[\begin{align}

& f(a)={{a}^{3}}+3p{{a}^{2}}+3qa+r \\

& f(a)=0 \\

& \Rightarrow {{a}^{3}}+3p{{a}^{2}}+3qa+r=0 \\

\end{align}\]

Substitute \[a=-p.\]

\[\begin{align}

& \therefore {{(-p)}^{3}}+3p{{(-p)}^{2}}+3q(-p)+r=0 \\

& -{{p}^{3}}+3{{p}^{3}}+3pq+r=0 \\

& \Rightarrow 2{{p}^{3}}-3pq+r=0 \\

\end{align}\]

Hence the condition for the given polynomial is \[2{{p}^{3}}-3pq+r=0\].

Note:

We might think that for taking conditions, you might want to do differentiation, but that’s not right. Taking f’(x) won’t give us the required condition. As we have found \[a=-p\], find\[f(a)=0\], and then substitute \[a=-p\] to get the required condition.

Last updated date: 31st May 2023

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