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If the zeros of the polynomial $f(x)\ =\ 2x^3\ –\ 15x^2\ +\ 37x\ –\ 30$ are in A.P., find them.
Given:
The zeros of the polynomial $f(x)\ =\ 2x^3\ –\ 15x^2\ +\ 37x\ –\ 30$ are in A.P.
To do:
Here, we have to find the zeros of the given polynomial.
Solution:
 Let the zeros of the given polynomial be α, β and γ.
Given that the zeros are in A.P.
So, consider the roots as,
$α = p – d$, $β = p$ and $γ = p +d$
where, $p$ is the first term and $d$ is the common difference.
Comparing $f(x) $ with the standard form of a cubic polynomial,
$a= 2$, $b= -15$, $c= 37$ and $d=-30$
Therefore,
Sum of the roots $= α + β + γ = (p– d) + p + (p + d) = 3p = \frac{-b}{a} =\frac{-(-15)}{2} = \frac{15}{2}$
$3p= \frac{15}{2}$
$p= \frac{5}{2}$
Product of the roots $= (p – d) \times (p) \times (p + d) = p(p^2 –d^2) = \frac{-d}{a} = \frac{-(-30)}{2} = 15$
$p(p^2 –d^2) = 15$
Substituting the value of p, we get
$\frac{5}{2}[(\frac{5}{2})^2 –d^2] = 15$
$\frac{25}{4}– d^2 = 3\times2$
$25 – 4d^2 = 6\times4$
$4d^2=25-24$
$d^2=\frac{1}{4}$
$d = \frac{1}{2} or \frac{-1}{2}$
Taking $d = \frac{1}{2}$ and $p = \frac{5}{2}$
The zeros are $\frac{5}{2}-\frac{1}{2}=2$, $\frac{5}{2}$ and $\frac{5}{2}+\frac{1}{2}=3$.
Taking $d = \frac{-1}{2}$ and $a = \frac{5}{2}$
The zeros are $3$, $\frac{5}{2}$ and $2$.
Updated on: 10-Oct-2022
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