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Verify that the numbers given alongside the cubic polynomial below are its zeroes. Also, verify the relationship between the zeros and coefficients:
$f(x)\ =\ 2x^3\ +\ x^2\ –\ 5x\ +\ 2;\ \frac{1}{2},\ 1,\ -2$
Given:
$f(x) = 2x^3 + x^2– 5x + 2$
To do:
We have to check whether $\frac{1}{2}, 1, -2$ are zeros of the given cubic polynomial.
Solution:
We know that,
The standard form of a cubic polynomial is $ax^3+bx^2+cx+d$, where a, b, c and d are constants and $a≠0$.
Comparing the given polynomial with the standard form of a cubic polynomial,
$a=2$, $b=1$, $c=-5$ and $d=2$
Also,
If α is a root of a given polynomial $f(x)$, then $f(α)=0$.
Therefore,
For $x = \frac{1}{2}$
$f(\frac{1}{2}) = 2(\frac{1}{2})^3 + (\frac{1}{2})^2 – 5(\frac{1}{2}) + 2$
$= 2(\frac{1}{8}) + \frac{1}{4} – 5(\frac{1}{2})+ 2 = 0$
$=\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2$
$=\frac{5}{2}-\frac{5}{2}$
$=0$
$f(\frac{1}{2}) = 0$, this implies, $x = \frac{1}{2}$ is a root of the given polynomial.
For $x = 1$
$f(1) = 2(1)^3 + (1)^2 – 5(1) + 2$
$= 2 + 1 – 5 + 2$
$= 0$
$f(1) = 0$, this implies, $x = 1$ is also a root of the given polynomial.
For $x = -2$
$f(-2) = 2(-2)^3 + (-2)^2 – 5(-2) + 2$
$= -16 + 4 + 10 + 2$
$= 0$
$f(-2) = 0$, this implies, $x = -2$ is also a root of the given polynomial.
Now,
Sum of zeros $= \frac{-b}{a}=\frac{-1}{2}$.
Sum of the zeros of $f(x)=\frac{1}{2} + 1 – 2 = – \frac{1}{2}$
Sum of the products of the zeros taken two at a time $=\frac{c}{a}=\frac{-5}{2}$.
Sum of the products of the zeros taken two at a time$=(\frac{1}{2} \times 1) + (1 \times -2) + (\frac{1}{2} \times-2) = \frac{-5}{ 2}$.
Product of the zeros $= \frac{– d}{a}=\frac{-2}{2}=-1$.
Product of the zeros$=\frac{}{2} \times 1 \times (– 2) = -1$.
Hence, the relationship between the zeros and the coefficients is verified.
Updated on: 10-Oct-2022
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