Question

Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and the coefficients in each case.
$2x^3+x^2-5x+2;\frac{1}{2},1,-2$

Answer 1

Step 1: Find the value of the given polynomial for the given numbers.

Given, $p\left(x\right)=2x^3+x^2-5x+2$
And the zeroes for $p\left(x\right)$ are $\frac{1}{2},1,-2$
$\therefore p\left(\frac{1}{2}\right)=2{\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{2}\right)}^2-5\left(\frac{1}{2}\right)+2$
$=2\left(\frac{1}{8}\right)+\left(\frac{1}{4}\right)-\left(\frac{5}{2}\right)+2$
$=0$
$p\left(1\right)=2(1{)}^3+(1{)}^2-5\left(1\right)+2=0$
$p(-2)=2(-2{)}^3+(-2{)}^2-5(-2)+2=0$

Hence, proved $\frac{1}{2},1,-2$ are the zeroes of $2x^3+x^2-5x+2$.

Step 2: Compare the given polynomial with general expression
Now, comparing the given polynomial with general expression, we get,
$a{x}^3+b{x}^2+cx+d=2x^3+x^2-5x+2$
$\therefore a=2,b=-1,c=-5$ and $d=2$

Step 3: Write down the relationship between the zeroes and the coefficients.
As we know, if $\alpha ,\beta ,\gamma$ are the zeroes of the cubic polynomial $a{x}^3+b{x}^2+cx+d$, then;
$\alpha +\beta +\gamma =-\frac{b}{a}$

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$

$\alpha \beta \gamma =-\frac{d}{a}$

Step 4: Verify the relatopnship between the zeroes and the coefficients.
Therefore, putting the values of zeroes of the polynomial,
$\alpha +\beta +\gamma =\frac{1}{2}+1+(-2)\Rightarrow -\frac{1}{2}=-\frac{b}{a}$

$\alpha \beta +\beta \gamma +\gamma \alpha =(\frac{1}{2}\times 1)+(1\times -2)+(-2\times \frac{1}{2})$
$\Rightarrow -\frac{5}{2}=\frac{c}{a}$

$\alpha \beta \gamma =\frac{1}{2}\times 1\times (-2)$

$\Rightarrow -\frac{2}{2}=-\frac{d}{a}$

Hence, the relationship between the zeroes and the coefficients are satisfied.