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: $\left(\mathrm{i}\right)2{\mathrm{x}}^3+{\mathrm{x}}^2-5\mathrm{x}+2;1/2,1,-2$

Answer 1

Step 1. Verify that given numbers are zeros or not of the given polynomial.

$\mathrm{p}\left(\mathrm{x}\right)=2{\mathrm{x}}^3+{\mathrm{x}}^2-5\mathrm{x}+2$

Given zeroes are $\frac{1}{2},1,-2$

Substitute $x=\frac{1}{2}$ in $p\left(x\right)$

$p\left(\frac{1}{2}\right)=2{\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{2}\right)}^2-5\left(\frac{1}{2}\right)+2$

$p\left(\frac{1}{2}\right)=2\left(\frac{1}{8}\right)+\left(\frac{1}{4}\right)-\frac{5}{2}+2$

$p\left(\frac{1}{2}\right)=\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2$

$p\left(\frac{1}{2}\right)=\frac{(1+1-10+8)}{4}$

$p\left(\frac{1}{2}\right)=0$

Substitute $\mathrm{x}=1\mathrm{in}\mathrm{p}\left(\mathrm{x}\right)$

$\mathrm{p}\left(1\right)=2\left(1\right)3+\left(1\right)2-5\left(1\right)+2$

$\mathrm{p}\left(1\right)=2+1-5+2$

$\mathrm{p}\left(1\right)=0$

Substitute$\mathrm{x}=-2\mathrm{in}\mathrm{p}\left(\mathrm{x}\right)$

$\mathrm{p}(-2)=2{(-2)}^3+{(-2)}^2-5(-2)+2$

$\mathrm{p}(-2)=-16+4+10+2$

$\mathrm{p}(-2)=-16+16$

$\mathrm{p}(-2)=0$

Therefore, $\frac{1}{2},1,-2$ are the zeroes of the polynomial.

Step 2. Verify the relationship between the zeroes and the coefficients

On comparison the given polynomial with general expression, we get;

$\therefore a{x}^3+b{x}^2+cx+d=2x^3+x^2-5x+2,a=2,b=1,c=-5\mathrm{and}d=2$

Now let $\alpha =\frac{1}{2},\beta =1and\gamma =2$

$$\begin{gathered} \mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=\frac{1}{2}+1+(-2) \\ \mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=-\frac{1}{2} \\ \mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=-\frac{b}{a} \end{gathered}$$

$$\begin{gathered} \mathrm{\alpha \beta }+\mathrm{\beta \gamma }+\mathrm{\gamma \alpha }=\frac{1}{2}\times 1+1\times (-2)+(-2)\times \frac{1}{2} \\ \mathrm{\alpha \beta }+\mathrm{\beta \gamma }+\mathrm{\gamma \alpha }=-\frac{5}{2} \\ \mathrm{\alpha \beta }+\mathrm{\beta \gamma }+\mathrm{\gamma \alpha }=\frac{c}{a} \end{gathered}$$

$$\begin{gathered} \alpha .\beta .\gamma =\frac{1}{2}\times 1\times (-2) \\ \alpha .\beta .\gamma =-\frac{2}{2} \\ \alpha .\beta .\gamma =-\frac{d}{a} \end{gathered}$$

Hence, the relation between zeroes and coefficient is verified.