Question

Verify that $\frac{1}{2}$, 1, -2 are zeros of cubic polynomial $2x^3+x^2-5x+2$. Also verify the relationship between, the zeros and their coefficients.

Answer 1

$f\left(x\right)=2x^3+x^2-5x+2$

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

$=\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2=0$

$f\left(1\right)=2(1{)}^3+(1{)}^2-5\left(1\right)+2=2+1-5+2=0$,

$f(-2)=2(-2{)}^3+(-2{)}_5^2(-2)+2=-16+4+10+2=0.$

Let $\alpha =\frac{1}{2},\beta =1$, and $\gamma =-2$

Now, Sum of zeros $=\alpha +\beta +\gamma =\frac{1}{2}+1-2=-\frac{1}{2}$

Also, sum of zero $=-\frac{\left(Coefficientof{x}^2\right)}{\text{Coefficient of x^3}}=-\frac{1}{2}$

So, sum of zeros $=\alpha +\beta +\gamma =-\frac{Coefficientof{x}^2}{\text{Coefficient of x^3}}$

Sum of product of zeros taken two at a time $=\alpha \beta +\beta \gamma +\gamma \alpha$

$=\frac{1}{2}\times 1+1\times (-2)+(-2)\times \frac{1}{2}=-\frac{5}{2}$

Also, $\alpha \beta +\beta \gamma +\gamma \alpha =\frac{\text{Coefficient of x}}{\text{Coefficient of x^3}}=\frac{-5}{2}$

So, sum of product of zeros taken two at a time $=\alpha \beta +\beta \gamma +\gamma \alpha =\frac{\text{Coefficient of x}}{\text{Cofficient of x^3}}$

Now, Product of zeros $=\alpha \beta \gamma =\left(\frac{1}{2}\right)\left(1\right)(-2)=-1$

Also, product of zeros $=-\frac{\text{Constant term}}{\text{Coefficient of x^3}}=\frac{-2}{2}=-1$

$\therefore$ Product of zeros $=\alpha \beta \gamma =-\frac{\text{Constant term}}{\text{Coefficient of x^3}}$