Question

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

Answer 1

$2x^3+x^2-5x+2,(\frac{1}{2},1.-2)$

$2x^3+x^2-5x+2$

Compare with this $a{x}^3+b{x}^2+cx+d$

If $x=\frac{1}{2}$

$p\left(\frac{1}{2}\right)=2{\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{2}\right)}^2-5\left(\frac{1}{2}\right)+2$ $=\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2$ $=\frac{1+1-10+8}{4}=\frac{0}{4}=0$

If $x=1$

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

if $x=2$

$P(-2)=2(-2{)}^3+(-2{)}^2-5(-2)+2=-16+4+10+2=0$

$\therefore \alpha =\frac{1}{2},\beta =1$ and $\gamma =-2$ are zeroes of $2x^3+x^2-5x+2$

$\therefore$a = 2,b = 1 , c = -5 , d= 2

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

$\alpha +\beta +\gamma =\frac{1}{2}+1+(-2)$$=\frac{1+2-4}{2}=-\frac{1}{2}=-\frac{b}{a}$

$\alpha \beta +\beta \gamma +\gamma \alpha =\left(1/2\right)\left(1\right)+\left(1\right)(-2)+(-2)\left(1/2\right)=\frac{-5}{2}=\frac{c}{a}$

$\alpha \beta \gamma =\left(1/2\right)\left(1\right)(-2)=-1=\frac{-2}{2}=\frac{-d}{a}$