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

(ii) $x^3-4x^2+5x-2;2,1,1$

Answer 1

(i) $2x^3+x^2-5x+2;\frac{1}{2},1,-2$
$p\left(x\right)=2x^3+x^2-5x+2$ .... (1)

Zeroes for this polynomial are $\frac{1}{2},1,-2$

Substitute the $x=\frac{1}{2}$ in equation (1)

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

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

$=0$

Substitute the $x=1$ in equation (1)

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

Substitute the $x=-2$ in equation (1)

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

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

Comparing the given polynomial with $a{x}^3+b{x}^2+cx+d$ we obtain,

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

Let us assume $\alpha =\frac{1}{2}$, $\beta =1$, $\gamma =-2$
Sum of the roots = $\alpha +\beta +\gamma =\frac{1}{2}+1=2=\frac{-1}{2}=\frac{-b}{a}$

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

Product of the roots = $\alpha \beta \gamma =\frac{1}{2}\times x\times (-2)=\frac{-2}{2}=\frac{d}{a}$

Therefore, the relationship between the zeroes and coefficient are verified.

(ii) $x^3-4x^2+5x-2;2,1,1$
$p\left(x\right)=$$x^3-4x^2+5x-2$ .... (1)
Zeroes for this polynomial are $2,1,1$

Substitute $x=2$ in equation (1)
$p\left(2\right)=2^3-4\times 2^2+5\times 2-2$
$=8-16+10-2=0$

Substitute $x=1$ in equation (1)

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

Therefore, $2,1,1$ are the zeroes of the given polynomial.

Comparing the given polynomial with $a{x}^3+b{x}^2+cx+d$ we obtain,

$a=1,b=-4,c=5,d=-2$
Let us assume $\alpha =2$, $\beta =1$, $\gamma =1$

Sum of the roots = $\alpha +\beta +\gamma =2+1+1=4=-\frac{-4}{1}\frac{-b}{a}$

Multiplication of two zeroes taking two at a time=$\alpha \beta +\beta \gamma +\alpha \gamma =\left(2\right)\left(1\right)+\left(1\right)\left(1\right)+\left(2\right)\left(1\right)=5=\frac{5}{1}=\frac{c}{a}$

Product of the roots = $\alpha \beta \gamma =2\times 1\times 1=2=-\frac{-2}{1}=\frac{d}{a}$

Therefore, the relationship between the zeroes and coefficient are verified.