Question

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

(2)
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 :
$\left(ii\right)x^3-4x^2+5x-2;2,1,1$

Answer 1

(1)
Find the value of the given polynomial for the given numbers.
Given, $p\left(x\right)=2x^3-5x+2$
And zeroes for $p\left(x\right)$ are $=\frac{1}{2},1,-2$

$\therefore p\left(\frac{1}{2}\right)=2{\left(\frac{1}{2}\right)}^3+\left(\frac{1}{2}\right)+2=\left(\frac{1}{4}\right)+\left(\frac{1}{4}\right)-\left(\frac{5}{2}\right)+2=0$
$p\left(1\right)=2\left(1\right)3+\left(1\right)2-5\left(1\right)+2=0$
$p(-2)=2(-2)3+(-2)2-5(-2)+2=0$
Hence, proved $\frac{1}{2},1,-2$ are the zeroes of $2x^3+x^2-5x+2.$

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

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

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

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

$\alpha \beta \gamma =\frac{1}{2}\times 1(-2)=-\frac{2}{2}=-\frac{d}{a}$
Hence, the relationship between the zeroes and the coefficients are satisfied.

(2)
Find the value of the given polynomial for the given numbers.
Given, $p\left(x\right)=x^3-4x^2+5x-2$
And zeroes for $p\left(x\right)$ are $2,1,1.$
$\therefore p\left(2\right)=23-4\left(2\right)2+5\left(2\right)-2=0$
$p\left(1\right)=13-(4\times 12)+(5\times 1)-2=0$
Hence proved, $2,1,1$ are the zeroes of $x^3-4x^2+5x-2$

Comparing the given polynomial with general expression
Now, comparing the given polynomial with general expression, we get ;
$\therefore a{x}^3+b{x}^2+cx+d=x^3-4x^2+5x-2$
$a=1.b=-4,c=5\text{and}d=-2$

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 ,
$\alpha +\beta +\gamma =-\frac{b}{a}$
$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$
$\alpha \beta \gamma =-\frac{d}{a}$

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

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

$\alpha \beta \gamma =2\times 1\times 1=2=-\frac{-2}{1}=-\frac{d}{a}$
Hence, the relationship between the zeroes and the coefficients are satisfied.