Question

Verify the 1/2 ,1, -2 are zeroes of cubic polynomial 2x3+x2 −5x +2 . Also verify
the relationship between the zeroes and their coefficients.

Answer 1

Answer:

sorry I don't know but I will give u all answer if I can give

Answer 2

Answer:

Hi there, here we have,

$p(x) = 2x^3+x^2-5x+2$

So, we need to verify whether, $\frac{1}{2},1,-2$ are zeroes of p(x).

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

So, $\frac{1}{2}$ is a zero of p(x)

$p(1) = 2(1)^3+(1)^2-5(1)+2\\=>2+1-5+2\\=>5-5\\=>0$

So, 1 is a zero of p(x)

$p(-2) = 2(-2)^3+(-2)^2-5(-2)+2\\=> 2(-8)+4+10+2\\=>-16+4+10+2\\=>16-16\\=> 0$

S0, 2 is a zero of p(x)

Now we need to verify the relationship between zeroes and coefficients:

Formulae :

α+β+γ =  $\frac{-b}{a}$

αβ+βγ+γα = $\frac{c}{a}$

αβγ = $\frac{-d}{a}$

Here we have a = 2, b =1, c =-5, d =2

so, α+β+γ =  $\frac{1}{2}+1+(-2) = \frac{-b}{a} = -(\frac{1}{2})$

        => $\frac{1+2-4}{2} = \frac{-1}{2}$

        => $\frac{3-4}{2} = \frac{-1}{2}$

        => $\frac{-1}{2} = \frac{-1}{2}$

And, αβ+βγ+γα =  $\frac{1}{2}*1+1*(-2)+(-2)*\frac{1}{2} = \frac{c}{a} = \frac{-5}{2}$

         $=>\frac{1}{2}-2-1=\frac{-5}{2}\\=> \frac{1-4-2}{2} = \frac{-5}{2}\\=>\frac{1-6}{2}=\frac{-5}{2}\\=>\frac{-5}{2}=\frac{-5}{2}$

And, αβγ =  $\frac{1}{2}*1*(-2) = \frac{-d}{a} = \frac{-2}{2}$

         => $-1 = -1$

So, We verified the relationship between the zeroes and the coefficients of the polynomial.

Hope it helps you,

All the best