Verify that the numbers given alongside the cubic polynomials below are their zeros. Also, verify the relationship between the zeros and the coefficients.fx=2×3+x2-5x+2; 12,1,-2

# Verify that the numbers given alongside the cubic polynomials below are their zeros. Also, verify the relationship between the zeros and the coefficients.

1. A
True
2. B
False

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### Solution:

Given: $f\left(x\right)=2{x}^{3}+{x}^{2}-5x+2$
For $x=\frac{1}{2}$,
$⇒f\left(\frac{1}{2}\right)=2{\left(\frac{1}{2}\right)}^{3}+{\left(\frac{1}{2}\right)}^{2}-5×\frac{1}{2}+2$
$⇒f\left(\frac{1}{2}\right)=2×\frac{1}{8}+\frac{1}{4}-\frac{5}{2}+2$
$⇒f\left(\frac{1}{2}\right)=\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2$
$⇒f\left(\frac{1}{2}\right)=\frac{1+1-10+8}{4}$
$⇒f\left(\frac{1}{2}\right)=\frac{0}{4}$
$⇒f\left(\frac{1}{2}\right)=0$
Now, for $x=1$,
$⇒f\left(1\right)=2{\left(1\right)}^{3}+{\left(1\right)}^{2}-5×1+2$
$⇒f\left(1\right)=2×1+1-5+2$
$⇒f\left(1\right)=2+1-5+2$
$⇒f\left(1\right)=0$
And, for $x=-2$,
$⇒f\left(-2\right)=2{\left(-2\right)}^{3}+{\left(-2\right)}^{2}-5×\left(-2\right)+2$
$⇒f\left(-2\right)=2×\left(-8\right)+4+10+2$
$⇒f\left(-2\right)=-16+4+10+2$
$⇒f\left(-2\right)=0$
Hence, are the zeros of the cubic polynomial.
Now, compare the given cubic polynomial with $a{x}^{3}+b{x}^{2}+\mathit{cx}+d$, we get,
$⇒a=2$
$⇒b=1$
$⇒c=-5$
$⇒d=2$
And, we know, the roots of the equation are given by , which are:
$⇒\alpha =\frac{1}{2}$
$⇒\beta =1$
$⇒\gamma =-2$
Therefore,
$⇒\alpha +\beta +\gamma =\frac{1}{2}+1+\left(-2\right)$
$=\frac{1}{2}+1-2$
$=\frac{1+2-4}{2}$
$=-\frac{1}{2}$
And, $-\frac{b}{a}=-\frac{1}{2}$
Hence, $\alpha +\beta +\gamma =-\frac{b}{a}$
Now, $\mathit{\alpha \beta }+\mathit{\beta \gamma }+\mathit{\gamma \alpha }=\frac{1}{2}×1+1×\left(-2\right)+\left(-2\right)×\frac{1}{2}$
$=\frac{1}{2}-2-1$
$=\frac{1-4-2}{2}$
$=-\frac{5}{2}$
And, $\frac{c}{a}=-\frac{5}{2}$
Hence, $\mathit{\alpha \beta }+\mathit{\beta \gamma }+\mathit{\gamma \alpha }=\frac{c}{a}$
Now, $\mathit{\alpha \beta \gamma }=\frac{1}{2}×1×\left(-2\right)=-1$
And, $-\frac{d}{a}=-\frac{2}{2}=-1$
Hence, $\mathit{\alpha \beta \gamma }=-\frac{d}{a}$

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