Answer
The prove is as shown below.
Work Step by Step
$\underline{\text{Direct Method:}}$
Let $\alpha$ and $\beta$ are roots for the polynomial $x^2+bx+c$, or, are solutions to the equation $x^2+bx+c=0$.
Using the quadratic formula, the roots are $x=\dfrac{-b\pm\sqrt{b^2-4c}}{2}$.
So, without loss of generality let $\alpha=\dfrac{-b+\sqrt{b^2-4c}}{2}$ and $\beta=\dfrac{-b-\sqrt{b^2-4c}}{2}$.
Now, sum of roots $=\alpha +\beta$
$=\dfrac{-b+\sqrt{b^2-4c}}{2}+\dfrac{-b-\sqrt{b^2-4c}}{2}$
$=\dfrac{-b+\sqrt{b^2-4c}-b-\sqrt{b^2-4c}}{2}$
$=\dfrac{-2b}{2}$
$=-b$.
$\Rightarrow b=-\alpha-\beta$.
Product of roots $=\alpha \beta$
$=\dfrac{-b+\sqrt{b^2-4c}}{2}\cdot \dfrac{-b-\sqrt{b^2-4c}}{2}$
$=\dfrac{(-b)^2-(\sqrt{b^2-4c})^2}{4}$ using identity $(x-y)(x+y)=x^2-y^2$.
$=\dfrac{b^2-(b^2-4c)}{4}$
$=\dfrac{b^2-b^2+4c}{4}$
$=\dfrac{4c}{4}$
$=c$
Therefore, $b=-\alpha-\beta$ and $c=\alpha \beta$. hence, proved.
$\underline{\text{Alternative Method:}}$
Let $\alpha$ and $\beta$ are roots for the polynomial $x^2+bx+c$, or, are solutions to the equation $x^2+bx+c=0$.
Thus, we have $\alpha^2+b\alpha+c=0$ and $\beta^2+b\beta+c=0$.
On subtracting both equations, we get
$(\alpha^2+b\alpha+c)-(\beta^2+b\beta+c)=0$
$\alpha^2-\beta^2+b\alpha-b\beta=0$
$(\alpha-\beta)(\alpha+\beta)+b(\alpha-\beta)=0$
$(\alpha+\beta)+b=0$, or, $b=-\alpha-\beta$.
Substitute this value in first equation, to get
$\alpha^2+(-\alpha-\beta)\alpha+c=0$
$\Rightarrow \alpha^2-\alpha^2-\alpha\beta+c=0$
$-\alpha\beta+c=0$, or, $c=\alpha\beta$.
Therefore, $b=-\alpha-\beta$ and $c=\alpha\beta$.
Hence, proved.
