Answer :
[tex]\large\underline{\sf{Solution-}}[/tex]
Given quadratic polynomial is [tex]\sf \: {x}^{2} - 2x - 8 [/tex]
[tex]\sf \:Let\:assume\:that\: f(x) = x^2 - 2x - 8 \\ [/tex]
[tex]\sf \: f(x) = {x}^{2} - 4x + 2x - 8\\ [/tex]
[tex]\sf \: f(x) = x(x - 4) + 2(x - 4)\\ [/tex]
[tex]\sf \: f(x) = (x - 4)(x + 2) \\ [/tex]
So, to find zeroes of f(x), we substitute
[tex]\sf \: f(x) = 0\\ [/tex]
[tex]\sf \: (x - 4)(x + 2) = 0\\ [/tex]
[tex]\sf\implies \sf \: x = 4 \: \: or \: \: x = -\:2 \\ [/tex]
Now, Consider
[tex]\sf \: Sum\:of\:zeroes \\ [/tex]
[tex]\qquad\sf \: = \: 4 - 2\\ [/tex]
[tex]\qquad\sf \: = \:2 \\ [/tex]
can be rewritten as
[tex]\qquad\sf \: =\: -\:\dfrac{( - 2)}{1} \\ [/tex]
[tex]\qquad\sf \: = \: - \: \dfrac{\: coefficient \: of \: x \: }{coefficient \: of \: {x}^{2}} \\ [/tex]
Hence,
[tex]\sf\implies \sf \: Sum\:of\:zeroes \: = \: - \: \dfrac{\: Coefficient \: of \: x \: }{Coefficient \: of \: {x}^{2}} \\ [/tex]
Now, Consider
[tex]\sf \: Product\:of\:zeroes \\ [/tex]
[tex]\qquad\sf \: = \:4 \times (- 2) \\ [/tex]
[tex]\qquad\sf \: = \: -\:8\\ [/tex]
[tex]\qquad\sf \: = \: \dfrac{-8}{1}\\ [/tex]
[tex]\qquad\sf \: = \: \dfrac{Constant \: term}{Coefficient \: of \: {x}^{2}}\\ [/tex]
Hence,
[tex]\sf\implies \sf \:Product\:of\:zeroes \: = \: \dfrac{Constant \: term}{Coefficient \: of \: {x}^{2}}\\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information:
[tex]\bf \:If\:\alpha, \beta \: are \: zeroes \: of \: {ax}^{2} + bx + c, \: then \\ [/tex]
[tex]\sf \: \alpha + \beta = - \dfrac{b}{a} \\ [/tex]
[tex]\sf \: \alpha \beta = \dfrac{c}{a} \\ [/tex]
[tex]\bf \:If\: \alpha, \beta, \gamma \: are \: zeroes \: of \: {px}^{3} + {qx}^{2} + rx + s, \: then\\ [/tex]
[tex]\sf \: \alpha + \beta + \gamma = - \dfrac{q}{p} \\ [/tex]
[tex]\sf \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{r}{p}\\ [/tex]
[tex]\sf \: \alpha \beta \gamma = - \dfrac{s}{p} \\ [/tex]
