Answer :

[tex]\large\underline{\sf{Solution-}}[/tex]

Given polynomial is [tex]\sf \: {x}^{2} + 8x + 7 [/tex]

[tex]\sf \:Let\:assume\:that\: f(x) = {x}^{2} + 8x + 7 \\ [/tex]

[tex]\sf \: f(x) = {x}^{2} + 7x + x + 7\\ [/tex]

[tex]\sf \: f(x) = x(x + 7) + 1(x + 7)\\ [/tex]

[tex]\sf \: f(x) = (x + 7)(x + 1) \\ [/tex]

So, to find zeroes of f(x), we substitute

[tex]\sf \: f(x) = 0\\ [/tex]

[tex]\sf \: (x + 7)(x + 1) = 0\\ [/tex]

[tex]\sf\implies \sf \: x = - 7 \: \: or \: \: x = - 1 \\ [/tex]

Now, Consider

[tex]\sf \: Sum\:of\:zeroes \\ [/tex]

[tex]\qquad\sf \: = \: - 7 - 1\\ [/tex]

[tex]\qquad\sf \: = \:- 8\\ [/tex]

can be rewritten as

[tex]\qquad\sf \: =\: -\:\dfrac{8}{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 \: = \:- 7 \times (-1) \\ [/tex]

[tex]\qquad\sf \: = \: 7\\ [/tex]

[tex]\qquad\sf \: = \: \dfrac{7}{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]

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