Find zeroesof polynomialx²+ 7 x+ 12and verify the relationship betwe

1.	Find zeroesof polynomialx²+ 7 x+ 12and verify the relationship betweenZeros andCoefficient
Answer» $\bf \huge \hookrightarrow \: \: Given:$ $\Large \mapsto \: {\textrm{{{\color{navy}{x² \: + <klux>7X</klux> \: + \: <klux>12</klux>}}}}}$ ☛Now , we find ZEROS of this polynomial. $\bf \Large \rightarrow \: \: {x}^{2} \: + 7x \: + 12 \\ \\ \bf \Large \rightarrow \: \: {x}^{2} \: + \: 4x \: + 3x \: + 12 \\ \\ \bf \Large \rightarrow \: \:x \: (x \: + \: 4) + \:3 \: (x + 4) \\ \\ \bf \Large \rightarrow \:(x + 3) \: \: \: (x + 4) \\ \\$ $\bf \Large \implies \: \: x \: + \: 3 \: = \: 0 \\ \\ \bf \Large \implies \: \: x \: = \: \: - 3 \\ \\ \\ \bf \Large \implies \: \: x \: + \: 4 \: = 0 \\ \\ \bf \Large \implies \: x \: = \: \: - 4$ $\bf \Large \hookrightarrow \: \: Then, \\ \\ \bf \Large \rightarrow \: \: \alpha \: = \: \: - 3 \\ \\ \bf \Large \rightarrow \: \: \beta \: = \: \: - 4$ ${\textrm{{{\color{navy}{<klux>GENERAL</klux> equation is ax² + bx + c}}}}}$ Now , we compare x² + 7x + 12 to the general equation. $\bf \large \therefore \: \: a \: \: = \: 1 \\ \\ \bf \large \therefore \: \: \: b \: \: = \: 7 \\ \\ \bf \large \therefore \: \:c \: \: = \: 12$ _______________________ Now , we find the RELATIONSHIP between Zeros and Coefficient. $\Large \mid \underline {\bf {{{\color{orange}{First \: \: Relation}}}}} \mid$ $\bf \Large \mapsto \:\: \: Sum \: \: of \: zeroes \\ \\ \bf \Large \mapsto \:\: \: \: ( \: \alpha \: + \: \beta \: ) = \frac{ - b}{a}$ $\bf \large \mapsto \:\: \: \: ( \: - 3 \: + \: - 4 \: ) = \frac{ - \: 7}{1} \\ \\ \bf \large \mapsto \:\: \: \: \: - 7 \: = \: - 7$ $\large {\textrm{{{\color{red}{First \: Relation \: is \: proved}}}}}$ $\Large \mid \underline {\bf {{{\color{orange}{Second \: \: Relation}}}}} \mid$ $\bf \large \mapsto \:\: \:Product \: \: of \: \: Zeros \\ \\ \bf \Large \mapsto \:\: \: \: ( \: \alpha \: \times \: \beta \: ) \: = \: \frac{ c}{a}$ $\bf \Large \mapsto \:\: \: \: ( \: - 3 \: + \: - 4 \: ) = \frac{ 12}{1} \\ \\ \bf \Large \mapsto \:\: \: 12 \: = \: 12$ $\large{\textrm{{{\color{red}{Second \: Relation \: is \: proved}}}}}$

Find zeroesof polynomialx²+ 7 x+ 12and verify the relationship betweenZeros andCoefficient

Answer»

$\bf \huge \hookrightarrow \: \: Given:$

$\Large \mapsto \: {\textrm{{{\color{navy}{x² \: + <klux>7X</klux> \: + \: <klux>12</klux>}}}}}$

☛Now , we find ZEROS of this polynomial.

$\bf \Large \rightarrow \: \: {x}^{2} \: + 7x \: + 12 \\ \\ \bf \Large \rightarrow \: \: {x}^{2} \: + \: 4x \: + 3x \: + 12 \\ \\ \bf \Large \rightarrow \: \:x \: (x \: + \: 4) + \:3 \: (x + 4) \\ \\ \bf \Large \rightarrow \:(x + 3) \: \: \: (x + 4) \\ \\$

$\bf \Large \implies \: \: x \: + \: 3 \: = \: 0 \\ \\ \bf \Large \implies \: \: x \: = \: \: - 3 \\ \\ \\ \bf \Large \implies \: \: x \: + \: 4 \: = 0 \\ \\ \bf \Large \implies \: x \: = \: \: - 4$

$\bf \Large \hookrightarrow \: \: Then, \\ \\ \bf \Large \rightarrow \: \: \alpha \: = \: \: - 3 \\ \\ \bf \Large \rightarrow \: \: \beta \: = \: \: - 4$

${\textrm{{{\color{navy}{<klux>GENERAL</klux> equation is ax² + bx + c}}}}}$

Now , we compare x² + 7x + 12 to the general equation.

$\bf \large \therefore \: \: a \: \: = \: 1 \\ \\ \bf \large \therefore \: \: \: b \: \: = \: 7 \\ \\ \bf \large \therefore \: \:c \: \: = \: 12$

_______________________

Now , we find the RELATIONSHIP between Zeros and Coefficient.

$\Large \mid \underline {\bf {{{\color{orange}{First \: \: Relation}}}}} \mid$

$\bf \Large \mapsto \:\: \: Sum \: \: of \: zeroes \\ \\ \bf \Large \mapsto \:\: \: \: ( \: \alpha \: + \: \beta \: ) = \frac{ - b}{a}$

$\bf \large \mapsto \:\: \: \: ( \: - 3 \: + \: - 4 \: ) = \frac{ - \: 7}{1} \\ \\ \bf \large \mapsto \:\: \: \: \: - 7 \: = \: - 7$

$\large {\textrm{{{\color{red}{First \: Relation \: is \: proved}}}}}$

$\Large \mid \underline {\bf {{{\color{orange}{Second \: \: Relation}}}}} \mid$

$\bf \large \mapsto \:\: \:Product \: \: of \: \: Zeros \\ \\ \bf \Large \mapsto \:\: \: \: ( \: \alpha \: \times \: \beta \: ) \: = \: \frac{ c}{a}$

$\bf \Large \mapsto \:\: \: \: ( \: - 3 \: + \: - 4 \: ) = \frac{ 12}{1} \\ \\ \bf \Large \mapsto \:\: \: 12 \: = \: 12$

$\large{\textrm{{{\color{red}{Second \: Relation \: is \: proved}}}}}$