Find the quadratic polynomial whose zeroes are -3 and 3

1.	Find the quadratic polynomial whose zeroes are -3 and 3
Answer» $\large\underline{\sf{Given- }}$ The zeroes of the quadratic polynomial as - 3 and 3. $\large\underline{\sf{To\:Find - }}$ The quadratic polynomial $\large\underline{\sf{Solution-}}$ Given that - 3 and 3 are the zeroes of quadratic polynomial. So, $\rm :\longmapsto\:Let \: \alpha \: = \: - \: 3 \: \: and \: \: \beta \: = \: 3$ Now, $\rm :\longmapsto\:Sum \: of \: zeroes = \alpha + \beta = - 3 + 3 = 0$ and $\rm :\longmapsto\:Product \: of \: zeroes = \alpha \beta = - 3 \times 3 = - 9$ We KNOW, $\sf \: If \: \alpha \: and \: \beta \: are \: the \: zeroes \: of \: quadratic \: polynomial \: then$ $\sf \: the \: <klux>REQUIRED</klux> \: polynomial \: f(x) \: is \:$ $\rm :\longmapsto\: \sf \: f(x) \: = k\bigg( {x}^{2} - ( \alpha + \beta )x + \alpha \beta \bigg) \: where \: k \ne \: 0$ Thus, The required Quadratic polynomial whose zeroes are- 3 and 3 is $\rm :\longmapsto\: \sf \: f(x) = k\bigg( {x}^{2} - 0x - 9 \bigg) \: where \: k \ne0$ $\rm :\longmapsto\:\sf \: f(x) = k\bigg( {x}^{2} - 9 \bigg) \: where \: k \ne0$ ADDITIONAL Information :- $\sf \: If \: \alpha \: and \: \beta \: are \: the \: zeroes \: of \: f(x) = a {x}^{2} + bx + c \: then$ $\boxed{{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$ Or $\boxed{{\tt Sum\ of\ the\ zeroes=\frac{-b}{a}}}$ And $\boxed{{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$ Or $\boxed{{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$ $\boxed{ \sf{ \: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta) }^{2} - 2 \alpha \beta }}$ $\boxed{ \sf{ \: { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta) }^{3} - 3 \alpha \beta( \alpha + \beta ) }}$

Answer»

$\large\underline{\sf{Given- }}$

The zeroes of the quadratic polynomial as - 3 and 3.

$\large\underline{\sf{To\:Find - }}$

The quadratic polynomial

$\large\underline{\sf{Solution-}}$

Given that

- 3 and 3 are the zeroes of quadratic polynomial.

So,

$\rm :\longmapsto\:Let \: \alpha \: = \: - \: 3 \: \: and \: \: \beta \: = \: 3$

Now,

$\rm :\longmapsto\:Sum \: of \: zeroes = \alpha + \beta = - 3 + 3 = 0$

and

$\rm :\longmapsto\:Product \: of \: zeroes = \alpha \beta = - 3 \times 3 = - 9$

We KNOW,

$\sf \: If \: \alpha \: and \: \beta \: are \: the \: zeroes \: of \: quadratic \: polynomial \: then$

$\sf \: the \: <klux>REQUIRED</klux> \: polynomial \: f(x) \: is \:$

$\rm :\longmapsto\: \sf \: f(x) \: = k\bigg( {x}^{2} - ( \alpha + \beta )x + \alpha \beta \bigg) \: where \: k \ne \: 0$

Thus,

The required Quadratic polynomial whose zeroes are- 3 and 3 is

$\rm :\longmapsto\: \sf \: f(x) = k\bigg( {x}^{2} - 0x - 9 \bigg) \: where \: k \ne0$

$\rm :\longmapsto\:\sf \: f(x) = k\bigg( {x}^{2} - 9 \bigg) \: where \: k \ne0$

ADDITIONAL Information :-

$\sf \: If \: \alpha \: and \: \beta \: are \: the \: zeroes \: of \: f(x) = a {x}^{2} + bx + c \: then$

$\boxed{{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\boxed{{\tt Sum\ of\ the\ zeroes=\frac{-b}{a}}}$

And

$\boxed{{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\boxed{{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\boxed{ \sf{ \: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta) }^{2} - 2 \alpha \beta }}$

$\boxed{ \sf{ \: { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta) }^{3} - 3 \alpha \beta( \alpha + \beta ) }}$