Question

Find a cubic function with the given zeros. (5 points) -6, 7, -4

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

A cubic function with the zeroes -6 , 7 , - 4

FORMULA TO BE IMPLEMENTED

$\sf{A \: cubic \: function \: with \: zeroes \: \: \alpha , \beta , \gamma \: \: is}$

$\sf{f(x) = {x}^{3} - ( \alpha + \beta + \gamma ) {x}^{2} + ( \alpha \beta + \alpha \gamma + \beta \gamma )x - \alpha \beta \gamma }$

CALCULATION

Here the zeroes of the required cubic

function is -6 , 7 , - 4

$\sf{ \alpha = -6 , \beta = 7 , \gamma = - 4 }$

So

$\sf{ \alpha + \beta + \gamma = - 6 + 7 - 4 = - 3\: }$

And

$\sf{ \alpha \beta + \alpha \gamma + \beta \gamma \: }$

$= \sf{ - 42 + 24 - 28 \: }$

$= \sf{ - 46}$

Also

$\sf{\alpha \beta \gamma =( - 6) \times 7 \times ( - 4) = 168 }$

Hence the required cubic function is

$\sf{f(x) = {x}^{3} - ( \alpha + \beta + \gamma ) {x}^{2} + ( \alpha \beta + \alpha \gamma + \beta \gamma )x - \alpha \beta \gamma }$

$\implies \sf{ f(x) = {x}^{3} - ( - 3) {x}^{2} + ( - 46)x - 168 \: }$

$\implies \sf{ f(x) = {x}^{3} + 3 {x}^{2} - 46x - 168 \: }$

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LEARN MORE FROM BRAINLY

Show that the square of any positive integer can't be of the form 5m+2or 5m+3 where 'm' is a whole number

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Answer 2

$Given \: -6,7 \:and \:-4 \:are \: zeroes$

$of \: a \:cubic \: function$

$Let \: cubic \: function = f(x)$

$= (x+6)(x-7)(x+4)$

$= [ x(x-7)+6(x-7)](x+4)$

$= (x^{2} -7x+6x - 42)(x+4)$

$= (x^{2} -x - 42 )(x+4)$

$= (x^{2}-x-42)x+(x^{2}-x-42)4$

$= x^{3}-x^{2}-42x+4x^{2}-4x-168$

$= x^{3} +3x^{2}-46x-168$

Therefore.,

$\red{ Required\:cubic \: function}$

$\green {= x^{3}+3x^{2}-46x-168}$

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