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Given that f(x) is a cubic function with zeros at -1  and -2i-3  , find an equation for f(x) given that  f(-9)= -6

 Jun 6, 2018
 #1
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Because of the conjugate property, the function will also have the zero  -3 + 2i

 

So....let the  function be  f(x)  and we have that

 

f(x)  = a( x + 1)( x - ( - 3 - 2i) ) ( x - (-3 + 2i) )   simplify

 

f(x)  = a ( x + 1)  ( x^2  -x (-3 + 2i)  -x(-3 -2i)  + ( -3 + 2i) (-3 - 2i) )

 

f(x)  = a  ( x + 1) ( x^2 + 3x - 2xi + 3x + 2xi + 9 - 4i^2 )

 

f(x)  = a ( x + 1) (x^2 + 6x + 9 + 4)

 

f(x)  = a (x + 1)(x^2 + 6x + 13)

 

f(x)  = a  ( x^3 + 6x^2 + 13x  + x^2 + 6x + 13 )

 

f(x)  = a  ( x^3 + 7x^2 + 19x + 13)

 

And  since  f(-9)  = -6, we can find  "a"  as

 

-6  = a ( (-9)^3 + 7(-9)^2 + 19(-9) + 13)

 

-6  = a ( -320)

 

a  =  6 / 320   =  3/160

 

So.....f(x)  =   (3/160) ( x^3 + 7x^2 + 19x + 13)

 

 

 

cool cool cool

 Jun 7, 2018
edited by CPhill  Jun 7, 2018

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