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Factor the polynomial function over the complex number.

 

f(x) = x^4 + 9x^3 + 15x^2 + 9x + 14

Guest Oct 27, 2017
 #1
avatar+92806 
+2

Factor the polynomial function over the complex number.

 

f(x) = x^4 + 9x^3 + 15x^2 + 9x + 14

 

any roots must be factors of 14    and when they are substituted they must make the function =0

By substitution I found that x=-7 and x=-2 are both roots of this function.

So two of the factors are (x+7) and (x+2)

(x+7)(x+2)=x^2+9x+14

 

I divided f(x) by x^2+9x+14 and found the answer to be (x^2+1)

 

so

\(f(x) = x^4 + 9x^3 + 15x^2 + 9x + 14\\ f(x) = (x+7)(x+2)(x^2+1)\\ f(x) = (x+7)(x+2)(x+i))(x-i)\\\)

Melody  Oct 27, 2017
 #2
avatar+9494 
+1

Factor the polynomial function over the complex number.

f(x) = x^4 + 9x^3 + 15x^2 + 9x + 14

laugh

Omi67  Oct 27, 2017
 #3
avatar+87342 
+1

 x^4 + 9x^3 + 15x^2 + 9x + 14

 

Split  15x^2  into  14x^2  +  x^2

 

 x^4 + 9x^3 + 14x^2 + x^2 + 9x + 14     factor as

 

x^2  [x^2 + 9x + 14 ]  +  1 [ x^2 + 9x + 14 ]   take out GCF

 

[ x^2 + 1 ] [ x^2 + 9x + 14 ]

 

[ x^2 + 1 ] [  x + 7 ] [ x + 2 ]

 

 

cool cool cool

CPhill  Oct 27, 2017

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