SOLUTION: f(x)=x^4+10x^3+27x^2+10x+26; if i is a zero, find the other three.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: f(x)=x^4+10x^3+27x^2+10x+26; if i is a zero, find the other three.       Log On


   



Question 316381: f(x)=x^4+10x^3+27x^2+10x+26; if i is a zero, find the other three.

Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=x%5E4%2B10x%5E3%2B27x%5E2%2B10x%2B26

i | 1 10    27      10      26
  |     +i  -1+10i -10+26i -26
    1 10+i  26+10i     26i   0

So we have factored f(x) this way

f%28x%29=%28x-i%29[x%5E3+%2B+%2810%2Bi%29x%5E2+%2B+%2826%2B10i%29x+%2B+26]

Since i, that is, (0+i), is a zero, so is its conjugate (0-i), or -i

So we factor further by using synthetic division to factor the
expression in brackets:

-i | 1  10+i  26+10i  26i
   |      -i    -10i -26i  
     1  10    26        0

So we have factored f(x) this way

f%28x%29=%28x-i%29%28x%2Bi%29%28x%5E2%2B10x%2B26%29

Set the last factor = 0

x%5E2%2B10x%2B26=0

That does not factor so we use the quadratic formula:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

x+=+%28-10+%2B-+sqrt%28+10%5E2-4%2A1%2A26+%29%29%2F%282%2A1%29+

x+=+%28-10+%2B-+sqrt%28+100-104%29%29%2F2+

x+=+%28-10+%2B-+sqrt%28-4%29%29%2F2+

x+=+%28-10+%2B-+2i%29%2F2+

x+=+%28-10%29%2F%282%29+%2B-+2i%2F2

x+=+-5+%2B-+i

So the four zeros are i, -i, -5%2Bi, and -5-i

Edwin