SOLUTION: z^(4)-z^(2)-2z+2 Please could someone help me with the last part of this question. Write the following polynomial as products of linear factors: z^(4)-z^(2)-2z+2 My try s

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: z^(4)-z^(2)-2z+2 Please could someone help me with the last part of this question. Write the following polynomial as products of linear factors: z^(4)-z^(2)-2z+2 My try s      Log On


   

Question 978698: z^(4)-z^(2)-2z+2
Please could someone help me with the last part of this question.
Write the following polynomial as products of linear factors:
z^(4)-z^(2)-2z+2
My try so far:
(z-1)(z^3+z^2-2)
(z-1)(z-1)(z^2+z+2)
From here I tried to use the quadratic formula and got:
Z=(-1+ sq rt 7)/2 or Z=(-1- sq rt 7)/2
The answer in the book is:
(z-1)^2 (z+1-i)(z+1+i)
Could anyone tell me where I went wrong?
Thank you for your time in advance.
Answer by josgarithmetic(39838) About Me  (Show Source):
You can put this solution on YOUR website!
First try check for root of 1 gives %28z-1%29%28z%5E3%2Bz%5E2-2%29. Be sure that your dividend of synthetic division shows ALL degrees of z.

continuing, again if root of 1 is checked, you find factorization %28z-1%29%28z-1%29%28z%5E2%2B2z%2B2%29. You might have made a visual omission on paper or through keyboard, or maybe forgot about accounting for degrees of z in the division process.

NOW to deal with the quadratic factor,
roots are %28-2%2B-+sqrt%284-4%2A2%29%29%2F2
-1%2B-+i.


To conitnue,
%28z-1%29%5E2%28z-%28-1-i%29%29%28x-%28-1%2Bi%29%29, and you simply simplify this obviously complex factorization or keep as linear and quadratic factorization...
%28z-1%29%5E2%28z%2B1%2Bi%29%28z%2B1-i%29
highlight%28%28z-1%29%5E2%28%28z%2B1%29%2Bi%29%28%28z%2B1%29-i%29%29
-
Either you want it that way, or you want ...
highlight%28%28z-1%29%5E2%28z%5E2%2B2z%2B2%29%29.