# Factoring equation with real coefficients

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1. Mar 18, 2015

### Nathew

1. The problem statement, all variables and given/known data
Find the roots of $$z^4+4=0$$ and use that to factor the expression into quadratic factors with real coefficients.

2. Relevant equations
DeMoivre's formula.

3. The attempt at a solution
I have been able to identify they are $$\pm 1 \pm i$$ but i have no idea how to factor the expression.
Thanks!!

2. Mar 18, 2015

### Dick

No, they aren't. None of those are roots. Try them. Use deMoivre!

3. Mar 19, 2015

### Nathew

$$(1+i)^4+4=0$$
$$(-1+i)^4+4=0$$
$$(1-i)^4+4=0$$
$$(-1-i)^4+4=0$$

Not sure what you're talking about.

4. Mar 19, 2015

### SammyS

Staff Emeritus
It is possible to do this in the reverse order. That is:
1. Factor $\ z^4+4\$ into quadratic factors with real coefficients.
then
2. find the roots of $\ z^4+4=0\ .\$​

That's not following the instructions, but it may give some insight.

Suppose $\ z^4+4=(z^2+az+2)(z^2+bz+2)\$.
Expand the right hand side & equate coefficients.

5. Mar 19, 2015

### Dick

Sorry! I read your post as saying the roots were $\pm 1$ and $\pm i$. If $r_1$ and $r_2$ are roots then $(z-r_1)(z-r_2)$ is a factor of your polynomial. Try multiplying that out when $r_1$ and $r_2$ are complex conjugates.

6. Mar 19, 2015

### HallsofIvy

Yes, 1+ i, 1- i, -1- i, and -1+ i are roots so the we can write $z^4+ 4= (z- (1+ i))(z- (1- i))(z- (-1+ i)(z- (-1- i))= (z- 1- i)(z- 1+ i)(z+ 1- i)(z+ 1+ i)$
Write $(z- 1- i)(z- 1+ i)= ((z- 1)- i)((z- 1)+ i)$ and $(z+ 1- i)(z+ 1+ i)= ((z+ 1)- i)((z+ 1)+ i)$. Now use the fact that $(a- b)(a+ b)= a^2- b^2$.

7. Mar 19, 2015

### Nathew

Thanks a lot this worked great!

8. Mar 19, 2015

### SammyS

Staff Emeritus
What did you get for the quadratic factors?

9. Mar 19, 2015

### Nathew

$$(z^2-2z+2)(z^2+2z+2)$$