complex zeros and factored form

Question

x⁴ +22x^{2 +} 21 find the complex zeros and put into factored form 

Answer 1

I'm going to break down Amol's correct answer, because how to get there is important.  

 

First, notice that there are no odd numbers exponents.  That tells you that in your first level of factoring, you will have only X^2 terms.  (X^2+a)(X^2+b)

 

You can treat this part as an algebra problem and think about it as solving y^2 + 22y + 21, where Y = x^2.

 

That means, you are looking for two factors of 21 that add to 22.   (1,3,7,21) quickly reduce to 1 and 21.

 

Which means you are factored to (y+21)(y+1).  

 

Substituting y=X^2, we get Amol's intermediate result

 

(X^2+21)(X^2+1)

 

Now, factors are +/-i and +/-i*sqrt(21), and the factored form is 

 

(X+i*sqrt(21))(X-i*sqrt(21))(X+i)(X-i)

 

Answer 2

 

x^4+22x^2+21=(x^2+21)(x^2+1)

 

Zeros at x^2=-21 and x^2=-1 so x=+-i sqrt(21) and x=+-i