find the complex zero of polynomial function f(x)=x^4+4x^2+3

Expert Answers
quirozd eNotes educator| Certified Educator

Answer: There are four complex zeros: `x=+-sqrt(3)i or x=+-i`

You can solve this a number of ways. The way I will show you is by using substitution and the quadratic formula.

Recall that the quadratic formula is based on the following quadratic formatted `ax^2+bx+c = 0`

Since we are looking for roots (when `f(x) = 0` ) then we are on the right track. However, this is a quartic not a quadratic! So, let's see if we can make it a quadratic using a substituion.

Let `u = x^2` Substitute `u` in for any instance of `x^2`

`f(x) = x^4+4x^2+3`


`f(u) = u^2+4u+3 `

Now we have a quadratic-looking equation where `a = 1``b=4` , and `c=3`


Using the quadratic formula:




`u=-3 or u=-1`

Now we have to substitute back

`x^2=-3 or x^2=-1`

Taking the squareroot of both sides, we end up with 

`x=+-sqrt(-3)or x=+-sqrt(-1)`

since `i=sqrt(-1)` , then the answer becomes


This function has four complex roots, as the largest polynomial exponent suggests.

Just in case, here's a graph:

pramodpandey | Student

Given polynomial function f(x)=x^4+4x^2+3

let y be zero of f(x) i.e.

`y^4+4y^2+3=0`      9i)

factorize  (i) 


`` `y^2(y^2+3)+1(y^2+3)=0`


`either (y^2+3)=0`

`or (y^2+1)=0`

If   `y^2+1=0`



If `y^2+3=0`



Thus roots are `+-i,+-isqrt(3)`