**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`

`0=u^2+4u+3`

Using the quadratic formula:

`u=(-b+-sqrt(b^2-4ac))/(2a)`

`u=(-4+-sqrt(4^2-4(1)(3)))/(2(1))`

`u=(-4+-sqrt(16-12))/2rArru=(-4+-sqrt(4))/2rArr(-4+-2)/2`

`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

**Answer:**

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

Just in case, here's a graph:

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^4+3y^2+y^2+3=0`

`` `y^2(y^2+3)+1(y^2+3)=0`

`(y^2+3)(y^2+1)=0`

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

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

If `y^2+1=0`

`y^2=-1`

`y=+-i`

If `y^2+3=0`

`y^2=-3`

`y=+-isqrt(3)`

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