Question

How many zeros does this function have? I thought it had 4 but I also think it might be 2 I'm not sure

Answer 1

a) 0

Step-by-step explanation

Clarification: this question asks how many x-intercepts this function has, which is not the same as zeros.

A zero is the solution to f(x) = 0, which can be either a real number or a complex number. If the zero is a complex number, it means that the graph of the function does not actually intersect the x-axis - this is an x-intercept.

In summary, a zero is an x-intercept only if the zero is real. If it is not real, then it is not an x-intercept.

Now, let's solve this. To find the x-intercepts we have to solve f(x) = 0,

`(x^2+1)(x^2+2)=0`

Since we have two factors, the function will be zero if any of the factors are zero, so,

`x^2+1=0`

To solve this we have to subtract 1 from both sides and take the square root,

`x=\pm√(-1)=\pm√(i^2)=\pm i`

Since the value under the radical is negative, the solution is not real, so there are no x-intercepts regarding this factor.

Let's solve the other factor,

`x^2+2=0`

The same process: subtract 2 and take the square root,

`x=\pm√(-2)=\pm√(i^2\cdot2)=\pm i√(2)`

Again, the value under the radical is negative and, therefore, the solution is not real either.

In summary, this function has four zeros, but the four of them are complex - i.e. not real. Hence, this function has no x-intercepts.