ANSWER
a) 0
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,
[tex](x^2+1)(x^2+2)=0[/tex]
Since we have two factors, the function will be zero if any of the factors are zero, so,
[tex]x^2+1=0[/tex]
To solve this we have to subtract 1 from both sides and take the square root,
[tex]x=\pm\sqrt{-1}=\pm\sqrt{i^2}=\pm i[/tex]
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,
[tex]x^2+2=0[/tex]
The same process: subtract 2 and take the square root,
[tex]x=\pm\sqrt{-2}=\pm\sqrt{i^2\cdot2}=\pm i\sqrt{2}[/tex]
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.
