For f(x)= x^4+25x^2 + 144 answer the following questions.

a) How many zeros will f(x) have?

b) Find all the zeros of f(x).

c) Factor f(x).

d) What are the x intercepts of f(x)?

Question

For f(x)= x^4+25x^2 + 144 answer the following questions.
 
a) How many zeros will f(x) have?
 
b) Find all the zeros of f(x).
 
c) Factor f(x).
 
d) What are the x intercepts of f(x)?

anonymous · Answer

factors are (x^2+9)(x^2+16) and when you write that you'll see f(x)=0 has no real solutions for x, since x^2 cannot be negative. If you are allowed to use complex numbers, you sure have x for which f(x)=0.

anonymous · Answer

So a would be zero?

anonymous · Answer

yeah, x is a real number right?

anonymous · Answer

right. So for the x - intercepts  would I have to set the factors to zero?

anonymous · Answer

@sarahjones667 sure -- but you will notice you get no real solutions! are you also considering complex zeros?

anonymous · Answer

I don't think so. And no real solution? So that means there's no solutionfor x

anonymous · Answer

well then :-p and the x-intercepts are the same things as the zeros

anonymous · Answer

f(x) will not cross the x axis but will cross the y axis at (0,144). You could easily find that by setting x=0