how do i determine the number of real zeros of the function

You can use factor theorem and synthetic division to answer the question.

f(x) = x

^{3}+ 4x^{2}+ x - 6Since f(1) = 0, by factor theorem x-1 is a factor.

Using synthetic division,

1 | 1 4 1 -6

........1 5 6

-----------------

....1 5 6 | 0

So, f(x) = (x-1)(x

^{2}+5x+6) = (x-1)(x+2)(x+3)Answer: There are three real zeros: x = 1, -2, -3.

## Comments

^{3}-(a+b+c)x^{2}+(ab+ac+bc)x-abc