f(x) = x.3 + 4x.2 + x – 6

Question

what are the number of real zeros of the function

Answer 1

Small note: people usually use ^ for exponents.

 

To start, list the factors of the of the "d" (x⁰) term. These are 1, 2, 3, and 6. Try plugging these into the formula to see if you get zero. It turns out the only one that works is +1. That means that is one of the roots, and there will be a (x - 1) term in the final answer.

 

The next step is to split the x² and x terms so that you can "recreate" this (x - 1) factor. For example, you have

 

x³ + 4x²

 

To get an (x - 1), turn this into

 

x³ - x² + 5x² = x²(x - 1) + 5x²

 

To the same thing on the other part of the formula:

 

x - 6 = 6x - 5x - 6 = 6(x - 1) - 5x

 

So the formula now is

 

x²(x - 1) + 5x² + 6(x - 1) - 5x

 

You can see that 5x² - 5x can be written as 5x(x - 1), so

 

x²(x - 1) + 5x(x - 1) + 6(x - 1) = (x - 1)(x² + 5x +6)

 

The quadratic term is pretty simple to factor: the factors are +3 and +2, so

 

(x - 1)(x + 3)(x + 2)

 

In this form, it is obvious what the roots are: +1, -3, and -2.

Answer 2

There are three real zeros x = -3, x = -2 and x = 1