Question

Without using a calculator, determine the number of real zeros of the function:

f(x)= x^3+ 4x^2+x-6

*the clue on pluto, synthetic division finds that there are three real zeros. as suggested by the highest degree in the function*

Answer 1

1 , -3, -2

It factors into:

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

Explanation:

First, we use the rational root theorem to determine any solutions of p(x). = x3 + 4x2 + x − 6

Factoring -6:

1

-1

2

-2

3

-3

6

-6

x = 1

p(1) = 1^3 + 4 * 1^2 + 1 - 6 = 6 - 6 = 0

x = 1 is a solution.

(x^3 + 4x^2 + x - 6) / (x - 1) =

x^3 / x = x^2

x^2 * (x - 1) = x^3 - x^2

x^3 + 4x^2 - x^3 + x^2 = 5x^2

5x^2 / x = 5x

5x * (x - 1) = 5x^2 - 5x

5x^2 + x - 5x^2 + 5x = 6x

6x / x = 6

6 * (x - 1) = 6x - 6

6x - 6 - 6x + 6 = 0

(x - 1) * (x^2 + 5x + 6)

x^2 + 5x + 6 factors to (x + 3) * (x + 2)

Factors:

(x - 1)

(x + 2)

(x + 3)

roots:

x = 1

x = -2

x = -3