help!!!! pre-cal?

Question

List the possible rational roots of x^3 + 6x^2 +10x + 3 = 0. Then determine the rational roots.

please explain as you're solving, thankj you very much

Answer 1

The possible rational roots are ±1, ±3.

x³ + 6x² + 10x + 3 = 0
(x + 3)(x² + 3x + 1) = 0

x = -3 is the only rational root.

Answer 2

First we use Descartes' rule of signs. Since there are no sign changes, there are no positive real roots of f (x).

Now use Descartes' rule of signs for negative roots. To do that solve f (-x).

(-x)^3 + 6(-x)^2 + 10(-x) +3 = 0 ===> -x^3 + 6x^2 - 10x +3 = 0

So there are 3, 1 or 0 negative real roots. So now we can just focus on negative values.

Find the rational roots:

f (x) = x^3 + 6x^2 + 10x + 3

Now find factors of 3:

-1 , -3

We take the negative because we found that there are no positive values using Descartes' Rule of Signs.

At this point you can use synthetic division to test the values. When you do, you'll find the -3 is the only real rational root of the function.

Answer 3

Dude, you've got to graph that bad boy. You got a Ti83+? If so just plug it in setting the equation to y and determine where the function intersects with the y axis.