Finding Zeros of f(x)=x3-3x2-6x+8

[math4life]

Homework Statement

f(x)=x³-3x²-6x+8

Homework Equations

p/q=+-(1,2,4,8)
f(0)=8

The Attempt at a Solution

What is the fastest way to find the zeros? I have to graph it. The back of the book shows the zeros looking to be 4, -2, and most likely 1.5. I am assuming we use the rational zeros theorum to divide by factors of 8 in a trial and error fashion. After I divided by 4 synthetically I got 1,1,-2,0 which I then divided by -2 to get 1, -1, 0. What does this mean? Is there a more efficient way to do this than randomly try all the p/qs?

 

[rochfor1]

f(1)=0, so you know x-1 is a factor. Either use synthetic division or regular old polynomial long division to get a quadratic. It should be pretty simple after that.

 

[Redbelly98]

After I divided by 4 synthetically I got 1,1,-2,0 which I then divided by -2 to get 1, -1, 0. What does this mean?

It means that:

• (x-4) and (1·x² + 1·x - 2) are factors (using the 1,1,-2 as the coefficients, with 0 as the remainder.
• (x+2) and (1·x - 1) are factors

Is there a more efficient way to do this than randomly try all the p/qs?

Well, you could combine that with the Rule of Signs to figure out something about the number of positive and negative roots. Also, it's probably easier to start with the smaller numbers (±1, ±2) since it's a little easier to check whether f(x)=0 for those, especially for x=±1. If you can find just one factor, the expression becomes a quadratic which is easier to solve.

 

[math4life]

Got it, thanks.