# Factoring problem

1. Feb 9, 2008

### torquerotates

Well, I admit, I looked at the solution manual. They apparantly solved this equation, x(cubed)-(3/4)x-(1/4)=0 by factoring it and making it look like this, (x-1)((
x+(1/2))squared. But they didn't show any steps. I'm completely confused. I have no idea how to factor cubic polynomials.

2. Feb 9, 2008

### rocomath

$$x^3 - \frac 3 4 x - \frac 1 4=0$$

Correct?

Here is a hint:

$$x^3-x^2+x^2-\frac 3 4 x - \frac 1 4=0$$

Complete the square, then you will notice that you have difference of squares. Continue simplifying, and once again, complete the square and it's solved.

Last edited: Feb 9, 2008
3. Feb 9, 2008

### Dick

Change it into an integer coefficient polynomial by multiplying by 4. So 4x^3-3x-1=0. Now look up the rational root theorem. If p/q is a rational root of the polynomial then p divides 1 and q divides 4. So the only choices for rational roots are 1,1/2 or 1/4 or their negatives. This is important since if p/q is a root of the polynomial then (x-p/q) is a factor.

4. Feb 9, 2008

### torquerotates

Wow, I don't think I could get used to that type of thinking. How did you even know that you should start by adding and subtracting 2x? Maybe I'm just not creative enough.

5. Feb 9, 2008

### rocomath

I think you meant $$x^2$$ but just making sure.

Algebra - https://www.amazon.com/Algebra-I-M-...bs_sr_1?ie=UTF8&s=books&qid=1202621463&sr=8-1

Gelfand is the man!

6. Feb 9, 2008

### rohanprabhu

the polynomial $$p(x) = x^3 - \frac{3}{4}x - \frac{1}{4}$$ is 0 at $x = 1$, so $(x - 1)$ is a factor of $p(x)$.

Now try solving this question...

7. Feb 9, 2008

### rohanprabhu

this seems like a pretty cool method.. could you give the general statement please.. or maybe a link somewhere??

8. Feb 10, 2008

### Dick

Just google 'rational root theorem'. Wikipedia has an entry.