Factoring Cubic Polynomials: A Confused Student's Guide

[torquerotates]

Well, I admit, I looked at the solution manual. They apparently 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.

 

[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.

 

[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.

 

[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.

 

[rocomath]

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.

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

Algebra - https://www.amazon.com/dp/0817636773/?tag=pfamazon01-20

Gelfand is the man!

 

[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...

 

[rohanprabhu]

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.

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

 

[Dick]

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