For polynomial equations of order above $2$, if a quadratic factor has no solution, does it only have one solution?

Question

I am to study the following equation for real solutions:

$$x^3 - 3x^2 + 4 = 0$$

I can see that $x = 2$ is a solution.

Then, using polynomial long division, I get the factor $x^2 - x - 2$.

Now, using the quadratic equation to solve this factor for solutions, I get:

$$b^2 - 4ac = 1^2 - (4 * 1 * 2) = -7$$

As I understand from my notes, this means $x^2 - x - 2 = 0$ has no real solution.

If this is correct, can I now rest happy that the only real solution to the original equation is $x = 2$?

You may want to double check your polynomial division. The factor you should get is $x^2-x-2$. — Ben Sheller, Dec 16 '15 at 05:34
No problem. That should also change your answer as far as the number of real solutions goes. — Ben Sheller, Dec 16 '15 at 05:39
Wonderful. I can see that now. However, can I please ask, if I have a similar question where b2 - 4ac < 0, does that definitely mean the equation just has the one solution? — Hanshan, Dec 16 '15 at 05:41
If it's a cubic equation like yours, then only one real solution. — SchrodingersCat, Dec 16 '15 at 05:42
@nulliusinverba Yes, if that were the case, then it would have only one real solution. However, the one above has three. — Ben Sheller, Dec 16 '15 at 05:45

score 1 · Accepted Answer · answered Dec 16 '15 at 06:03

The factor $f_2(x) = x^2-x-2$ grows to $+\infty$. And it possesses potentially negative terms $-x$ and $-2$. So it might take negative values as well, for instance $f_2(0) = -2$. Since this is a continuous function with positive and negative values, it should have roots too.

This is a hint your discriminant is not correct: $\Delta = (-1)^2-4(1\times -2)=9$. From that you can get two roots, $-1$ and $2$, the latter being a double root of the original cubic equation.

For polynomial equations of order above $2$, if a quadratic factor has no solution, does it only have one solution?

1 Answers1