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

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 5:34
• Yes, I see. Thanks, Ben. – Hanshan Dec 16 '15 at 5:37
• No problem. That should also change your answer as far as the number of real solutions goes. – Ben Sheller Dec 16 '15 at 5:39
• If it's a cubic equation like yours, then only one real solution. – SchrodingersCat Dec 16 '15 at 5:42
• @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 5:45

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.