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I just did a national exam and this question was in it; I am convinced this does not work:

Given that $(x - 1)$ is a factor of $x^3 + 3x^2 + x - 5$, factorize this cubic fully.

My attempt

1 | 1  3  1 -5
  |    1  4  5
    1  4  5  0

$$(x - 1)(x^2 + 4x + 5) = 0$$

That's all I got. I gave up and couldn't factorize further. Am I missing something?

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up vote 13 down vote accepted

The discriminant of your polynomial is $$\Delta=4^2-4\cdot 5<0$$ so this has no real roots, that is, factorization over $\Bbb R$ is finished. On the other hand, factorization over $\Bbb C$ is possible, yielding


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You've gotten as far as you can in factoring:

$$x^3 + 3x^2 + x - 5 = (x - 1)(x^2 + 4x + 5)$$

When the discriminant is less than zero, as in the case of the quadratic $x^2 + 4x + 5$, there are no real roots. Recall, the $\color{blue}{\bf \text{discriminant}}$ is the portion of the quadratic formula inside the square root $$\sqrt{\Delta} = \sqrt{\color{blue}{b^2 - 4ac}} $$

In your case, we have $a = 1, \; b = 4, \; c = 5,$ so $\Delta = b^2 - 4ac = 16 - 20 = -4 \lt 0$.

Whenever you need to factor a quadratic over the reals, take the time to check the value of the discriminant $\Delta$. If it evaluates to less than 0, don't waste time trying to factor it, especially when you are taking an exam and time is limited!

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Perhaps worth it to add that this is true when factoring over the reals. For complex numbers, it's a different story... – gt6989b May 22 '13 at 19:12
@gt6989b agreed! – amWhy May 22 '13 at 19:22
@amWhy: deserves another thumbs up! +1 – Amzoti May 23 '13 at 1:24

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