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How do you factor $x^3-2x-4$?

To me, this polynomial seems unfactorable.

But I check my textbook answer key, and the answer is $(x-2)(x^2+2x+2)$.

So I got to solve backward:

$$x(x^2-2)-4$$ And add some terms and subtract them later, $$x(x^2+2x+2-2x-4)-4=0$$ $$x(x^2+2x+2)-4-2x^2-4x=0$$ $$x(x^2+2x+2)-2(x^2+2x+2)=0$$

But I would have thought of this way to factor if I didn't look at the answer key.

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  • $\begingroup$ Did you check for rational roots? $\endgroup$ Dec 15, 2017 at 1:13

6 Answers 6

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Rational root theorem might be helpful, try to check whether factors of $-4$ satisfy the equation.

By doing so, you would have encountered that $2$ is a root of the polynomial.

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Hint: if a cubic (degree 3) polynomial has a factor, it must have a linear (degree 1) factor, say $x - c$ where $c$ is a root of the polynomial. If you sketch the graph of $x^3 -2x -4$, you will see that it must have a root near $x = 2$. You can now test whether $x=2$ is a root and ...

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It is better to go with Rational root theorem as suggested. I just want to write another way to see but clearly will not always work out that nicely. \begin{equation} x(x^2-2)-4 \end{equation} instead of $x^2-2$, write $x^2-4$(because it is $(x-2)(x+2)$) and see if it is possible to factor: \begin{equation} x(x^2-4) + 2x -4 = x(x-2)(x+2) +2(x-2) = (x-2)(x^2+2x+2) \end{equation} There is a guess work here, but for example if you start with $x(x^2-2)-1$, you might wanna consider $x^2-1$ instead of $x^2-4$.

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The way I factor

$x^3 - 2x - 4 \tag 1$

is to notice that $2$ is a root; this I do by intuition and a good amount of experience. $2$ looks like a good guess to me since the non-leading coefficients are divisible by $2$; once I find that $2$ is a zero of (1), I use synthetic division to find $q(x)$ such that

$x^3 - 2x - 4 = (x - 2)q(x), \tag 2$

thus: $x - 2$ into $x^3$ yields $x^2$;

$x^3 - 2x - 4 - x^2(x - 2) = 2x^2 - 2x - 4; \tag 3$

$x -2$ in to $2x^2$ yields $2x$:

$2x^2 - 2x - 4 - 2x(x - 2) = 2x - 4; \tag 4$

$x - 2$ into $2x - 4$ yields $2$, and since

$2x - 4 - 2(x - 2) = 0, \tag 5$

we are done. Gathering the partial quotients yields $x^2 + 2x + 2$, and it is easily checked that

$x^3 -2x -4 = (x - 2)(x^2 + 2x + 2); \tag 6$

now if we want to go further we can use the quadratic formula to find the roots $\mu_\pm$ of $x^2 + 2x + 2$:

$\mu_\pm = \dfrac{-2 \pm \sqrt{-4}}{2} = \dfrac{-2 \pm 2i}{2} = -1 \pm i; \tag 7$

we easily check

$x^2 + 2x + 2 = (x + (1 + i))(x + (1 - i)); \tag 8$

therefore

$x^3 - 2x - 4 = (x - 2)(x + (1 + i))(x + (1 - i)). \tag 9$

And that's the way I factor it!

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You could go for polinomial division. You know that $2$ is an root, so, use the Ruffini's rule: $$ \begin{matrix} 2 \mid& 1&0&-2&-4\\ \hline \,\,\,\mid & 1 &2&2 & 0\end{matrix} \\ \therefore x^3-2x-4=(x-2)(x^2+2x+2)$$

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hint: $x-2$ is a factor. Can you finish it? (didn't look at your post)

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