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How would I factor $-x^3+x^2+12$? Grouping doesn't work and I tried synthetic division but none of the possible zeros worked for me. I had $\pm1,\pm2, \pm4, \pm6, \pm12 $. What am I missing?

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So probably you need to know its real/complex roots to be able to factor, can't do much without that. I don't see any other way to approach this. – Mark Fantini Nov 23 '13 at 22:59
WolframAlpha gives approximately $-(x-2.67589) (x+(0.837944-1.94482 i)) (x+(0.837944+1.94482 i))$. – MasterOfBinary Nov 23 '13 at 23:00
$x$ is not an integer, one real root is $x = 2.67588866957270$, the other two are $x = -0.83794433478635 \pm 1.94482410365933 i$ – K. Rmth Nov 23 '13 at 23:01
We get questions like these all the time. Not every polynomial is factor-able into integers. – Sekots Nov 23 '13 at 23:06
I knew that because I couldn't solve it that it must have been unfactorable. I figured it would be a complex number. So basically theres no way of solving this by hand right? – Ruth Gutierrez Nov 23 '13 at 23:10

You cannot factor this to find the roots. One way to solve the cubic is to first convert it into a depressed cubic (without the $x^2$ term). To do this, we have to make the substitution $x=y-\dfrac{b}{3a}$. $$-x^3+x^2+12=0$$ $$-(x^3-x^2-12)=0$$ $$=-\left[\left(y-\left(\dfrac{1}{3}\right)^3\right) - \left(y-\left(\dfrac{1}{3}\right)^2\right) + 12\right]=0$$ Expanding the expression and simplifying gives us: $$-\left(y^3 + \dfrac{1}{3}y + \dfrac{326}{27}\right)=0$$ $$y^3+\dfrac{1}{3}y + \dfrac{326}{27}=0$$ $$y^3+\dfrac{1}{3}y = -\dfrac{326}{27}$$ The depressed cubic is now in the form $y^3 + Ay = B$. To solve this we have to assume two variables, $s$ and $t$, with the condition that: $$3st = A$$ $$s^3 - t^3 = B$$ $y = s-t$ will be a root of the cubic. This is proved by substituting $(s-t)$ into $y$, $3st$ into $A$, and $s^3-t^3$ into $B$. You can do this on your own time; I will not write the proof here because I am lazy.

Let's derive a formula to convert our cubic into a quadratic. We first solve $3st = B$ for $s$, and substitute it into $s^3-t^3=B$ $$s = \dfrac{A}{3t}$$ $$\left(\dfrac{A}{3t}\right)^3 - t^3 = B$$ $$\dfrac{A^3}{27t^3} - B - t^3 = 0$$ Multiply by $-t^3$ on both sides. $$-\dfrac{A^3}{27} + Bt^3 + t^3\cdot t^3 = 0$$ $$t^6 + Bt^3 - \dfrac{A^3}{27}$$ Finally, we make the substitution $u=t^3$ to obtain: $$u^2 + Bu - \dfrac{A^3}{27} = 0$$ Using our formula, we get: $$u^2 - \dfrac{326}{27}u - \dfrac{\frac{1}{9}}{27} = 0$$ Multiply by 27 on both sides. $$27u^2 - 326u - \dfrac{1}{9} = 0$$ Multiply by 9 on both sides. $$243u^2 - 2934u - 1 = 0$$ I can tell that the solution is not going to be pretty. Let's find the solution. $$u = \dfrac{2934 \pm \sqrt{2934^2 - 4(243)(-1)}}{2\times 243}$$ $$u = \dfrac{2934 \pm \sqrt{8609328}}{486}$$ $$u = \dfrac{2934 \pm 4\sqrt{556833}}{486}$$ $$u = \dfrac{1467 \pm 2\sqrt{556833}}{243}$$ Remember $u=t^3$. Also discard the negative square root. $$t^3 = \dfrac{1467 + 2\sqrt{556833}}{243}$$ $$t = \sqrt[3]{\dfrac{1467 + 2\sqrt{556833}}{243}}$$ Substitute the value of t^3 into the equation $s^3 + t^3 = B$ to obtain $s$. $$s^3 + t^3 = -\dfrac{326}{27}$$ $$s^3 = -\dfrac{326}{27} - t^3$$ $$s^3 = -\dfrac{326}{27} - \dfrac{1467 + 2\sqrt{556833}}{243}$$ $$s = \sqrt[3]{-\dfrac{326}{27} - \dfrac{1467 + 2\sqrt{556833}}{243}}$$ $$s = \sqrt[3]{-\dfrac{326 + 1467 + 2\sqrt{56833}}{243}}$$ $$s = \sqrt[3]{-\dfrac{1793 + 2\sqrt{56833}}{243}}$$ Remember $y = s-t$: $$y = s-t$$ $$y= \sqrt[3]{-\dfrac{1793 + 2\sqrt{56833}}{243}} - \sqrt[3]{\dfrac{1467 + 2\sqrt{556833}}{243}}$$ Finally, remember $x = y - \dfrac{b}{3a}$: $$x = y - \dfrac{b}{3a}$$ $$x = \sqrt[3]{-\dfrac{1793 + 2\sqrt{556833}}{243}} - \sqrt[3]{\dfrac{1467 + 2\sqrt{556833}}{243}} - \dfrac{1}{3}$$ The solution to the cubic equation $-x^3 + x^2 + 12$ is (hopefully): $$\boxed{x = \sqrt[3]{-\dfrac{1793 + 2\sqrt{556833}}{243}} - \sqrt[3]{\dfrac{1467 + 2\sqrt{556833}}{243}} - \dfrac{1}{3}}$$ By the way, the mathematician Scipione del Ferro was the first person to invent a "cubic formula" using the method described above. In case you were wondering, the cubic formula is what it sounds like: a formula that can be used to solve cubic equations.

The roots of any cubic equation $ax^3 + bx^2 + cx + d = 0$ is: $$x = \sqrt[3]{\left(\dfrac{-b^3}{27a^3} + \dfrac{bc}{6a^2} - \dfrac{d}{2a}\right) + \sqrt{\left(\dfrac{-b^3}{27a^3} + \dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)^2 - \left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}\right)^3}} + \sqrt[3]{\left(\dfrac{-b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a}\right) + \sqrt{\left(\dfrac{-b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)^2 - \left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}\right)^3}} - \dfrac{b}{3a}$$ I really, really do not think that you should memorize this formula because even if you do, it will be very, very inefficient.

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The polynomial is irreducible over the rationals. Why do you want to factor this equation? Can you post the problem in which you believe you have to factor this equation?

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