## anonymous 5 years ago x^3-34x-12 find the solution set and an actual root

1. anonymous

There's a theorem called The Rational Root Theorem that says, if a polynomial is to have rational roots, then they will be of the form,$x=\frac{p}{q}$where p and q are co-prime (i.e., p and q have no common factors other than 1), and p is a plus/minus factor of the constant (here, -12) and q is a plus/minus factor of the coefficient of the highest powered x (here, +1). So here, the +/- factors of 12 are$\pm \left\{ 1,2,3,4,6,12 \right\}$and the +/- factor of 1 is, funnily enough,$\pm \left\{ 1 \right\}$So if your polynomial has rational roots, they will be from the set of all combinations of $\frac{\pm 1,2,3,4,6,12}{\pm 1}$If you try +6, you'll find your polynomial ends up being zero. This means$(x-6)$is a factor of your cubic.

2. anonymous

Now that you have one factor of your cubic, you can use synthetic or polynomial division to find the quadratic, Q(x), where$x^3-34x-12=(x-6)Q(x)$That Q(x) is found to be,$Q(x)=\frac{x^3-34x-12}{x-6}=x^2+6x+2$That is,$x^3-24x-12=(x-6)(x^2+6x+2)$

3. anonymous

You can find the solutions of the quadratic using the quadratic formula or by completing the square. Using the quadratic formula gives$x=\frac{-6 \pm \sqrt{6^2-4.1.2}}{2.1}=\frac{-6\pm \sqrt{28}}{2}=-3 \pm \sqrt{7}$The solution set is therefore,$\left\{ 6, -3+\sqrt{7},-3-\sqrt{7} \right\}$