# Finding roots of the fourth degree polynomial: $2x^4 + 3x^3 - 11x^2 - 9x + 15 = 0$.

My son is taking algebra and I'm a little rusty. Not using a calculator or the internet, how would you find the roots of $2x^4 + 3x^3 - 11x^2 - 9x + 15 = 0$. Please list step by step. Thanks, Brian

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Guessing one root sometimes opens up the whole equation for you.

First notice that $\displaystyle x=1$ gives 0. So $\displaystyle x-1$ is a factor.

Next, rewrite as

$\displaystyle 2x^4 - 2x^3 + 5x^3 - 5x^2 -6x^2 + 6x - 15x + 15$

This is to try and get $x-1$ as a factor.

This gives us

$\displaystyle 2x^3(x-1) + 5x^2(x-1) - 6x(x-1) - 15(x-1) = (x-1)(2x^3 + 5x^2 - 6x - 15)$

Now notice that $\displaystyle 2x^3 - 6x = 2x(x^2-3)$ and $5x^2 - 15 = 5(x^2 - 3)$

Thus

$\displaystyle (x-1)(2x^3 + 5x^2 - 6x - 15) = (x-1)(2x(x^2 - 3) + 5(x^2 - 3)) = (x-1)(x^2-3)(2x+5)$

and so the roots are $\displaystyle 1, \pm\sqrt{3}, -\frac{5}{2}$

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If you're familiar with it, the process of "getting $x-1$ as a factor" can be done as polynomial long division: en.wikipedia.org/wiki/Polynomial_long_division . Personally I find this to be a bit cleaner than just mushing around the polynomial, even though of course they amount to exactly the same thing. –  Aaron Mazel-Gee Dec 3 '10 at 7:50
Note that what Moron did here is a very clever application of Horner's rule for dividing out linear factors. –  J. M. is back. Dec 3 '10 at 9:13
You pretty much start with the rational root test: The roots are p/q where p=factor of 15, not necessarilty positive and q=factor of 2, same rules. Edit: The other user posted this. –  user43400 Dec 10 '13 at 21:19

You can try first finding the rational roots using the rational root theorem in combination with the factor theorem in order to reduce the degree of the polynomial until you get to a quadratic, which can be solved by means of the quadratic formula or by completing the square.

For example, to complement a little bit on Aryabhata's answer, the first solution he found $x = 1$ can be guessed by using the rational root theorem since the theorem tells you to look for rational solutions only in the set of fractions $\frac{\pm a}{b}$ where $a, b \in \mathbb{Z}$ are integers such that $a$ divides $15$ and $b$ divides $2$. Thus you list all divisors of $15$, which are $\pm 1, \pm 3, \pm 5, \pm 15$ and the divisors of $2$ are $\pm 1, \pm 2$. then your list of possible rational roots would be $\pm \frac{1}{1}, \pm \frac{1}{2}, \pm \frac{3}{1}, \pm \frac{3}{2}, \pm \frac{5}{1}, \pm \frac{5}{2}, \pm \frac{15}{1}, \pm \frac{15}{2}$, and you have to start proving to see if any of those is a root of your polynomial.

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