# [SOLVED] Rational Roots and Factoring

• Jan 19th 2010, 09:52 PM
Noxide
[SOLVED] Rational Roots and Factoring
This is my polynomial:

5x^5 - 6x^4 - 24x^3 + 20x^2 + 7x - 2 = 0

I found the rational roots and from that found some factors:

(x-1)(5x-1)(x +2)

I don't know how to find the rest of the solutions...
Please don't just give me an answer, tell me what you did to get from the factors found from rational roots to the solutions of the equation.

Thanks
• Jan 19th 2010, 10:22 PM
Jhevon
Quote:

Originally Posted by Noxide
This is my polynomial:

5x^5 - 6x^4 - 24x^3 + 20x^2 + 7x - 2 = 0

I found the rational roots and from that found some factors:

(x-1)(5x-1)(x +2)

I don't know how to find the rest of the solutions...
Please don't just give me an answer, tell me what you did to get from the factors found from rational roots to the solutions of the equation.

Thanks

assuming what you have so far is correct, just divide (via long division or synthetic division) the polynomial by each of the factors you found (divide it by one, then divide the answer by the next, then that answer by the last). this will result in a quadratic that you can factor over the reals (if there are real solution(s)) to get the remaining two factors.
• Jan 19th 2010, 10:23 PM
Soroban
Hello, Noxide!

Quote:

Solve: . $f(x)\;=\;5x^5 - 6x^4 - 24x^3 + 20x^2 + 7x - 2 \:=\: 0$

I found some rational roots and from that found some factors:
. . $(x-1)(5x-1)(x +2)$

I don't know how to find the rest of the solutions.

Evidently, you weren't told about "dividing".

You found that $x = 1$ is a root, so $(x-1)$ is a factor.

Use long division (or synthetic division) to find both factors:

. . $f(x) \;=\;(x-1)(5x^4 - x^3 - 25x^2 - 5x + 2)$

Now use the Rational Roots Test on that quartic factor.

You found that $x = -2$ is a root, so $(x+2)$ is a factor.

Divide the quartic: . $5x^4 - x^3 - 26x^2 - 5x + 2 \;=\;(x+2)(5x^3-11x^2 - 3x + 1)$

Hence: . $f(x) \;=\;(x-1)(x+2)(5x^3 - 11x^2 - 3x + 1)$

Now use the Rational Roots Test on that cubic.

You found that $\tfrac{1}{5}$ is a root, so $(5x-1)$ is a factor.

Divide the cubic: . $5x^3 - 11x^2 - 3x + 1 \;=\;(5x-1)(x^2-2x-1)$

Hence: . $f(x) \;=\;(x-1)(x+2)(5x-1)(x^2-2x-1)$

. . and we get: . $x \:=\:1 \pm\sqrt{2}$
Therefore, the five roots are: . . $1,\;\;-2,\;\;\tfrac{1}{5},\;\;1 + \sqrt{2},\;\;1 - \sqrt{2}$