Posted by **Nick** on Thursday, April 10, 2014 at 11:31am.

5x^4-41x^3-87x^2-31x+10=0

- Math -
**Steve**, Thursday, April 10, 2014 at 11:39am
I assume you want values of x which solve the equation.

Well, you know the rational roots will be of the form 1,2,5,10 or 1/5,2/5

A little synthetic division shows that x+1 divides twice, so you have

(x+1)(x+1)(5x^2-51x+10)

Now you can easily see that it is

(x+1)(x+1)(x-10)(5x-1) = 0

## Answer This Question

## Related Questions

- chemistry - For the equation, SO2(g) + 1/2 O2(g)-> SO3(g) Kp=9.5 EQUAL ...
- Precalculus(URGENT) - solve the problem 2xcubed-17xsquared+31x+20 given that 5 ...
- MATH 101 - Factor: 15x2(2x + 3) + 41x(2x + 3) + 28(2x + 3)
- algebra - (36x^3-60x^2-87x-21)divided by 9x+3
- Math - Find a polynomial of degree 3 with roots 3, 4 - i. Which one is it? A.x3...
- Math - X^x-31x+108=0
- Cara-Chem - I made a bobo, which may lead to some confusion. Here is my post ...
- Algebra 2 - I also need help with this two. 1.0.5=41x-2(1.3x-4) 2. 8.7=3.5m-2.5(...
- algerbra 2 - what are the other roots of 3x^4-12x^3+41x-8x+26 that include 2-3i
- Math - X^4+9x^3+31x^2+49x+3=0 find all solutions.

More Related Questions