Saturday

April 18, 2015

April 18, 2015

Posted by **Jessica** on Thursday, January 27, 2011 at 5:07pm.

Given that one root is 2+i, solve the equation.

So far I have it set up like this:

2X^4-9X^3+3X^2-X-5/X^2-4X+5

I don't know how to get the answer, I know that I have to use long division but when I started to try it I kept getting weird answers. Please Help

- algebra 2 -
**Reiny**, Thursday, January 27, 2011 at 5:48pmI think there is something wrong here,

either you have a typo in the opening equation, or the question is faulty.

If one root is 2+i, then 2-i must be another root

You correctly determined that

x^2 - 4x + 5 must then be a factor

When I divided that into the original, I got an answer of 2x^2 - x -11 with a remainder of -40x + 50

- algebra 2 -
**Jessica**, Thursday, January 27, 2011 at 5:55pmmy bad the original equation is:

2X^4 - 9X^3 + 13X^2 - X - 5 = 0

- algebra 2 -
**Reiny**, Thursday, January 27, 2011 at 6:02pmOk, this time my division was exact, as expected,

and I got 2x^2 - x - 1, which factors to

(2x + 1)(x - 1)

so the roots are

2+i, 2-i, -1/2, and 1

- algebra 2 -
**Jessica**, Thursday, January 27, 2011 at 6:12pmoh I get it thanks

**Answer this Question**

**Related Questions**

algebra 2 - 2X^4-9X^3+3X^2-X-5=0 Given that one root is 2+i, solve the equation...

algebra 2 - 2X^4-9X^3+13X^2-X-5=0 Given that one root is 2+i, solve the equation...

Algebra - Solve 2+ the square root of 4-x=x what is your question? ok lets start...

Algebra - I have some problems that I am having trouble with and I have some ...

college algebra - Solve: sq root of (x+1) minus 3 = sq root of (x+4) when I ...

math/algebra - I have a triangle one side = 3 side 2 = square root5 - square ...

Math - The graph of the equation y=4-x consists of all the points in the ...

MATH/ALGEBRA - SOLVE: (4x+ 5) + 1 = 0 (4x+ 5)= -1 ((4x+ 5))squared= (-1)...

College Algebra--Still Confused - I have a few problems I need help with and ...

Geometry - how do you solve using midpoint equation if you only have one pair of...