solve the equation 2x^4+x^3-11x^2+x+2=0?

2x^4+x^3-11x^2+x+2=0

using synthetic division

2|....2.....1...- 11...1...2
.............4......10..-2..-2
-------------------------------------
......2......5....-1...-1...0

(2x^3 + 5x^2 - x - 1)(x - 2) = 0

1/2 |....2....5....- 1...-1
................1......3....1
---------------------------------
...........2...6.....2.....0

(2x^2 + 6x + 2)(x - 2)(2x - 1) = 0
x^2 + 3x + 1 = 0
x^2 + 3x = - 1
x^2 + 3x + 9/4 = - 1 + 9/4
(x + 3/2)^2 = 5/4
x + 3/2 = ± √(5/4)
x = - 3/2 ± √(5)/2

x = 1/2[- 3 + √(5)], 1/2[-3 - √(5)], 2, 1/2 answer//

a million. sparkling up for x interior the quadratic equation, ( x - 3 ) ( x - 2 ) = 0? ( x - 3 ) = 0 x=3 ( x - 2 ) = 0 x=2 2. sparkling up for x interior the quadratic equation, ( 3x - 9 ) ( x + 5 ) = 0. ( 3x - 9 ) = 0 3x = 9 x=3 ( x + 5 ) = 0 x=-5

Rational factors must be of the form (x+/-1), (x+/-2), (2x+/-1) so possible roots are
x=+/-1, x=+/-2, x=+/-(1/2). Try x=1 NO, x=-1 NO, x=2 YES so (x-2) is a factor and divide by (x-2)
to give (2x^3+5x^2-x-1) and you try to solve 2x^3+5x^2-x-1=0 and possible rational roots are
x=+/-1, x=+/-(1/2) and you find x=1/2 is a solution so a factor is (2x-1) and dividing (2x^3+5x^2-x-1)
buy (2x-1) gives x^2+3x+1 and you solve this=0 to get the other two solutions.

First you must expand the polynomial factorization. To do this, you can divide this polynomial by x-2.
So 2x^4+x^3-11x^2+x+2=(x-2)(2x^3+5x^2-x-1)=0
x=2 - the first root of this equation
Further we must expand the polynomial factorization for 2x^3+5x^2-x-1. For this you must divide 2x^3+5x^2-x-1 by x-1/2. 2x^3+5x^2-x-1=(x-1/2)(2x^2+6x+2)=0
x=1/2 - the second root of this equation
2x^2+6x+2=0
x^2+3x+1=0
D=9-4=5
x3=(-3-sqrt(5))/2
x4=(-3+sqrt(5))/2
Answer: x1=2, x2=1/2, x3=(-3-sqrt(5))/2, x4=(-3+sqrt(5))/2