We have to find the solutions of 2x^4 - 11x^3 + 23x^2 - 19x + 5 = 0.

2x^4 - 11x^3 + 23x^2 - 19x + 5 = 0

=> 2x^4 - 11x^3 + 9x^2 + 14x^2 - 19x + 5 = 0

=> 2x^4 - 9x^3 - 2x^3 + 9x^2 + 14x^2 - 14x - 5x + 5 = 0

=> x^3(2x - 9) - x^2( 2x - 9) + 14x(x - 1) - 5( x - 1) = 0

=> (x^3 - x^2)(2x - 9) + (14x - 5) ( x - 1) =0

=> x^2 ( x-1)(2x - 9) + (x-1)(14x - 5) = 0

=> (x - 1)[ x^2(2x - 9) + 14x - 5] =0

=> (x - 1)[ 2x^3 - 9x^2 + 14x - 5] = 0

This gives x = 1, x = 0.5, x = 2 + i, x = 2 - i

**So the solutions of x are 1, 0.5, 2+ i and 2- i.**

We notice that **x = 1** is the one of the 4 roots of the equation. That means that substituted in the equation, it cancels the equation.

2*1^4 - 11*1^3+ 23*1^2 - 19*1 + 5 = 0

2 - 11 + 23 - 19 + 5 = 0

-9 + 4 + 5 = 0

0 = 0

Using Horner's table, we'll get the quotient:

1*2 - 11 = -9

1*(-9) + 23 = 14

1*14 - 19 = -5

1*(-5) + 5 = 0

The quotient is:

2x^3 - 9x^2 + 14x - 5

The equation could be written as:

2x^4- 11x^3 + 23x^2 - 19x + 5 = (x-1)(2x^3 - 9x^2 + 14x - 5)