# Solve the equation:Find all solutions of the equation 2x4-11x3+23x2-19x+5 = 0

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The equation to be solved is 2x^4 - 11x^3 + 23x^2 - 19x + 5 = 0

We can see that for x = 1, 2 - 11 + 23 - 19 + 5 = 0

This gives us one solution as 1.

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

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

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

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

=> (x - 1)(2x^3 - x^2 - 8x^2 + 4x + 10x - 5) = 0

=> (x - 1)(2x^2(x - 0.5) - 8x( x - 0.5) + 10(x - 0.5)) = 0

=> (x - 1)(x - 0.5)(2x^2 - 8x + 10) = 0

The roots of 2x^2 - 8x + 10 = 0 are 2 + i and 2 - i

**All the solutions of the equation 2x^4 - 11x^3 + 23x^2 - 19x + 5 = 0 are (1 , 0.5 , 2 + i, 2 - i)**

Since **x = 1** is one of the 4 roots of the equation, then 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

**According to reminder theorem, we'll get: 2x^4- 11x^3 + 23x^2 - 19x + 5 = (x-1)(2x^3 - 9x^2 + 14x - 5)**