# Solve the equation 2x^4-3x^3-9x^2+15x-5=0.

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### 2 Answers

We have to solve the equation: 2x^4-3x^3-9x^2+15x-5=0

2x^4-3x^3-9x^2+15x-5=0

=> 2x^4 - 2x^3 - x^3 + x^2 - 10x^2 + 10x + 5x - 5 = 0

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

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

=> (x^2( 2x - 1) - 5(2x - 1))(x - 1) = 0

=> (x^2 - 5)(2x - 1)(x - 1) = 0

x - 1 = 0

=> x = 1

2x - 1 = 0

=> x = 1/2

x^2 - 5 = 0

=> x^2 = 5

=> x = sqrt 5 and -sqrt 5

**The solution of the equation are {1/2, 1, sqrt 5, -sqrt 5}**

how about the rational roots of the polynomial equation?