I have prob getting other two factors.
Had solved that (x-1) is a factor of 2X^3+5X^2-11X+4.
2X^3+5X^2-11X+4
=2(1)^3+5(1)^2-11(1)+4
=0
You appear to be assuming that the other factor is $\displaystyle 2x^2+ x- 4$. Why?
If you insist upon that method (though ordinary division or synthetic division is simpler) you need to write $\displaystyle 2x^3+ 5x^3- 11x+ 4= (x-1)(ax^2+ bx+ c)$. Do the multiplication and combine like powers and you will get three equations for a, b, and c. (Actually, you get 4 equations but they are not "independent" since x-1 is a factor of $\displaystyle 2x^3+ 5x- 11x+ 4$.
Let $\displaystyle q(x)=Ax^2+Bx+C$ (it's a method I prefer)
$\displaystyle (x-1)(Ax^2+Bx+C) = 2x^3+5x^2-11x+4$
Compare coefficients
- $\displaystyle x^3 \: \rightarrow \: 2 = A$
- $\displaystyle x^2 \: \rightarrow \: 5 = -A+B \: \: \therefore \: B = 7$
- $\displaystyle x^0 \: \rightarrow \: 4 = -C $
$\displaystyle (x-1)(2x^2+7x-4) = 2x^3+5x^2-11x+4$
$\displaystyle \Delta = 7^2-(4 \times 2 \times -4) = 81$. As this is a perfect square the quadratic has rational solutions and can be factorised.
$\displaystyle 2x^2+7x-4 = (2x-1)(x+4)$
Therefore $\displaystyle 2x^3+5x^2-11x+4 = (x-1)(x+4)(2x-1)$