Since x-2 and x+3 are factors of f(x) = 2x^4-ax^3-10x^2+bx-54, we can write
2x^4-ax^3-10x^2+bx-54 = (x-2)(x+3)Q(x)......(1).
Putting x=2 in (1) we get: 2*2^4-a(2)^3-10(2)^2+b(2)-54 = (2-2)(2+2)Q(2) =0. Or
32-8a-40+2b-54 =0. Or
Put x=-3. Then
2(-3)^4-a(-3)^3-10(-3)^2+b(-3)-54 = (-2-2)(-2+2)Q(-2) =0. Or
162+27a-90-3b-54 =0. Or
27a-3b= -18. Or dividing by 3,
9a-b = -6........(3).
Adding eq(2) and eq(3) we get: -4a+9a = 31-6 =25. Or
5a=25. So, a = 5.
Substituting in eq (3), we get: 9*5-b = -6. Or
b = 9*5+6 = 51.
We'll apply the division of polynomials. Considering the fact that x-2 and x+3 are factors, that means that 2 and -3 are the roots of the polynomial 2x^4-ax^3-10x^2+bx-54. We know that we could write a polynomial as a product of linear factors, depending on it's roots.
We've noticed that multiplying (x-2)(x+3), we'll obtain a second degree polynomial. But the given polynomial has the fourth degree, so we have to multiply (x-2)(x+3) with another polynomial of second degree.
We'll do the math, to the right side and the result will be:
(x-2)(x+3) = x^2+x-6
(x-2)(x+3)(cx^2+dx+e) = (x^2+x-6)(cx^2+dx+e)
(x^2+x-6)(cx^2+dx+e) = cx^4+dx^3+ex^2+cx^3+dx^2+ex-6cx^2-6dx-6e
If 2 polynomials are identically, that means that the corresponding coefficients are equal.
2x^4= cx^4, so c=2
-ax^3=(d+c)x^3, so d+c=-a, where c=2, d+2=-a
e+d=2, where e=8, so d=2-8, d=-6
d+2=-a, a=-2-d, where d=-6
b=e-6d, where d=-6 and e=8