Given that (x-2) and (x+3) are factors of 2x^4-ax^3-10x^2+bx-54, find a and b.

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.

2x^4-ax^3-10x^2+bx-54=(x-2)(x+3)(cx^2+dx+e)

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

-10x^2=(e+d-6c)x^2

-10=e+d-6c

e+d=-10+12

e+d=2,
where e=8, so d=2-8,
d=-6

bx=(e-6d)x

b=e-6d

-6e=-54

e=8

d+2=-a,
a=-2-d, where
d=-6

a=-2-(-6)

a=6-2

a=4

b=e-6d,
where d=-6 and
e=8

b=8-6(-6)

b=8+36

b=44