Given that (x-2) and (x+3) are factors of 2x4-ax3-10x2+bx-54, find a and b.HELP!

The given expression is:

2x^4
- ax^3 -10x^2 + bx - 54

As (x - 2) is a factor of the given
expression:

For x = 2 the value of factor (x - 2) = (2 - 2)
= 0

As one factor is 0 the value of the complete expression
is also zero. Therefor substituting value x = 2 in the given expression and equating it
to 0 we get:

2(2^4) - a(2^3) -10(2^2) + b*2 - 54
=54

32 - 8a - 40 + 2b - 54 =
0

- 8a + 2b = 62   ...  
(1)

Similarly:

As (x + 3) is a
factor of the given expression:

For x = -3 the value of
factor (x + 3) = (- 3 + 3) = 0

As one factor is 0 the value
of the complete expression is also 0. Therefore substituting value x = - 3 in the given
expression and equating it to 0 we get:

2(3^4) - a(3^3)
-10(3^2) + b*(-3) - 54 =54

162 + 27a - 90 - 3b - 54 =
0

27a - 3b = - 18

9a - b = -
6   ...   (2)

Multiplying equation 2 by 2 we
get:

a + 2b = 62

18a - 2b = -
12   ...   (3)

Adding equations (1) and (3) we
get:

- 8a + 18a + 2b - 2b = 62 -
12

10a =
50

Therefore:

a = 50/10 =
5

Substituting this value of a in equation (2) we
get:

9*5 - b = - 6

45 - b = -
6

b = - 6 - 45 = -
51

Answer:

a = 5, and  b = -
51