To solve this:
Using long division divide the polynomial by (x - 2) and (x - 3).
The result of the final subtraction (which will be some expression involving p and q) will be the same as the remainders.
When the divisor is a factor, the remainder is 0.
Using the results of the final two subtractions gives two simultaneous equations in p and q which can then be solved.

Hint on how to do the long division:

As you are dividing by x - 2 and x - 3, at each stage find by what you need to multiply the x which when subtracted will remove the highest power of x remaining. Multiply the whole x-2 or the x-3 by this multiplier and subtract; each subtraction will involve x to some power and x to one less than that power.

Now you know the method, have a go before reading the solution below:

---------------------------------------------------------------------------------------------------------

Divide x³ + px² + qx + 6 by (x - 2) which is a factor

________________x²_+___(p+2)x_+___(2p+q+4)
______------------------------------------------------
(x-2)_|_x³_+__px²_+_______qx_+___________6
________x³_+_-2x²
________------------
__________(p+2)x²_+_______qx
__________(p+2)x²_+_-2(p+2)x
__________-------------------------
___________________(2p+q+4)x_+___________6
___________________(2p+q+4)x_+_-2(2p+q+4)
___________________--------------------------------
__________________________________4p+2q+14

As (x - 2) is a factor, this final subtraction must result in 0

→ 4p + 2q + 14 = 0
→ 2p + q + 7 = 0
→ 2p + q = -7

Divide x³ + px² + qx + 6 by (x - 3) with remainder 3:

________________x²_+___(p+3)x_+___(3p+q+9)
______------------------------------------------------
(x-3)_|_x³_+__px²_+_______qx_+___________6
_________x³_+_-3x²
_________-----------
___________(p+3)x²_+_______qx
___________(p+3)x²_+_-3(p+3)x
__________-------------------------
____________________(3p+q+9)x_+___________6
____________________(3p+q+9)x_+_-3(3p+q+9)
____________________--------------------------------
___________________________________9p+3q+33

As dividing by (x - 3) leaves a remainder of 3
→ 9p + 3q + 33 = 3
→ 3p + q + 11 = 1
→ 3p + q = -10

There are now two simultaneous equations in p and q which can be solved:

2p + q = -7
3p + q = -10

(2) - (1) gives:
3p - 2p + q - q = -10 - (-1)
→ p = -3
Substituting in (a) gives:
2×-3 + q = -7
→ q = -1

→ p = -3
& q = -1

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