Answer:

Step-by-step explanation:
Here, we want to find the value of m and n
If we substituted a supposed root into the parent polynomial, the value after evaluation is the remainder. If the remainder is zero, then the value substituted is a root.
for x+ 3
x + 3 = 0
x = -3
Substitute this into the first equation as follows:

We do this for the second value as follows:
x-2 = 0
x = 2
Substitute this value into the polynomial:

Now, we have two equations so solve simultaneously:

Add both equations:

To get the value of n, we simply susbstitute the value of m into any of the two equations. Let us use the second one:
