Good evening Mirabelle!

When a polynomial is divided by a divisor, the degree of the resulting remainder is at least 1 lower than the degree of the divisor (or even lower). You would probably have seen this when you did manual long division.

So, if our divisor is x² + x - 2 (and is therefore of degree 2), our remainder would have a degree of up to 1, so we let our remainder take the form Ax + B where A and B are constants to be found.

Let P(x) = D(x) Q(x) + R(x)

where P(x) is our original polynomial (which we will never be able to know), D(x) is our divisor x² + x - 2, Q(x) is our quotient polynomial (which we will never be able to know) and R(x) is our remainder polynomial which is Ax + B.

P(x) = (x² + x - 2) Q(x) + Ax + B
P(x) = (x + 2) (x - 1) Q(x) + Ax + B

P(x) leaves a remainder of 80 when divided by x + 2 and a remainder of -10 when divided by x - 1, so by the Remainder Theorem,

P(-2) = 80 and P(1) = -10

P(-2) = 80
0 + A (-2) + B = 80
-2A + B = 80
B = 80 + 2A

P(1) = -10
0 + A (1) + B = -10
A + B = -10

A + 80 + 2A = -10
3A + 80 = -10
3A = -90
A = -30

B = 80 + 2 (-30)
B = 20

P(x) = (x² + x - 2) Q(x) - 30x + 20

so the remainder when P(x) is divided by x² + x - 2 is -30x + 20.