Question

How do I solve for a, b and c?

Answer 1

Firstly expand (x - 3)(ax^2 + bx + c):
ax^3 + bx^2 + cx - 3ax^2 - 3bx - 3c

Now rearrange it so that it's in the form ax^3 + bx^2 + cx + d:
ax^3 + (b-3a)x^2 + (c-3b)x - 3c

Now we can compare both equations:
ax^3 + (b-3a)x^2 + (c-3b)x - 3c = 2x^3 - x^2 - 19x + 12

We get:
(1) a = 2
(2) b - 3a = -1
(3) c - 3b = -19
(4) -3c = 12

If we substitute (1) into (2) we get:
b - 3*2 = -1
b - 6 = -1
b = 5

Now if we solve (4) we get:
-3c = 12
c = -4

Therefor a = 2, b = 5 and c = -4