Answer:
Step-by-step explanation:
We want to find the constants 'a', 'p', and 'q' such that ...
(x -a)(x -1)(x +2) ≡ x³ +px +q
(x -a)(x² +x -2) ≡ x³ +px +q
x³ +x² -2x -ax² -ax +2a ≡ x³ +px +q
This gives rise to three equations:
1 -a = 0 . . . . coefficient of x²
-2 -a = p . . . coefficient of x
2a = q . . . . . constant
The first of these tells us a=1. Then, p=-3 and q=2.
The values of p and q are -3 and 2, respectively.
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The remainder from division by (x+1) is the value of the cubic for x=-1. That value is ...
(-1)³ -3(-1) +2 = -1 +3 +2 = 4
The remainder from division by (x+1) is 4.
