Answer:
When a polynomial is divided by a linear factor, the remainder is equal to the value of the polynomial at the root of the factor. Therefore, we have:
y - 2 = 0 => y = 2
y + 3 = 0 => y = -3
The remainder when the polynomial is divided by y - 2 is 9, so we have:
(2)^3 + a(2)^2 + b(2) + 1 = 9
8 + 4a + 2b + 1 = 9
4a + 2b = 0
2a + b = 0
Similarly, the remainder when the polynomial is divided by y + 3 is 19, so we have:
(-3)^3 + a(-3)^2 + b(-3) + 1 = 19
-27 + 9a - 3b + 1 = 19
9a - 3b = 45
3a - b = 15
We now have two equations with two variables:
2a + b = 0
3a - b = 15
Solving for b in the first equation and substituting into the second equation, we get:
b = -2a
3a - (-2a) = 15
5a = 15
a = 3
Substituting a = 3 into either of the equations above, we get:
2a + b = 0
2(3) + b = 0
b = -6
Therefore, the values of a and b are a = 3 and b = -6.