<h3>
Answer: 7x - 5</h3>
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Explanation:
We're dividing some unknown polynomial P(x) over (x+2) to get a quotient Q1(x) and remainder -19. That must mean
P(x)/(x+2) = Q1(x) - 19/(x+2)
P(x) = Q1(x)*(x+2) - 19
Note that plugging in x = -2 leads to P(-2) = -19. This is an example of the remainder theorem.
Similarly,
P(x) = Q2(x)*(x-1) + 2
since dividing by (x-1) leads to a remainder of 2. We have P(1) = 2.
We'll use P(-2) = -19 and P(1) = 2 in the later sections below.
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We want to find the value of r such that
P(x) = Q3*(x+2)(x-1) + r
Let's say that r = ax+b
Meaning we now have
P(x) = Q3*(x+2)(x-1) + r
P(x) = Q3*(x+2)(x-1) + ax+b
The goal from here is to find 'a' and b so we can get the remainder ax+b.
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Plug in x = -2 and solve for b.
P(x) = Q3*(x+2)(x-1) + ax+b
P(-2) = Q3*(-2+2)(-2-1) + a(-2)+b
-19 = Q3*(0)(-3)-2a+b
-19 = -2a+b
b = 2a-19
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Plug in x = 1 to get...
P(x) = Q3*(x+2)(x-1) + ax+b
P(1) = Q3*(1+2)(1-1) + a(1)+b
P(1) = a+b
2 = a+b
a+b = 2
a+(2a-19) = 2 .... plug in b = 2a-19
3a-19 = 2
3a = 19+2
3a = 21
a = 21/3
a = 7
Use that value to find b
b = 2a-19
b = 2(7)-19
b = 14-19
b = -5
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We found that a = 7 and b = -5.
The remainder ax+b updates to the final answer 7x-5
