if f(x)= ax² + bx + c is divided by x+2, x-1, and x-3, it leaves a remainder of 64, 1, and 22, respectively.
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The remainder theorem tells us that if we divide a polynomial f(x) by
x-a, the remainder will be f(a).
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Therefore:
The remainder theorem tells us that if we divide f(x) = ax² + bx + c by
x+2, the remainder will be f(-2), which is a(-2)²+b(-2)+c or 4a-2b+c and
since that's given to be 64 we have the equation:
4a-2b+c = 64
The remainder theorem tells us that if we divide f(x) = ax² + bx + c by
x-1, the remainder will be f(1), which is a(1)²+b(1)+c or a+b+c and
since that's given to be 1 we have the equation:
a+b+c = 1
The remainder theorem tells us that if we divide f(x) = ax² + bx + c by
x-3, the remainder will be f(3), which is a(3)²+b(3)+c or 9a+3b+c and
since that's given to be 22 we have the equation:
9a+3b+c = 22
So we have the system of three equations:
4a-2b+c = 64
a+ b+c = 1
9a+3b+c = 22
Solve that system and get a=, b= and c=
Those are terrible fractions, but they are correct.
Edwin