Question 639899

if f(x)= ax˛ + 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) = ax˛ + bx + c by
x+2, the remainder will be f(-2), which is a(-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) = ax˛ + bx + c by
x-1, the remainder will be f(1), which is a(1)˛+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) = ax˛ + bx + c by
x-3, the remainder will be f(3), which is a(3)˛+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={{{63/10}}}, b={{{-147/10}}} and c={{{47/5}}}

Those are terrible fractions, but they are correct.

Edwin</pre>