Asked by Jack
When P(x) is divided by (x-1) and (x+3), the remainders are 4 and 104 respectively. When P(x) is divided by x^2-x+1 the quotient is x^2+x+3 and the remainder is of the form ax+b. Find the remainder.
Answers
Answered by
oobleck
(x^2-x+1)(x^2+x+3) = x^4 + 3x^2 - 2x + 3
so that means that
P(x) = x^4 + 3x^2 + (a-2)x + (b+3)
Now, by the Remainder Theorem, you know that
P(1) = 4
P(-3) = 104 so now you just have to solve for a and b in
1+3+(a-2)+(b+3) = 4
81+27 -3(a-2) + (b+3) = 104
So the remainder is 3x-4
so that means that
P(x) = x^4 + 3x^2 + (a-2)x + (b+3)
Now, by the Remainder Theorem, you know that
P(1) = 4
P(-3) = 104 so now you just have to solve for a and b in
1+3+(a-2)+(b+3) = 4
81+27 -3(a-2) + (b+3) = 104
So the remainder is 3x-4
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