Let \(P(x)\) be a cubic polynomial such that \(P(0) = -3\) and \(P(1) = 4\) . When \(P(x)\) is divided by \(x^2 + x + 1\) the remainder is \(2x-1\) . What is the quotient when \(P(x)\) is divided by \(x^2+x+1\)?
Being a cubic polynomial, it has the form: P(x) = ax3 + bx2 + cx + d
Since P(0) = -3 ---> P(0) = a(0)3 + b(0)2 + c(0) + d = -3 ---> d = -3
Since P(1) = 4 ---> P(1) = a(1)3 + b(1)2 + c(1) + -3 = a + b + c - 3 = 4 ---> a + b + c = 7
Using long division, I divided ax3 + bx2 + cx - 3 by x2 + x + 1 and I got
a quotient of ax + (b - a) and a remainder of (c - b)x + (a - b - 3)
Since the problem said that the remainder was 2x - 1 ---> (c - b)x + (a - b - 3) = 2x - 1
which means that c - b = 2 and a - b - 3 = -1 ---> c = b + 2 and a = b + 2
Combining a + b + c = 7 and c = b + 2 and a = b + 2
---> a = 3 and b = 1 and c = 3
Since the quotiet was ax + (b - a) and a = 3 and b = 1,
replace to determine the answer.
Being a cubic polynomial, it has the form: P(x) = ax3 + bx2 + cx + d
Since P(0) = -3 ---> P(0) = a(0)3 + b(0)2 + c(0) + d = -3 ---> d = -3
Since P(1) = 4 ---> P(1) = a(1)3 + b(1)2 + c(1) + -3 = a + b + c - 3 = 4 ---> a + b + c = 7
Using long division, I divided ax3 + bx2 + cx - 3 by x2 + x + 1 and I got
a quotient of ax + (b - a) and a remainder of (c - b)x + (a - b - 3)
Since the problem said that the remainder was 2x - 1 ---> (c - b)x + (a - b - 3) = 2x - 1
which means that c - b = 2 and a - b - 3 = -1 ---> c = b + 2 and a = b + 2
Combining a + b + c = 7 and c = b + 2 and a = b + 2
---> a = 3 and b = 1 and c = 3
Since the quotiet was ax + (b - a) and a = 3 and b = 1,
replace to determine the answer.