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 $x^2 + x + 1$?
+128408 ax^3 + bx^2 +cx + d
P(0) = -3 indicates that d = - 3
And
a + b + c - 3 = 4
a + b + c = 7
ax + (b - a)
x^2 + x + 1 [ ax^3 + bx^2 + cx - 3 ]
ax^3 +ax^2 + ax
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(b - a) x^2 + ( c - a)x - 3
(b - a)x^2 + (b - a)x + ( b - a)
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( - b + c))x + ( a - b - 3) = 2x - 1
Then...we have this system
- b + c = 2 c = 2 + b (1)
a - b - 3 = -1 ⇒ a - b = 2 ⇒ a = b + 2 (2)
a + b + c = 7 (3)
Sub (1) and (2) into (3)
(b + 2) + b + ( b + 2) = 7
3b = 3
b = 1
a = 3
c = 3
So
ax + ( b - a) =
3x + ( 1 - 3) =
3x - 2 = the quotient
\(\displaystyle \frac{P(x)}{(x^{2}+x+1)}=Q(x)+\frac{2x-1}{(x^{2}+x+1)}. \)
(The quotient Q(x) will be linear since it's a cubic divided by a quadratic.)
From that,
\(\displaystyle P(x)=Q(x)(x^{2}+x+1)+(2x-1).\)
\(\displaystyle P(0)=-3, \text{ and } P(1)=4, \text{ gets you}\\ -3=Q(0)-1, \text{ so }Q(0)=-2, \\ \text{and} \\ 4=Q(1).3+1, \text{ so }Q(1)=1. \)
If we now let
\(\displaystyle Q(x)=ax+b, \text{ then} \\ Q(0)=b = -2, \text{ and} \\ Q(1)=a+b=1, \text{ so } a = 3,\\ \text{ and therefore} \\ Q(x)= 3x-2.\)
