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# Halp Plez

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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 $x^2 + x + 1$?

Jan 5, 2021

#1
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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

____________________________

(b - a) x^2 + ( c - a)x  -  3

(b - a)x^2   + (b - a)x  + ( b - a)

___________________________

( - 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

Jan 5, 2021
#2
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$$\displaystyle \frac{P(x)}{(x^{2}+x+1)}=Q(x)+\frac{2x-1}{(x^{2}+x+1)}.$$

(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

$$\displaystyle Q(x)=ax+b, \text{ then} \\ Q(0)=b = -2, \text{ and} \\ Q(1)=a+b=1, \text{ so } a = 3,\\ \text{ and therefore} \\ Q(x)= 3x-2.$$

Jan 5, 2021