SOLUTION: When polynomial P(x) is divided by x + 1, x + 2, and x + 3, the remainders are 2, 3, and 6, respectively. Find the remainder when P(x) is divided by (x + 1)(x + 2)(x + 3).

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: When polynomial P(x) is divided by x + 1, x + 2, and x + 3, the remainders are 2, 3, and 6, respectively. Find the remainder when P(x) is divided by (x + 1)(x + 2)(x + 3).      Log On


   



Question 1086582: When polynomial P(x) is divided by x + 1, x + 2, and x + 3, the remainders are 2, 3, and 6, respectively. Find the remainder when P(x) is divided by (x + 1)(x + 2)(x + 3).
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
When polynomial P(x) is divided by x + 1, x + 2, and x + 3,
the remainders are 2, 3, and 6, respectively.

Let P%28x%29+=+x%5E3%2BAx%5E2%2BBx%2BC to see if that is a possibility.

Then, using the remainder theorem: 

P%28-1%29+=+%28-1%29%5E3%2BA%28-1%29%5E2%2BB%28-1%29%2BC+=+2
P%28-1%29+=+-1%2BA-B%2BC+=+2
A-B%2BC+=+3

P%28-2%29+=+%28-2%29%5E3%2BA%28-2%29%5E2%2BB%28-2%29%2BC+=+3
P%28-2%29+=+-8%2B4A-2B%2BC+=+3
4A-2B%2BC+=+11

P%28-3%29+=+%28-3%29%5E3%2BA%28-3%29%5E2%2BB%28-3%29%2BC+=+6
P%28-3%29+=+-27%2B9A-3B%2BC+=+6
9A-3B%2BC+=+33

Solve this system by elimination or other method:

system%28A-B%2BC=3%2C4A-2B%2BC=11%2C9A-3B%2BC=33%29

Solution:  A=7, B=13, C=9

So

P%28x%29+=+x%5E3%2B7x%5E2%2B13x%2B9

Find the remainder when P(x) is divided by (x + 1)(x + 2)(x + 3).

When %28x+%2B+1%29%28x+%2B+2%29%28x+%2B+3%29 is multiplied out, 
we get x%5E3%2B6x%5E2%2B11x%2B6

Dividing by long division:
                        1 
x³+6x²+11x+6)x³+7x²+13x+9
             x³+6x²+11x+6
                 x²+ 2x+3  

Answer: x²+2x+3

Edwin

Answer by ikleyn(52847) About Me  (Show Source):
You can put this solution on YOUR website!
.
When polynomial P(x) is divided by x + 1, x + 2, and x + 3, the remainders are 2, 3, and 6, respectively.
Find the remainder when P(x) is divided by (x + 1)(x + 2)(x + 3).
~~~~~~~~~~~~~~~~~

We are given that 

     "when polynomial P(x) is divided by x + 1, x + 2, and x + 3, the remainders are 2, 3, and 6, respectively."


According to the Remainder theorem, it is equivalent to these equalities:

    P(-1) = 2,    (1)
    P(-2) = 3,    (2)
    P(-3) = 6.    (3)


Now, the question is to find a remainder polynomial R(x) after dividing P(x) by (x+1)*(x+2)*(x+3):

    P(x) = g(x)*(x+1)*(x+2)*(x+3) + R(x).     (4)


It is clear that the polynomial R(x) has the degree <= 2, so we can write

   R(x) = Ax^2 + Bx + C.                      (5)


Substituting x= -1, x= -2 and x= -3 into (4), from (1), (2) and (3) we have 

    P(-1) = g(-1)*0 + R(-1) = 2,   i.e.   R(-1) = 2;     (6)
    P(-2) = g(-2)*0 + R(-2) = 3,   i.e.   R(-2) = 3;     (7)
    P(-3) = g(-3)*0 + R(-3) = 6,   i.e.   R(-3) = 6.     (8)


So, we need to find coefficients A, B and C of the remainder polynomial R(x) from conditions (6), (7) and (8).


Equation (6) gives

    A*(-1)^2 + B*(-1) + C = 2,   or   A - B + c = 2;     (9)
   
Equation (7) gives

    A*(-2)^2 + B*(-2) + C = 3,   or   4A - 2B + c = 3;   (10)

Equation (8) gives

    A*(-3)^2 + B*(-3) + C = 6,   or   9A - 3B + c = 6.   (11)


Thus you have this system of 3 equations to find A, B and C:

 A -  B + c = 2,
4A - 2B + c = 3,
9A - 3B + c = 6.


Solve it by any method you want/you know (Substitution, Elimination, Determinanf (= Cramer's rule) ). You will get  A = 1,  B= 2  and  C = 3.


So, the remainder, which is under the question, is R(x) = x^2 + 2x + 3.

Answer. The remainder under the question is R(x) = x^2 + 2x + 3.


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   Theorem   (the remainder theorem)
   1. The remainder of division the polynomial  f%28x%29  by the binomial  x-a  is equal to the value  f%28a%29  of the polynomial.
   2. The binomial  x-a  divides the polynomial  f%28x%29  if and only if the value of  a  is the root of the polynomial  f%28x%29,  i.e.  f%28a%29+=+0.
   3. The binomial  x-a  factors the polynomial  f%28x%29  if and only if the value of  a  is the root of the polynomial  f%28x%29,  i.e.  f%28a%29+=+0.


See the lesson
    - Divisibility of polynomial f(x) by binomial x-a
in this site.


Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic
"Divisibility of polynomial f(x) by binomial (x-a). The Remainder theorem".