Question 1086582
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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).
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<pre>
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.
</pre>

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



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&nbsp;&nbsp; <B>Theorem</B> &nbsp;&nbsp;(the <B><I>remainder theorem</I></B>)

&nbsp;&nbsp; <B>1</B>. The remainder of division the polynomial &nbsp;{{{f(x)}}}&nbsp; by the binomial &nbsp;{{{x-a}}}&nbsp; is equal to the value &nbsp;{{{f(a)}}}&nbsp; of the polynomial. 

&nbsp;&nbsp; <B>2</B>. The binomial &nbsp;{{{x-a}}}&nbsp; divides the polynomial &nbsp;{{{f(x)}}}&nbsp; if and only if the value of &nbsp;{{{a}}}&nbsp; is the root of the polynomial &nbsp;{{{f(x)}}}, &nbsp;i.e. &nbsp;{{{f(a) = 0}}}.

&nbsp;&nbsp; <B>3</B>. The binomial &nbsp;{{{x-a}}}&nbsp; factors the polynomial &nbsp;{{{f(x)}}}&nbsp; if and only if the value of &nbsp;{{{a}}}&nbsp; is the root of the polynomial &nbsp;{{{f(x)}}}, &nbsp;i.e. &nbsp;{{{f(a) = 0}}}.



See the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Divisibility-of-polynomial-f%28x%29-by-binomial-x-a.lesson?content_action=edit_dev>Divisibility of polynomial f(x) by binomial x-a</A>

in this site.



Also, &nbsp;you have this free of charge online textbook in ALGEBRA-II in this site

&nbsp;&nbsp;&nbsp;&nbsp;<A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A>.


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