Question 1111313
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<pre>
I will use the Remainder theorem. I will make all necessary explanations and references, but will not go in details.


Based on the Remainder theorem, from the given part you have these equations

f(1)  =  2,   or  1^3    + a*1^2    + b*1    + c =  2    (1)

f(-2) = -1,   or  (-2)^3 + a*(-2)^2 + b*(-2) + c = -1    (2)

f(2)  = 15,   or  2^3    + a*2^2    + b*2    + c = 15    (3)


Simplifying

 1 +  a +  b + c =  2       (1')
-8 + 4a - 2b + c = -1       (2')
 8 + 4a + 2b + c = 15       (3')


Simplifying one more time

      a +  b + c = 1        (1'')
     4a - 2b + c = 7        (2'')
     4a + 2b + c = 7        (3'')

Subtract (2'') from (3''). You will get  4b = 0  ====>  b = 0.

Now substitute this value of b into eqs (1'')  and (2'').  You will get

     a + c = 1        (4)
    4a + c = 7        (5)

--------------------------------------- Subtract (4) from (5)

          3a = 6  ====>  a = 2


Then  from (4)  c = 1 - 2 = -1


Thus I restored the 3-rd degree polynomial. It is

f(x) = {{{x^3 + 2x^2 - 1}}}.


The rest is pure mechanical work:


f(x) = (x+1)*(x^2 + x -1).


<U>Answer</U>.  The quotient under the question is  (x^2 + x - 1).  The remainder is  0.
</pre>

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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 lessons

&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>Divisibility of polynomial f(x) by binomial (x-a) and the Remainder theorem</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Solved-problems-on-the-Remainder-theorem.lesson>Solved problems on the Remainder thoerem</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 lessons are the part of this online textbook under the topic 
"<U>Divisibility of polynomial f(x) by binomial (x-a). The Remainder theorem</U>".


Save the link to this online textbook together with its description


Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson


to your archive and use it when it is needed.