Question 1205720
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what is the remainder when P(x)=x^(2)+5 is divided by (x+1)
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        I will show you two ways of solving this problem.



<pre>
       <U>First way: using the remainder theorem</U>


According to the Remainder theorem, the remainder of division any polynomial P(x)
by a binomial (x-a), where "a" is a real (or integer) number, is the value 
of the polynomial P(x) at x= a, i.e. P(a).


In this problem, a= -1, so, according to the Remainder theorem, the remainder 
of division P(x) = x^2+5 by (x+1) is P(-1) = (-1)^2 + 5 = 6.


<U>ANSWER</U>.  The remainder of division P(x) = x^2+5  by  (x+1)  is  6.



       <U>Second way: using explicit division via grouping</U>


Using grouping, we can write

    P(x) = {{{x^2+5}}} = {{{x^2+x)}}} + {{{(-x+5)}}} = x(x+1)  + (-x-1) + 1 + 5 = x(x+1) - (x+1) + 6 = 

         = (x+1)*(x-1) + 6.


It shows that when P(x) is divided by (x+1), the quotient is (x-1) and the remainder is 6.


Thus the <U>ANSWER</U> is the same as in the first solution above.
</pre>

Solved in two ways for your better understanding.


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Surely, there is third way, which is direct long division of polynomials.


This method is straightforward and purely mechanical procedure - therefore, 
I do not say more about it and even do not show it here.