Using the remainder theorem to find a remainder.

Using the remainder theorem to find a remainder.

Question 

The polynomial {\color{Blue} x^{55}-6x^{42}-6x^{29}-3x^{23}+x^4-2} is divided by {\color{Blue} x+1}. What is the remainder?

Solution

Do you want to try this problem using polynomial long division? That could be a nightmare. How about synthetic division? I didn’t think so. You need the remainder theorem.

The remainder theorem states that when a polynomial (like, for example, the one in this problem) is divided by a binomial {\color{Blue} x-c}, the remainder is equal to the value of the polynomial at point {\color{Blue} x=c}.

Consider a simple example: When polynomial {\color{Blue} x^2-2x+2} is divided by {\color{Blue} x-1}, the remainder is equal to the value of {\color{Blue} x^2-2x+2} at point {\color{Blue} x=1}. Since 1 is an easy number at which to evaluate a function, it’s not too hard to see that at {\color{Blue} x=1}, {\color{Blue} x^2-2x+2}=1.  Thus when {\color{Blue} x^2-2x+2} is divided by {\color{Blue} x-1}, the remainder is 1.

You can confirm this result with polynomial long division or synthetic division.

Now try the question. To find the remainder when {\color{Blue} x^{55}-6x^{42}-6x^{29}-3x^{23}+x^4-2} is divided by {\color{Blue} x+1}, evaluate {\color{Blue} x^{55}-6x^{42}-6x^{29}-3x^{23}+x^4-2}  at {\color{Blue} x=-1.} That’s

{\color{Blue} (-1)^{55}-6 \cdot (-1)^{42}-6 \cdot (-1)^{29}-3 \cdot (-1)^{23} + (-1)^4-2}

Note that -1 taken to an even power is 1 and -1 taken to an even power is 1. Thus {\color{Blue} (-1)^{55}=-1}, {\color{Blue} (-1)^{29}=-1}, and {\color{Blue} (-1)^{23}=-1}; while {\color{Blue} (-1)^{42}=1} and {\color{Blue} (-1)^{4}=1}. The whole polynomial equals

{\color{Blue} -1-\left (6 \cdot 1 \right )-\left (6 \cdot -1 \right )-\left (3 \cdot -1 \right )+1-2}

{\color{Blue} -1-6+6+3+1-2=1}

The remainder is 1.

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