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Long Division with Polynomials

05 July 2014 By Gideon In Polynomials
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How do we divide polynomials?

Suppose, for example that you are given the problems:


Express \(\frac{x^{2} - 1 } {x - 1 } \) in terms of \(x\).


Find \(\frac{x^{3} + 2x^{2} + x}{x + 1}\) in terms of \(x.\)

What do we do?


First, to get a feel for dividing polynomials, lets plug in some numbers. Note that this will not give us answers in terms of \(x,\) so we will still have some work to do.

However, plugging in numbers makes the problems look more familiar and less intimidating.


Note that there is one important restriction when plugging in numbers for \(x'\): we must not allow the denominator to equal zero. Therefore, in the first problem we can plug in any value for \(x\) other than \(1\), and in the second problem we can plug in any value other than \(-1.\)

Suppose we take the value \(x = 0\) in the first problem. Then we obtain:


\[ \frac{x^{2} - 1 } {x - 1 } = \frac{ 0^{2} - 1}{0 -1} = \frac{ -1}{-1} = 1.\]


If we plug in the value \(x = 2,\) we obtain:


\[ \frac{x^{2} - 1 } {x - 1 } = \frac{ 2^{2} - 1}{2 -1} = \frac{ 3}{1} = 3.\]


If we plug in the value \(x = 5,\) we obtain:

\[ \frac{x^{2} - 1 } {x - 1 } = \frac{( 5^{2} - 1)}{(5 -1)} = \frac{(25 -1)}{4} = \frac{ 24}{4} = 6.\]


Note that in all cases the answer is \(x +1.\) This suggests that the answer in terms of \(x\) is probably \(x +1,\) but we would have to check an infinite number of values of \(x\) in order to prove this.

We will soon see how to solve this problem and get the answer: \(x +1,\) (without checking an infinite number of possible values) but first lets look at some simple problems where we can leave the variable as \(x\) and still get the answer by inspection.

Let's start with some easy problems:


We know that \( 2 \times x = 2 x, \) therefore:


\[ \frac{2x}{x} = 2.\]


Similarly, we know that \(x \times x = x^{2},\) therefore:

\[ \frac{x^{2}}{x} = x.\]


What about \( \frac{2x +1}{x} ?\)


We will set this up just like a division problem for numbers.


First we write:

poly1

 

Now multiply \(x\) by \(2\) to obtain \(2x,\) then subtract this from \(2x + 1\) to find \(1.\)

So the answer is \(2\) with remainder \(1,\) or:

\( 2 + \frac{1}{x}.\)


We write this whole process as:

poly2

Now we are ready to go back to our original problems.


First express \(\frac{ x^{2} - 1}{ x -1}\) in terms of \(x.\)

There are actually two ways to solve this problem:

First, we can factor \(x^{2} -1:\)

\[ x^{2} -1 = (x + 1)(x-1).\]

Then

\[ \frac{ x^{2} -1}{x -1 } = \frac{( x+ 1)( x-1)}{(x -1)} = (x + 1).\]


Without factoring \(x^{2} -1\) we can write:

poly3

 

We want to cancel out the highest power of \(x.\) In this case this is \(x^{2}.\)

Therefore multiply \(x -1\) by \(x\) to obtain \(x^{2} - x.\)

Next we subtract this from \(x^{2} -1,\) to obtain:


\[ \left( x^{2} -1\right) - (x)(x -1 ) = (x^{2} -1) - (x^{2} + x) = -1 - x = (-1)(x +1).\]

Note that the highest power of \(x\) is now gone (there are no \(x^{2}\) terms).


Finally, we see that the remaining term is just minus one times \(( x + 1).\)

Therefore we multiply \(( -1)(x +1)\) and subtract this from \( -1 -x \) to obtain zero.


Therefore the answer is \(x -1.\)


We can write all these steps as:

poly4




The next problem is:

Express \(\frac{ x^{3} + 2x + x}{x +1}\) in terms of \(x.\)

We write:

poly5

 

Again, we want to cancel the highest power of \(x,\) in this case \(x^{3},\) and so we multiply \(x + 1\) by \(x^{2}\) to obtain:


\[ (x + 1)(x^{2}) = x^{3} + x^{2}.\]


Then we subtract this from \(x^{3} + 2x^{2} + x\) to obtain:


\[ x^{2} + x.\]

Now we multiply \((x +1)\) by \(x\) to obtain \(x^{2} + x\) and we are done.

We write all these steps as:

poly6

 

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