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Main Topic # 1: [Long Division with polynomials] The basic concept of worksheet is write rational expressions as as a polynomial plus a simpler rational ...
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Long Division
Main Topic # 1: [Long Division with polynomials] The basic concept of worksheet is write rational expressions as as a polynomial plus a simpler rational expression. This idea was seen with numbers when you were younger. More specifically we can find a q called the quotient and r the remainder so that
a b
= q +
r b
so that r < b. This is not as tricky as it might seem to the mathematical adverse student. It is only a tricky way of writing that there is a number q such that a = q · b + r
that is q is the most amount of times b divides into a an r is the left over, that is why it is less than b if it was greater we could get another factor of b. Now we want to do the same process with polynomials specifically lets look at (x^2 + 2x − 1) ︸ ︷︷ ︸ dividend
÷ (x − 1) ︸ ︷︷ ︸ divisor
We begin very much the same as long division of numbers but this time we only look at the highest term in each. The highest term in x − 1 is just x^1 and the highest term in x^2 + 2x − 1 is x^2. We even write the set up the same!
x − 1
x^2 + 2x − 1
So we first ask the question what can we multiply x (i.e. our highest term in our divisor) to get x^2 (i.e. our highest term in our dividend) well that is of course x or to put it in an equation
x · x = x^2
We collect this data just as before by writing it above the line
x x − 1
x^2 + 2x − 1
We then proceed like with numbers and multiply our new factor x by (x − 1) and get
x · (x − 1) = x^2 − x
and again we must subtract (don’t forget to distribute the subtraction sign i.e. −(x^2 − x) = −x^2 + x
x x − 1
x^2 + 2x − 1 − x^2 + x
and then performing the subtraction by subtracting like terms we get
x x − 1
x^2 + 2x − 1 − x^2 + x 3 x − 1
Now we repeat the process with the left over polynomial in this example it is 3x − 1. Notice the highest term of 3 x − 1 is 3x and we ask what do we need to multiply x (the highest term of x − 1) by to get 3x? We see here it is more simple we need only multiply by 3! That is
3 · x = 3x
Now to collect this information we add the new term 3 above the line
x + 3 x − 1
x^2 + 2x − 1 − x^2 + x 3 x − 1
And again multiply 3 by (x − 1) we get 3x − 3 and when we subtract we get −(3x − 3) = − 3 x + 3 and we collect this data as follows x + 3 x − 1
x^2 + 2x − 1 − x^2 + x 3 x − 1 − 3 x + 3
again combining like terms we get (^) x + 3
x − 1
x^2 + 2x − 1 − x^2 + x 3 x − 1 − 3 x + 3 2
and now we see that we are at a degree lower than that of the divisor so we are done! this is our remainder 2! So we have the following: x^2 + 2x − 1 x − 1
= (x + 3) +
x − 1
Before you move on make sure you can follow these next examples!
Examples: x + 2 x^2 − x + 1
x^3 + x^2 + 2x − 1 − x^3 + x^2 − x 2 x^2 + x − 1 − 2 x^2 + 2x − 2 3 x − 3
i.e.
x^3 + x^2 + 2x − 1 x^2 − x + 1
x + 2
3 x − 3 x^2 − x + 1 3 4 x^ +^
3 16 4 x^2 − x + 1
3 x^3 + 2x − 1 − 3 x^3 + 34 x^2 − 34 x 3 4 x
4 x^ −^1 − 34 x^2 + 163 x − 163 23 16 x^ −^
19 16
i.e.
3 x^3 + 2x − 1 4 x^2 − x + 1
3 4 x^ +^
3 16
23 16 x^ −^
19 16 4 x^2 − x + 1