Long Division with polynomials, Slides of Algebra

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 qcalled the quotient and rthe 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 qsuch that
a=q·b+r
that is qis the most amount of times bdivides into aan ris the left over, that is why it is less than bif it was
greater we could get another factor of b. Now we want to do the same process with polynomials specifically lets
look at
(x2+ 2x1)
| {z }
dividend
÷(x1)
| {z }
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 x1 is just x1and the highest term in x2+ 2x1 is x2. We even write the set up the same!
x1x2+ 2x1
So we first ask the question what can we multiply x(i.e. our highest term in our divisor) to get x2(i.e. our highest
term in our dividend) well that is of course xor to put it in an equation
x·x=x2
We collect this data just as before by writing it above the line
x
x1x2+ 2x1
We then proceed like with numbers and multiply our new factor xby (x1) and get
x·(x1) = x2x
and again we must subtract (don’t forget to distribute the subtraction sign i.e. (x2x) = x2+x
x
x1x2+ 2x1
x2+x
and then performing the subtraction by subtracting like terms we get
x
x1x2+ 2x1
x2+x
3x1
Now we repeat the process with the left over polynomial in this example it is 3x1. Notice the highest term of
3x1 is 3xand we ask what do we need to multiply x(the highest term of x1) by to get 3x? We see here it is
more simple we need only multiply by 3! That is
3·x= 3x
1
pf3

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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