HELLO!!!, Mr. Tarrou. Today we are talking about long division. This skill is going
to be used in my Precalculus class and probably your as well for working through the solutions
to higher oder degree polynomials. Solving polynomials that don't just have a degree
of one or two. We will be solving for degrees of three, polynomials with a degree of four.
But we first need to know how to do this long division before we solve or work on any more
complicated questions. So, let's see what long division looks like with polynomials.
Don't forget, polynomials are a collection of terms that are added or subtracted together
all of which have exponents that are positive integers. So we are going to take two x to
the third plus nine x squared minus nineteen x plus seven and divide that by two x minus
one. This is probably how your textbook has the division written. I am going to change
this into a more old fashion, an old school division symbol which is a big old division
symbol and 2x - 1. We are going to divide 2x - 1 into this four term polynomial. If
you are learning long division, you are probably also learning something called synthetic division.
Synthetic division is a short cut version of this. It only works, synthetic division
only works when you are dividing by a binomial with a degree of one, which I am, and the
leading coefficient is equal to one. So I will not be able to short cut this with synthetic
division because my leading coefficient is equal to two, clearly not equal to one. So
when you do long division, you are attempting to make this first term equal to the first
term of the polynomial inside the division symbol. I am going to stack mine right above
what I am trying to get rid of. A textbook probably has your answers shifted over to
the right a column. It is just my habit. You don't have to exactly my placement, but I
think it helps. I want to put something up here where I can multiply 2x and get it to
equal 2x to the third. So what am I going to multiply this by to get it to equal this?
Well I already have the two's, so I just have to multiply by x squared. Once you think you
know what you need to multiply by, go ahead and distribute that and make sure that these
first terms match. X squared times 2x is, good it is 2x cubed. That is what I had in
the first place, what I need to match. Now x squared times negative one is negative x
squared. That became a three of course because when you multiply like bases you add the exponents.
Alright, wrap that in parenthesis and subtract. A lot of students will forget this subtraction
symbol is out here, so you may want to distribute it through the parenthesis before you continue.
I am going to leave it out there and just keep pointing to it, but don't forget it is
out there. Two minus two is equal to zero. Nine minus negative is addition, so that is
nine minus a negative one which is nine plus one which is 10x squared. I am going to bring
everything else down. Now I want to make this first term match this first term. Well how
do you get 2x to equal 10x squared? Multiply it by five and another x to get the exponent
of one to be an exponent of two. So multiply by 5x. Alright, so 5x times 2x is 10 x squared.
5x times negative one is negative 5x. Again with the parenthesis and wrap that line and
put a minus sign out front. You may want to distribute before you subtract but then you
would add. But I am going to keep pointing to the minus sign. Ten minus ten is zero.
Negative nineteen minus a negative five. Well that is negative nineteen plus five and that
is negative fourteen x plus seven. Now how do I get 2x to again equal equal the first
term, get 2x to become negative fourteen x? Multiply by negative seven. The degree on
the x, the exponent on the x, and the exponent here are already matched. So I just have to
make two equal to negative fourteen, and I am going to that my multiplying by negative
seven. So we get negative seven times two is negative fourteen. Negative seven times
negative one is positive seven. You are going to subtract those. Again, negative minus a
negative is plus, so negative fourteen plus fourteen is zero. Seven minus seven is zero.
We got a remainder of zero, and this is how you should write your remainder, of zero or
zero over 2x - 1. Your teacher could very well not even want you to write this remainder
because it is zero. Adding by zero does not do anything, but I am just showing you how
to write remainders. So take it or leave it. But that is the remainder of 0 over 2x-1.
What does the fact that we got a remainder of zero actually mean? Well if you are not
sure, let's think of just a regular arithmetic problem. 8 divided by 2. Well that is 4 of
course. We know that. 4 times 2 is 8. Then when you subtract you get a remainder of zero
just like we did here. Why did we get a remainder of zero? Of course two divides into eight
evenly, but what that really means is that two is a factor of eight. Ok, I am going to
get ahead of myself a little bit to make you understand why you even care about this. If
you were doing a... You are going to be doing this soon in your math book. You are going
to be getting questions like if f of x equals 2 x to the third plus 9 x squared minus 19x
plus seven, of course that is what I just divided into, you will be asked to solve this
higher degree equation. Well when you solve an equation, that means set y equal to zero
and solve it. Now I know that 2x-1 is one of the factors of this higher degree polynomial.
I can write zero equals (2x-1) times, well clearly we just divided so, (2x-1) times our
answer of (x^2+5x-7). If we were to multiply all this back together, it would give you
this third degree polynomial. This here is a partial factored version of this polynomial.
Now is it fully factored or partially factored? Well do you have factors of seven that subtract
to five? The answer to that would be no, so this is fully factored. Now as far as long
division, this is done. What you will be doing soon is using that fact that 2x - 1 is factor
because we got a remainder of zero. Going ahead and finishing these higher order problems.
In this particular case it will be 2x minus 1 is equal to zero, and then you will finish
this one using the quadratic formula which is x is equal to negative b plus or minus
the square root of b squared minus 4ac all over 2a. Now you will be seeing some videos
of me solving these higher order degree problems later because that is where my textbook is
going. But for right now, we are just focusing on the long division and that question is
done. WHOOO!!! Moving on. Moving on. Moving on. Ok, what about this problem? y minus 2
divided into y to the fifth minus three y squared plus twenty. Well a couple of things.
I am not going to show you synthetic division in this video, but we have a first degree
binomial with a leading coefficient of one. So this could be short cutted with synthetic
division. What I am really showing you this for is because these exponents need to count
down by one if you want to properly do your long division. So I need to rewrite this.
I did not have a fourth degree term, so I am going to do plus 0 y to the fourth. I did
not have a third degree term, so I going to do plus 0 y to the third. I did have a second
degree term, minus three y squared. I didn't have a first degree term, so plus 0 y to the
first minus twenty. So my deceptively small looking division problem all of a sudden got
much much longer will all of my terms with leading coefficients of zero. You are going
to want those as you finish this long division process and you try to subtract all your like
terms. Ok. How do we get y to become y to the fifth. Multiply y by y to the fourth and
then distribute. So y to the fifth minus 2 y to the fourth. Ah, do you see what just
happened? I needed that zero column and that exponent of four so I could line up my like
terms. Otherwise long division will become pretty ugly and messy and you probably will
get it wrong. One minus one is zero and zero minus a negative two is positive two. I don't
need that plus sign. We have positive 2 y to the fourth and I am going to bring down
zero y to the third. How do I get y to equal 2 y to the fourth? Well multiply by two y
to the, I already have one y, so multiply by 2 y to the third. 2y^3 times y is 2y^4.
And 2y^3 times negative 2 is negative 4y^3. And again get your subtraction set up. I think
I am going to run out of room. Two minus two is zero and zero minus negative four is positive
four y to the third minus 3 y squared. Ok. How do I get once again this leading term
to equal this leading term? I have to multiply it by four y to the second. Ok, 4y^2 times
y is 4y^3. And 4y^2 times negative two is negative 8y^2. Again with the subtraction.
Four minus four is zero. Negative three minus negative eight is negative three plus eight
which is five y squared. Now I am going to bring down the zero y. I think I am going
to have to erase my board and.... I am going to just squeeze this in here. Sorry for the
mess I am making here. How do I make y equal to 5 y squared? Multiply by five y. 5y times
y is 5y^2. 5y times negative two is negative 10y. Subtract those. I am going to write my
answer over here. Five minus five is zero. zero minus negative ten is positive ten y
minus twenty. Now we are going to get a remainder again of zero. I am going to write that remainder
of zero in proper remainder form. That is not necessary if the remainder is zero. But
if you have a remainder of anything other than zero, you have got to do that. So how
do we get 1y to equal 10y? How do we get those leading terms to match? Well, multiply by
ten. We are just about done. Ten times one y is 10y. Ten times negative two is negative
twenty. Subtract those. Again get your answer of zero. If you are having to write your remainders
even if they are zero, it is y^4 plus 2y^3 plus 4y^2 plus 5y plus 10 and our remainder
of zero over y minus 2. Do you see how much work that was with long division? That would
have been a mess, I am sure I would have gotten the wrong answer if I had not put all those
zeros to keep my work in nice columns. All the exponents of five, all the exponents of
four are lined up, all the exponents of three, and two, and then of course I ran out of room
and so on. But you want those zeros in there as your placeholders to make sure that you
don't loose track of what you are doing, you get all of your like terms together, and you
get the right answer. That is two examples of how to do long division. WHOOO!!! Go do
your homework!