BAM!!! Mr. Tarrou What are we going to take a look at today? If you have been following
my videos and you saw my last post for PreCalculus you might have an idea. Synthetic Division.
Ok. Long division is monotonous and it is easy to lose your place, you make sign errors,
you always have to check out for that little minus sign as you are working with your polynomials
as you work your way down. So if there is ever an opportunity to take a short cut, why
not do it? And that is what synthetic division. Synthetic division is possible if you are
dividing by a binomial in the form of x minus c. We are about to do an example and I am
going to show you the Remainder Theorem using the same example. With Synthetic Division
as long as you are dividing by a binomial with a leading coefficient of one and the
binomial has a degree of one, you can do Synthetic Division to shorten the problem and not go
through all the steps of long division. Synthetic division only works if your leading coefficient
is one and the degree of that binomial is one, otherwise you are stuck doing long division.
So synthetic division, let's do an example. We are going to divide x plus one into five
x cubed plus six x plus eight. I have written this in long division format just to tie these
concepts together. Also, we are going from a third degree to a first degree term. If
you want to do this properly whether it is long division or synthetic division, you need
to put the zeros in there as place holders for the exponents so it all lines up in nice
columns. That means that this need to be rewritten as five x to the third plus zero x squared
plus six x plus eight. But we are not doing long division. We have a binomial. We have
a leading coefficient of one. We have a degree of one, or an exponent of one and we are dividing
by a binomial in the form of x minus c. Do you see that minus sign in the form? That
means that for this x plus one it must be x minus a negative one. We are going to do
synthetic division with negative not positive one. When pull these numbers out of the binomials
to do synthetic division you must remember to change the sign. Alright, so here is what
we have got. We have got negative one and little square bracket to indicate synthetic
division. I am going to write this polynomial out with no variables, just the coefficients,
Five, zero, six, and eight. I am going to draw a base line and unlike with long division
where you are subtracting your way down as you do all your steps of long division, with
synthetic division you are going to be adding. Now, here is the process. Bring the first
number down. Ok, nothing happened. Whooptie Doo. What you do now is you take this number
that is in the red box, that I have now made red if you can see that, and you are going
to come through here and you are going to multiply to whatever number is down here.
So negative one times five is negative five and then add. Zero plus negative five is negative
five. Now multiply negative one times negative five is positive five. Bam! Now you are on
this line, add, six plus five is eleven. And now back to your synthetic division, negative
one times eleven is going to be negative eleven. Now add and we get negative three. What the
heck does all this mean? It is just a bunch of numbers. Well, when you have a polynomial
to the third degree and you do long division and you divide by the first degree, you are
going to subtract one from all of the exponents. If you watched my video on long division you
would notice that pattern. So this was originally five x cubed, and now dividing by a degree
of one this is five x squared. The exponent of one subtracting a one from this exponent
of three. This was zero x squared and now it is negative five and the exponents are
counting down, still counting down, so negative five x plus a constant of eleven. Now your
three. Remember your original exponents were 3,2,1 and then a constant. Now it is not three
but two, it has been reduced by one, two, one, constant so this must be the remainder.
So plus, excuse me, minus.... I can write that. Plus negative three, the remainder,
over what you were dividing by which was....technically I am dividing by negative one but the original
binomial was x plus one. Voila. You would gotten the same answer if you had done long
division but that was so much quicker!!! AH, I love it. Now, I want to give you an example
of something called the Remainder Theorem. nananana Remainder Theorem is always in the
textbook with Synthetic Division and I am not going to teach you the Remainder Theorem
by showing you the Remainder Theorem. I am just going to ask you to do this, F of x equals
five x cubed plus six x plus eight. Does that sound familiar? It is right there. It is what
we divided into. I want to find out what is F of negative one. Does the negative one look
familiar? The negative one is what we did our synthetic division by. I am just giving
you a polynomial and saying HEY, PLUG IN NEGATIVE ONE! Let's see how bad the maths are if you
were to do this by just plugging in the number. You do have to remember I am teaching this
in front of a camera that putting over the internet. So I am going to pick some numbers
that are relatively easy to figure out. However, this process that I am showing you where you
just plug in the number, that very often is a very cumbersome process that you would not
want to do without the aid of a calculator. But let's see what it looks like anyway. Negative
one to the third is negative one plus six times negative one plus eight. Negative one
times five is negative five, oops that is negative, minus six plus eight. So negative
five and negative six is negative eleven plus eight, and negative eleven plus eight is negative
three. Where I have seen that answer before? That was my remainder. So, what is the mean?
That means that when you are given a function and you say "HEY, find F of ten." I can plug
it into the function and work it out. Now negative one is a simple number and these
were very simple to come up with even with doing this in my head. But a lot of times
it is very hard to do this where you just say with a lot of high exponents, go tell
me what F of ten is or F of five. What you can do instead is set up synthetic division
and do the process with that synthetic division. And synthetic division.... Let me erase this
because I don't need this for the remainder theorem. I did for the original problem:)
But with synthetic division, you can say I want F of negative one so you synthetically
divide by negative one. The remainder is the answer you would get from plugging in that
number into the function. Also note that if I say what is F of negative one, I used negative
one. I did not change the sign. And I think I have time, I might change a number here.
Ohhh Yeah! So let's just do that again. Let's say that I want... It is probably just easier
to rewrite it all. Let's say I want F of three. Well, that is five times three to the third
power plus six times three plus eight. Let's see if I mess this video up because I cannot
do this in my head. Three cubed, 3, 9, 27... so five times twenty-seven plus three times
six is eighteen plus eight. Five times twenty-seven. Well, mental math...if you are trying to do
mental math you should work left to right and not right to left like you do on your
paper. What is five times twenty? Five times twenty is a hundred and five times seven is
thirty-five. So five times twenty-seven is a hundred plus thirty five so one hundred
thirty-five plus...what is eighteen plus eight...that is twenty-six. Now what is 135 plus 26. Well
it is 135 plus 20 which is 155 and 155 plus 6 is 161. So we got through that without the
aid of a calculator. What would that look like synthetic division? Well I am asked to
figure out what F of three is so I am going to synthetically divide by three. Five for
the third degree, zero for the non-existent second degree, six for the first degree, and
eight. Let's see if we can do this. Remember it is addition here and multiplication here.
So bring down the five and multiply. Three times five is fifteen and add on this line.
Zero and fifteen is fifteen. Now multiply, three times fifteen, that is forty-five. And
add, that is now fifty one. Now multiply. Three times fifty-one. Three times fifty is
150 and three times one is three so 153. Three times fifty-one is a hundred and fifty-three.
Finally 8 plus 153 is 161. WHOO!! Alright, the synthetic division gave us the same answer
as when we plugged in the three. That is synthetic division. That is the remainder theorem. BAM!!!
Thank you very much:)