Hello.
I am Mr. Tarrou.
Today we are going to be doing a relatively long problem.
Certainly you can see a lot of notes up here which is not normal for my PreCalc lessons.
We are going to solve some higher order equations.
Those equations that don't just have exponents of one and two, but exponents of maybe three
and four.
So the example we are going to eventually do has a degree of three.
But first there is a lot of steps that you need to take to solve higher order equations
without the aid of a graphing calculator or computer.
We will cheat a little bit on this Descartes Rule and just get a little tip on how to start
these though.
First thing that we want to do, is we are going to look at an equation which I don't
have any room to put up here on the board right now.
But...
And we are going to identify the highest exponent is bigger than two.
So how do you solve that?
Well the first thing is going to be put in General Form.
Get all of those exponents to count down by one.
The highest degree first, in my case it will be an exponent of 3 in this example, and the
constant out at the end.
And you want to list all of the possible rational zeros.
Now this is called the Rational Zero Theorem in your textbook.
And what you do is, you look at the constant and you break down all of factors of the constant.
That is the number at the end of the equation with no variable next to it and you take factors
of the leading coefficient, that is the number in front of the term with the highest exponent,
and you start making a list of possible rational zeros.
Remember rational number is a fraction like 7/2, or 5/2, or 5.
But rational is not imaginary and rational is not numbers like the square root of seven.
It is a fraction.
Don't forget to put a plus or minus.
So, you can end up with a long list of possible zero right off the bat at the beginning of
the problem.
That is where the time consuming part of this section if you are not getting help using
a graphing calculator to aid with your solutions.
By the way, this material was really relevant before graphing calculators.
Now days we can find solutions of any degree problem very very easily with the aid of a
graphing calculator.
Descartes Rule, this is a rule to help you figure out, if you are solving these questions
purely with no graphing calculator and it is all going to be trial and error with these
possible rational zeros, then the Descartes Rule is going to help you limit the number
of guess and checks that you have to do with synthetic division.
When you look at an original equation, the number of sign changes or that number reduced
by multiples of 2 is going to be the number of possible positive real zeros that the equation
has.
So if look through your equation and you notice that there is four sign changes going from
negative to positive and positive back to negative and so on, if you notice that there
is four sign changes then there is going to be possible four, two, or zero positive real
solutions.
Then you are going to take your equation and plug in negative x, the opposite of x, and
then re-simplify.
It is kind of like you are checking for even and odd functions.
If you plug in -x into your equation and simplify and look for your sign changes again, the
number of negative real zeros equals the number of sign changes or again that number reduced
by multiples of 2.
So if you plug in -x, you simplify your equations and you have only one sign change, then you
know that you only have one negative real solution.
That is where the graph crosses the x axis.
Then you want to...
Hopefully this Descartes Rule will limit the number of guess and checking that you have
to do.
You start doing synthetic division over and over and over again until you find that you
have a remainder of zero.
If you do synthetic division and have a remainder of zero, that is a place... if you do synthetic
division of three and you get a remainder of zero, well that graph is crossing the x
axis at 3 or whatever number you did synthetic division with.
Remember with synthetic division you were dividing by the form of x-c.
So you're synthetically dividing by a binomial with a degree of one.
If your original equation has a degree of three and you do synthetic division and you
find a zero remainder, then what is left at the bottom of your synthetic division will
not be a third degree like you started with but a second degree.
That would be nice, because we can solve second degree equations.
If the remaining factor, that is from the numbers at the bottom of the synthetic division...
the new polynomial, if the remaining factor has a degree greater than 2 then you are going
to have to keep doing synthetic division to try and eventually find another number that
gives you a remainder of zero... another x intercept, which then eventually one by one
you will bring that degree down until you get a degree of two.
And once you break down to a degree of two, then you can finish that with Quadratic or
factoring to finish solving the equation.
Well, that is a lot of words, but how do you actually apply it to a problem.
I am only going to have time to do one example in this video.
I will be doing other examples though, I just know I am going to run out of time for this
one.
Ok, so our equation of interest here that we want to solve is f(x) equals x cubed minus
3 x squared minus 13 x plus 15.
Ok, well our leading coefficient is one.
That is pretty easy.
Factors of one, you only have one choice.
Factors of the constant at the end of the equation is 15.
We have 1, 3, 5, and 15.
So I am going to make a list of all my possible rational zeros.
My step one of possible rational zeros are plus or minus 1/1, 3/1, 5/1, and 15/1.
Now if my leading coefficient was 2 and it was 1 and 2, then it would be 1/1 and 1/2,
3/1 and 3/2, 5/1 and 5/2, and then 15/1 and 15/2 again that would be if the leading coefficient
were 2.
You need all of the factors of 15 to be paired up over all of the factors of your leading
coefficient.
Of course mine is one so I have a relative short list.
But look in my relatively short list I still have a possibility of guess and checking 8
times.
I am not going to do all of that guessing and checking on the video because I would
never finish with the example.
Ok, what does Descartes Rule say?
Descartes Rule says to look at the original f(x) and look at the sign changes.
I go from positive to negative, so that is one sign change, I stay negative and I go
back to positive.
There are two sign changes.
That means that there is either 2 or 0 positive real solutions.
Ok.
What about f(-x)?
Well that is going to be negative x to the third power minus three times negative x to
the second power minus thirteen times negative x plus fifteen.
That is going to be -x^3-3x^2+13x+15.
Now look at what we have got.
We have got negative to negative to positive to positive.
There is only sign change here which means that we have negative real solution.
That means that... you know... maybe I just want to do synthetic division with the negative
numbers...
Negative one, negative three, negative five, and negative fifteen.
Because as soon as I find one I know I have found the only one and I can throw out all
of the negative numbers in that list if I had to do more synthetic division.
That would really cut down the amount of times I have to do trial and error with my synthetic
division looking for a remainder of zero.
Now of course with graphing calculators as an aid, we don't really have to do a lot of
guess and checking and plus for time I am going to say that the first number that works
is going to be negative three.
So let's do synthetic division with negative three.
I am just going to be really lucky and get the right number to do synthetic division
by the first time.
Start doing step number three where I said do all of this trial and error...
Maybe your teacher is going to go ahead and let you figure out what the first or second
solution is so that you can do synthetic division and you can get down to the second degree
faster and just finish it with Quadratic Formula or factoring if they are trying to save you
a bit of busy work in your homework.
So we are going to synthetically divide by negative three.
Our factors here are....
Our numbers when we do synthetic division, and your exponents need to count down by one...
If there is a missing exponent don't forget to fill it in with zero.
So I have got 1, -3, -13, and 15.
Don't forget with synthetic division the second line is addition and the last line, when you
get to that last line you are multiplying.
We are going to bring down the one with addition.
1 plus 0 is 1.
Then multiply through the bottom, -3 times 1 is -3, then add.
-3 plus -3 is -6.
And then multiply and we get 18.
-13 and 18 is 5.
And multiply, -3 times 5 is -15.
Viola, we have a remainder of zero.
Again this remainder of zero means that -3 is a solution or an x intercept.
That what solutions, zeros, roots are... they are x intercepts.
So this graph, this third degree polynomial is crossing the x axis at negative three.
Ok, so now how do you finish it.
We had a third degree.
Synthetic division again is dividing in the form of x-c, x to the first degree.
That means that this is 1x^2-6x-5.
Again it is equalling zero, not because of that remainder but because we are finding
solutions to the higher order equation.
Those are x intercepts, that is when y equals zero.
Now this is a second degree.
I am going to solve this with factoring here in a second.
If it were a third degree I would have to set up synthetic division again and check
another rational zero until I got it down to a second degree.
(same number may work more than once) In other words if this started off as a fourth degree,
this would be a third degree and we would have to do synthetic division again until
we got to a degree of two.
Well now we have the degree of two.
This one will factor.
It becomes (x-5...
no it does not.
That is a plus.
I made a sign error.
(x-5) and (x-1).
Then the solutions are going to be 5 and 1.
Or a lot of times with these problems because we are getting kind of advanced in math, they
are not going to give you ones that factor easily so you might have to do the quadratic
formula.
That is x equals negative b plus or minus the square root of b squared minus 4ac, a
is 1 and c is 5, all over 2 times a.
When you finish with this, this is going to be 6 plus or minus... this is 36 minus 20...
all over 2.
And 36-20 is 16.
It is going to be 6 plus or minus the square root of 16 over 2.
That is 6 plus 4 over 2 and 6 minus 4 over 2.
We are going to get the answers that we just saw a second ago.
That is 5 and 1.
The other solution came out from the synthetic division, and that other solution was -3.
So the three places where this polynomial is crossing the x axis is -3, 5, and 1.
You can check it with your graphing calculator.
WHOOO!
I got one question done [laughing] Kind of long, kind of long.