BAM!!! Mr. Tarrou. In this video we are going to once again continue our discussion of factoring
polynomials. Specifically we are going to be doing quadratic trinomials in this video
where we have a quadratic equation or expression that has a degree of two or the highest exponent
is two. In this specific video we are just going to be paying to leading coefficients
that equal one. That really simplifies these problems. In the next video we do these where
the leading coefficient is something greater than one or just not equal to one. When you
are factoring polynomials, when you are factoring trinomials that have a degree of two, and
the leading coefficient is equal to one, you want to look at the patterns of the signs.
If the constant has a positive sign, then you are going to be looking for factors of
the constant that add to the middle term of b. If you do that, you automatically get these
signs correct anyways, but let me just point this out to you any way. If your middle term
is positive and your constant is positive when you get your trinomial factored, if indeed
even factors, then you are going to have two positives inside your two factors...because
this trinomial is going to factor into two binomials. If you have x squared minus bx
plus c, then you are going to of course, to get that constant that is positive when you
distribute these binomials together to check your work you need to be multiplying numbers
that have the same sign, so they are both going to be negative when your middle term
has a negative sign in front. So, constant is positive, then your factors are going to
have the same sign and they are going to match whatever the middle term's sign is. Now if
your constant is negative, your constant is simply the term with no variable in it...it
is just a number, if your constant is negative then when you.... I hate to say FOIL, but
when you FOIL together those binomials or when you distribute together those binomials
the constant which are going to be... I guess in harder in problems they don't have to be
constants...but when you multiply whatever your second terms are in your binomials together
they need to have different signs to get that negative constant at the end of your trinomial.
We are going to have examples where the last term really is not a constant, they are more
advanced problems. You want those to have different signs in each of your factors. We
will be saying that the factors of the constant, that you are going to be looking for factors
of the constant that subtract to the middle term of b. Now you can also add a negative,
so really could be saying that, you know, when you are factoring quadratic trinomials
that you are always looking for factors of c that add to the middle term if you are saying
a plus negative, but plus negative is the same as subtraction. Your teacher may be using
slightly different language. I might even vary my language a little bit from these notes.
This is a nice little framework of what we will be doing. We have x squared plus two
x minus fifteen. We do have a constant in that last term, so it is a very standard problem.
This is factorable with the leading coefficient that is equal to one. It is factorable if
there factors of fifteen that subtract to give you a value of two, or add...adding a
negative that comes out to be the middle coefficient of two. What are your factors of fifteen?
Your factors of negative fifteen are going to be one and negative fifteen but those subtract
to be negative fourteen. If I were to reverse those signs I am going to get positive fourteen
and my middle term has a two in it, so I don't want that. Another set of factors for negative
fifteen are three and negative five. Now, three minus five is equal to negative two
and my middle term is positive two. I don't want to be negative, I want the three to be
negative. This is just academic. Here is negative three times five which is negative fifteen
and negative three plus five is equal to two. There I am saying the negative and plus, but
you are adding a positive and a negative number, so you know... It is like whatever right?!
You need a negative three and a positive five, those come together if you add them or subtract,
add the negative or subtract, whatever you want to say. Negative three and five comes
out to be two, unless of course you are multiplying and then it is negative fifteen. That means
that we are going to factor this like this, we are going to have a couple sets of parenthesis
and factors of your first term go into the first two blanks and factors of your constant
go in the last two blanks, if you will, of your binomials. The factors of x squared,
we are just going to square root that first term as far as the variable is concerned.
We are going to have x and x. Really what I am doing as well, is that I am looking for
factors of one which is just one and one. We will do some more advanced problems in
the next video. Then factors of fifteen, well the ones that we want are negative three and
five. Now, you really want to make sure that you check these answers. It is very easy to
get a sign error with these problems. I am going to go ahead, and let's see..what color
have I not used yet, we will do our check in green. Let's distribute this together.
X times x is x squared. X times positive five is 5x. Negative three times x is -3x. Then
negative three times five is negative fifteen. Look at this middle term, really that is...unless
you have really messed up...that is the only place you really have to worry about checking.
We have x squared and we have the negative fifteen, and positive five x minus three x
does indeed equal two x. This was our correct answer. I have lots of lines all over the
place now. X squared minus ten x plus twenty-one. We have another very straight forward problem.
Our last term is just a constant. It is just now my last term, or may last constant or
my constant, is positive as opposed to being negative. And do note by the way, with a negative
constant we did need those different signs. Now we are going to find factors of 21. The
factors of 21 are 1 and 21, we have 3 and 7 and that is pretty much it. Now I am writing
all positive numbers and when I said my constant was positive we need a set of factors and
they also need to add to that middle term. Well, I can't add two positive numbers and
get a negative ten. I have the ten, but it is negative. Well, to do multiplication and
get a positive number you need to have those two factors to have the same sign. I don't
need that positive three and positive seven, I need that three to be negative and I need
that seven to be negative. That is still going to be positive 21 and then when I add them,
negative three and negative seven comes out to be negative ten which matches our middle
term...at least the middle coefficient. Let's go ahead and get that in factored form. The
factors of your first term go in your first two blanks. Factors of your last term in go
in your last two blanks. We have factors of 1 are one and one, so we have 1x and 1x. The
factors of twenty-one that we need is negative three and negative seven. Again, if you want
to make sure that your answer is right before you have your teacher grade your work, distribute
it together and make sure that it equals the original problem. Now I have got three more
examples. They start to get trickier. I have two examples where my last term is not just
a constant. Then I have got another example where we have, well higher degrees. We are
going to do a fourth degree and see what is up with that. BAM!!! Keep on watching. For
our next two examples we are getting a little bit more complicated. We again have a trinomial
here, but you will notice that there is a variable in the first term and in the third
term which is different as far as the examples that I have done. Before we have had that
x squared term in the first term and then we showed up the square root of x or just
a single x or x to the first degree I mean in the middle term. Now, we have a y squared
in the last term and the square root of y squared is y and we do see that in the middle
terms as well. So, the fact that this last term has a y in it is not really a problem.
It is y squared and the square root of y squared is y and we see that in the middle term, just
like we have been seeing with the x. We are going to let that look like this... The square
root of x squared is x, so we are going to see that in the first two blanks or the first
two term in each of the binomials that we have here that our factors. The square root
of y squared is y. Then we are just going to continue factoring like I did with all
of the other problems. What factors of 24, this is a positive now again, so what factors
of 24 add to be fourteen. Well, we have 1 and 24, 2 and 12, 3 and 8, and 4 and 6. The
pair of factors that add to 14, the middle coefficient or b of 14, is going to be 2 and
12. Then, if the problem bothers you that the last term has a y in it, you are not really
sure that this is right, then of course again we can distribute this back together and make
sure that we get that middle term that we were looking for. X times x is x squared.
X times 12y is positive 12xy. Then 2y times x is positive 2xy. Then finally 2y times 12y
is two times twelve is there is the 24 that we needed and y times y which is y squared.
Look, there is 12xy and 2xy, and the last time I checked that adds up to be 14xy. Now,
the next example I have up here, is 3 x cubed minus 24 x squared minus 99 x. This is not
a quadratic, and that is what I said that we were factoring. However again, the first
step of factoring is to see if there is a GCF, a greatest common factor, that you can
pull out of all the terms in your function or expression you are being asked to factor.
Each of these terms have a factor of three, that alone is not really an issue except that
I have not done leading coefficients that are not equal to one in this video, it is
that x cubed. They all do have an x and we can take out the lowest exponent, the lowest
number of x's that are in one of the terms. That is x to the first power. We are going
to take a 3x out of these three terms. I am not going to show the division like I did
when we did the video about factoring monomials. But, three divided by three is one. You don't
need to write that if you don't want to. We are factoring out an x from each of these
terms. We have three x's here and we are taking one of them out and we are left with x squared.
24 divided by three is eight, so minus 8x because there is x squared and we are taking
one of those out. Following that, 99 divided by 3 is 33, and the only x that is in this
term I have now taken, or pulled out, or undistributed, or divided out with our 3x. Now, we can factor,
you know, x squared minus eight x minus 33. 33 is a little bit of a large number, but
hopefully you are not that bad at doing mental math. Some of my students are that are too
dependent on their calculators. We want factors of 33 that subtract to give you negative eight.
Well, this last term is positive, excuse me, negative. That means that one of my factors
and one of my factors needs to be negative. We are going to be looking for negative eleven
times positive three. Negative eleven times positive three is negative 33, and -11 plus
3 is equal to negative eight...that middle coefficient that we need to match up with.
In factored form we have factors of the first term going to your first two blanks which
are going to be x and x, and factors of negative thirty-three that we want is negative eleven
and positive three. That is that example in factored form. I got just one more example,
and we are going to be done with this video. So, for my last example is x to the fourth
minus x squared minus twelve. Now, this looks different of the other problems because it
is not a quadratic again like title of this section says. It is x to the fourth, so is
that really an issue? Well, I am going to factor it in this format, but I am going to
temporarily say let u equal x squared. So, I see an x squared here, but I am going to
rewrite this a little bit and write this x to the fourth as x squared raised to the second
power. Is that the same thing? When we have stacked exponents we are supposed to multiply
them. That is the power rule or the power property of exponents, and x squared to the
second power is x to the fourth minus x squared minus twelve. Now if I just temporarily let
this other variable represent x squared, we have u squared minus, well x squared I am
letting temporarily equal u, which is going to be u minus twelve. So by doing this temporary
substitution of letting u equal x squared, I can rewrite the equation, or excuse me expression
because there is no equal sign, I can rewrite this expression to be in that quadratic form
which is what we are factoring at the moment. Then factors of twelve that subtract to be
negative one are negative four and positive three. We can write this as u minus four times
u plus three. Then, remember that u was not in our original problem, it has variables
of x's and it was x to the fourth minus x squared minus twelve. I let u equal x squared
to rewrite the question to make it look simpler. I can take that x squared now and plug it
back in and get to this answer. Now, this may or not
be in your textbook, I am not sure. And, I am showing the substitution of u to make the
problem look different and you can still see that we are factoring a quadratic. I would
like to show you what this looks like without doing the substitution before I finish this
lesson. It is not really just a video right, I am trying to make lessons. I am not just
out dancing around, well.... ok, anyway:) You don't want to see my bad dancing. Another
way that you can identify that this is factorable, that this is still ok as far as this quadratic
type setting, is the first degree is four. If you take four and you divide it by two,
that is what square rooting does to your exponent, you get a degree of two. Four divided by two
is two. If you take that once again away you just get your constant at the end, where your
degree...you have an original degree in your first term, take that and you reduce it by
half, and then it just complete disappears. Like here four, two, and the variable is gone...or
if it was six, three, and the variable was gone and it was a constant at the end....or
ten, five, and then no variable in the last term, then you are still fitting that quadratic
pattern that you can use and just factor whatever your problem is. You will see that here. We
have a degree of two, and here we have a degree of one. Half of two is one. I am trying to
talk you many ways of seeing how this is still going to follow this pattern. I am going to
just write the answer. The square root of x to the fourth is x squared. The factors
of twelve, we have a lot of them...we have 1 and 12, 2 and 6, 3 and 4, but the factors
that subtract to equal negative one and give you a middle value of negative one and not
positive one is when four is negative and three is positive which is the same answer
that we have here from that temporary substitution of letting u equal x squared. Well, that is
how you factor quadratic polynomials, actually let's just call them trinomials since these
all had three terms when the leading coefficient is equal to one. What if it a number that
is not equal to one? That is coming up in the next video. But for now... BAM!!! Go Do
Your Homework:)