BAM!!! Let's do some more factoring. We are
going to continue our factoring of quadratics,
specifically trinomials. We now have the form
of ax^2 + bx + c. Now notice that I have a
variable at the start of this trinomial is
a. That means that we are not just factoring
quadratic trinomials that have a leading coefficient
equal to one. That means that, well when I
went to school it was factoring by trial and
error. That way stinks!!! What we are going
to do instead is a little process, a little
memorization that you need to do to make these
questions much much simpler. When you get
done, it is always good to check your work,
but when you are done and you get your factored
answer it is going to be correct. We are going
to multiply the first value, the leading coefficient,
by the constant and just get some number.
I am not going to bother giving it some specific
variable, just a number. We are going to find
factors of that number that add to the middle
coefficient, that add up to the middle term.
We are looking to match, that is almost the
same color and I want to make sure the camera
picks this up. We want to match that middle
term. When we figure out what are the products
are, or the factors are of that product that
matches or adds up to the middle term, we
are going to rewrite this trinomial into four
terms. We are going to do that by taking the
middle term and writing it as two terms being
added together using the coefficients that
we got from step two. Then we are going to
factor by grouping. When we are done, if our
trinomials...our quadratic trinomials are
factorable, we will have a correct answer.
Quadratic means of course that the degree
is two. Again, remember that you are learning
all of these factoring processes initially
because you are going to use them to solve
equations. That means that you are going to
find out where functions cross the x axis,
find the roots, find the zeros, find the solutions.
We have 2 x squared minus 7x minus 15. We
are going to factor this, because our leading
coefficient is not equal to one, we are going
to take the leading coefficient and the constant
and multiply them together. This is going
to be a little bit of scratch work on the
side. Thank you for letting me get that interruption
there. We are going to take the first and
last number, we are going to take two times
negative fifteen, take the sign along with
it, that is equal to negative thirty. Now
we want factors of negative thirty that are
going to add to the middle term. So we have
one and negative thirty, negative one and
thirty, we have two and negative fifteen and
negative two and fifteen, we have five and
negative six... Let's see here. We have one
and negative thirty, two and negative fifteen,
three and negative ten. Ahh... three and negative
ten. We need to have negative.... What do
we have here. We have a middle term which
is negative, so we want a negative ten and
a positive three. That equals negative seven
when added. So, let's go. We are going to
take that and use it to rewrite the middle
term. We have two x squared minus ten x plus
three x minus fifteen. Notice that I am using
those scratch numbers on the side there. Now,
negative ten and three does equal negative
seven. So, we have not change the value of
our expression, we have just made it looked
different. Now we have to factor by grouping.
We want to take something out of the first
two terms, and we want to take something out
of the last two terms, and the greatest common
factor is what we are looking for. Between
2x^2 and negative 10x, two and ten are both
divisible by two and 2x^2 and -10x both have
an x. so we are going to factor out a 2x.
Now, 2x^2... We will show this step for now,
for the first example. 2x^2 divided by 2x
minus 10x divided by 2x. Then we need to take
something out of the last two terms. Well,
three and fifteen both divisible by three.
We are going to take that positive three out
and get 3x divided by 3 minus 15 divided by
3. We get 2x times, two divided by two is
one and x squared divided by x is equal to
one x, so we have x, minus ten divided by
two is five and the x's are going to cancel
out. This all plus three times, three divided
by three is one x, minus fifteen divided by
three is equal to five. Now, I am not going
to show this middle step of actually doing
the long written out division for the next
of the examples. You will see as I go through
those is after we factor by grouping, we are
looking for a perfect match, a perfect...the
same binomial that this in these parenthesis.
That is where your common factors should be
showing up if you have done this correctly
and this is factorable. We do now have these
parenthesis locking all of this together,
and the 2 and the x and the parenthesis all
being multiplied together. We have one big
term plus another big term so there is two
terms here and they both have a common factor
now of x-5. If I take that common x-5 out
through division, then this term is only going
to have 2x left and this term here is only
going to have a positive three. If you stop
and need to, or your teacher would like you
to distribute this back together, you will
see that it equals the original problem. This
is a somewhat complicated problem and there
was no trial and error. As soon as I am done,
I have the right answer. x times 2x is 2x^2.
x times positive three is plus 3x. Negative
five times two is negative ten, so negative
10x. Negative five times positive three is
negative fifteen. Guess what the two middle
terms add up to? Negative seven! So, we are
going to do two more examples I think. Yes,
two more examples and we will completely understand
and compared this concept. So here they come:)
This one except for some sign changes is very
similar to the previous one. Nothing really
fancy going on. So, we are going to take the
first and last numbers and multiply those
together. Let's get our green chalk out of
the scratch work. Four times negative three
is negative twelve. All of our factors of
negative twelve, we have one and negative
twelve, negative one and twelve, we have two
and negative six, we have three and negative
four, and we can turn all of those signs around.
What set of those factors add up to the middle
term of positive one? We want...oh. Actually
the problem gave us the correct set of factors
that we needed. Four times is negative three
not only equals negative twelve, but four
plus negative three is one. That is the middle
coefficient that we need to rewrite this trinomial
into four terms instead of just three terms.
So we have 4x^2, now look at your scratch
work plus four x minus three x.... That again
is equal to one, our middle coefficient or
b. That all minus three. What comes out of
the first two terms? A four and an x. We have
4x, and that is going to equal x plus one.
Four x squared divided by four x is one x.
Four x divided by four x is one. Now, these
two terms, they both have a three in them
so they can of course give that up. They are
both also negative. Inside our first grouping
have a positive x and a positive one, so I
need the signs to change. I am going to do
that by factoring out a negative three. Negative
three divided by negative three is equal to
one x. Negative three divided by negative
three is again equal to positive one. There
is our matching factors in our two terms.
We are going to pull that out and we are going
to get x plus one times four x minus three.
Now, I am still going to look at my notes
because I really wanted something to happen
in this example and the way that I wrote it,
I did not see it. This is done and this is
correct. But, what if you had written these
two terms in the opposite order. What if you
wrote for some reason negative three x and
positive four x? Who cares? What comes out
of the first two terms? Well now you can't
take a four out, but you still have an x that
is in each of these two terms. So, the x comes
out and 4x^2 divided by x is 4x, and -3x divided
by x is minus three. Now, here is why I wanted
to change those around besides the fact that
I wanted you to see that it does not really
matter what order you write those two middle
terms. I have a 4x-3 here. Ok, so what? Well,
I have also have a 4x-3 as my last two terms,
they already match. But, when you factor by
grouping, you have to show that you are taking
something out of the first two terms, and
something out of the last two terms. So, I
can't just write 4x-3 again. What I am going
to pull out of those last two terms? What
am I going to divide out of those two terms
that is not going to change them because I
still need to see 4x-3? Divide by one. So,
we are going to take a positive one out of
those two terms. Plus one times and I have
to show again that I am dividing out something
that is in common from those last two terms.
I am going to divide out a positive, but of
course dividing by one does not do anything...
like change negative three to negative four!
When I do that, now I have got a term here
plus another term here, both of those terms
have a common factor of 4x-3. So, I am going
to factor that out. x+1, x will be left here
in the first term and a positive one there.
I do believe that is the same answer that
I got a second ago. I have one more example
that has a few more variables in it so it
will be a bit more complicated. BAM!!! Are
you still with me? I have one more example.
We have 4x^2 minus 17xy plus 15y^2. The only
difference of course is that I have a variable
in my last term. Everything is going to work
the same. Just let that float along and we
should not have any issues as we go through
this problem. That means first and last coefficients,
so we are going to do four times fifteen which
is equal to sixty. Now we want factors of
60 that add to the middle term. I don't know
why I pointed to that. We are looking for
factors of 60 that add to -17. Now the middle
term is negative, so that means that I am
looking for...I need factors that are both
going to be negative and they need to add
up to negative seventeen. I have got... What
do I have in 60. I have one and sixty, I have
got two and thirty, I have three and twenty.
None of those add up to seventeen. But, I
also have five and twelve. That adds up to
seventeen. But, of course I don't want positive
seventeen, I want negative. Whenever your
last term has a positive sign on it, you do
need to use the same signs twice. That is
the only way that you multiply and get a positive
sign, if they are both positive or both negative
like they will be here. We have four x squared
minus five, five what... The middle term has
an x and a y in it, and it still will. Minus
5xy minus, using this over here remember,
minus 12xy plus 15y^2. What comes out of the
first two, not a number, but they both do
both have an x. So, x comes out. 4x^2 divided
by x is 4x. -5xy divided by x, the x's are
going to cancel giving you -5y. In the last
two terms, the GCF...the greatest common factor
between twelve and fifteen is three, and both
of these terms also have a y and you can take
the lowest exponent one more time. Not only
are we going to take out a three and a y,
but I need my first term to be what... Right
now it is negative, but in here I have a positive
4x not a negative 4x. I want to take out a
negative three, because I want that twelve
to become positive. So, negative twelve divided
by negative three is four. That is good. I
have x and a y, and I just took the y out
so that leaves me with just x. I almost wrote
a plus sign. Remember don't just running your
signs through. DO THE MATH! Fifteen divided
by negative three is negative five and y squared
divided by y is equal to y. Also, don't just
look over here and go, well that is 4x-5y
so that is 4x-5y. Actually do the math again
because, you sort of checking your own work.
If you make a mistake over here and you just
copy it over here, then you made the same
mistake twice and you messed up the whole
problem. Lordy Lordy, what is left. Well what
is left is that I am going to drop the chalk.
4x-5y and 4x-5y, again that comes out. This
term then will only have the x as you divide
out the 4x-5y, so x minus 3y. I am done. BAM!!!
Go Do Your Homework:)
