BAM!!!
I'm Mr. Tarrou.
In this video we are going to learn how to factor by grouping.
Not the most common type of factoring.
We have factoring sum and difference of cubes, the difference of squares, and we do a lot
of factoring with regular quadratic equations.
The first step of factoring though is to try and factor out that monomial.
I have a video on that as well if you are struggling with it.
But, with this case all four of our examples that we are going to work on are going to
be factoring by grouping.
Now problems that involve factoring by grouping, generally have more than just two or three
terms.
They usually have four.
So, all of my examples are going to have four terms and we are going to just factor by grouping.
Now, it is kind of hard to describe steps to do this process so I am going to just hopefully
show you with these examples and help you understand how to do this process.
We are going to be looking at these expressions, or I can even write an equation at the end
of this or set this expression equal to zero and make be an equation, and we are going
to go ahead and solve this equation by factoring.
Again, that will be factoring by grouping.
Factoring by grouping is going to need to recall your skills of factoring out a monomial,
one of the very first things we teach you when we introduce factoring.
I am drawing this line down here because we are going to look at the first two and the
last two terms separately.
Kind of forgetting the rest of the equation is here, just looking at the 2 x cubed minus
2 x squared, what can be divided out or pulled out of those first two terms?
Well, they have coefficients which are both even so we can divide out a two, and I am
getting ahead of myself there, so two will divide into two and four evenly.
Now, that is for the coefficients.
Do these two terms have any common variables?
Clearly they both have an x, and how many can you take out of both?
How many x's can you ask both of these terms to give up?
This has three of them, as our exponent shows, and this has two.
So, I can take out an x squared from each of these terms.
Now, I am assuming if we are doing factoring by grouping that you have become comfortable
factoring out a monomial so I am not actually going to show the step where we are dividing
out this 2x^2.
I am going to just verbally do this.
Two x cubed, well two divided by two is one, and we can write that coefficient there or
not...it is not really going to do anything.
We have x cubed which means that there are three x's in this term, and we are taking
these three x's and we are divided out two of them, so x cubed divided by x squared is
going to leave you with just one of those x's left in that term.
Now going up to the second term, we have negative four divided by two is going to be negative
two, and x squared divided by x squared is one.
So...
This is bothering me that this one is in here.
So, we have 2x^2 times x minus two.
Now, this four terms expression, or now this equation that I have set equal to zero, if
it is truly factorable by grouping, then I need to be able to factor out something from
the last two terms and I need this x-2 to show up again.
If I cannot take a common factor out these last two terms and make the x minus two show
up again, then either I have a mistake...certainly if you are in a section in your textbook that
is, you know, factoring by grouping section, then you probably made a mistake.
If you are in a higher level class, then maybe this just does not factor by grouping.
So, we have a coefficients again of three and six, both of these numbers are divisible
by three.
See my x here is positive and my next term is positive, my number...my constant is negative
two...and over here my constant is negative six.
So, my signs are good.
I just need to take the three and factor it out of those terms.
I am going to factor out a positive three, and 3x divided by...
I am looking over here... 3x divided by 3 is 1x, and negative six divided by three is
negative two.
Excellent.
I have again, pulled the 2x^2 out of the first two terms, I have pulled a three or factored
out a 3 from my last two terms, and here I have got a matching factor.
I have this x-2.
Going back to the original problem, I have 1, 2, 3, four terms on the left side of this
equation.
Well, now I have got a lot of plusses and minuses still going on in my second line,
but these parenthesis are locking up my x-2 together.
So, this is has become one big rather complicated term.
This is not like two or three terms in here.
This is two times something times something with the parenthesis locking that x-2 together.
Addition and subtraction separates terms.
Again with the parenthesis locking this together, this is all acting as one term plus this second
term here.
Now, these two big terms separated by that addition sign, these to big terms both have
a common factor of x-2.
So, just like I factored out 2 x squared from these two terms, just like I factored out
three from these two terms, I am going to factor out an x-2 from these two terms.
So, we are going to see that common factor and divide it out.
2 x squared times x-2 divided by x-2 is just going to leave us with a two x squared.
Three times x-2 divided by x-2 is just going to leave us the plus three.
That is equal to zero.
Now, for these two factors to, and I am just...I just finished factoring by grouping on my
first example, for these two factors to multiply together to equal zero then one of these two
factors must be equal to zero.
So, either x-2 equals zero, or 2x^2 plus 3 equals zero.
Well, if I try to solve this we are going to subtract both sides by three, we are going
to need to divide both sides by two, and I can go ahead and stop.
Because, after I divide both sides by two, I am going to get x squared to equals NEGATIVE
three over two.
Well, unless you are studying complex numbers or imaginary solutions, you cannot square
root a negative number.
So this is done.
If I come over here and I add two to both sides, I have the answer of x equals two.
So, I found a real solution.
If I want to verify it, I can take this value of two and plug it into my original equation
and make sure that I get that answer of zero.
Over here I have got x cubed plus twenty x squared minus two x minus five.
We are not going to take that extra step and setting this equal to zero and solving the
equation.
I just wanted to show you that is what we generally do factoring for, solving equations.
Here are going to just factor this four term polynomial, this expression.
If you feel like you got that, you might want to stop the video and practice this on your
own and make sure that you can get it right.
What common factors do eight and twenty have?
They are both even so they are both divisible by two, but I need to take out the biggest
number.
These two are both divisible by four.
We are going to take out that four.
They both have x's and we can take out the lowest exponent, so we can take out an x squared.
8x^3 divided by 4x^2, eight divided by four is two and if we have three x's and we are
taking two of them out through division that is going to leave us with just one.
Twenty divided by positive four is positive five and x squared divided by x squared is
well...one...it just cancels out and we are left with 2x+5.
Now, see here we have a positive two x and a positive five, but my last two terms and...you
know...you might want to visually separate those two terms, we have negative 2x and a
negative five.
We need those two terms to be positive, so I am going to divide out a negative one.
Negative 2x divided negative one is positive 2x.
Negative five divided by negative one is positive five.
Sorry for the interruption there.
So, now we have again a minus sign separating these two big terms.
We have again the negative two divided by negative one is positive 2x and negative five
divided by negative one is positive five.
Now, these two terms, both have common factors of 2x+5 so we are going to factor out that
common factor of 2x+5.
Let's just show that we are doing that.
Four x squared times 2x+5 divided by 2x+5 is going to be 4x^2.
Negative one times 2x+5 divided by 2x+5 is negative one.
Now, we are not done factoring this.
You probably have also recently not only talked about factoring out a monomials, but you have
probably should have recently discussed or learned how to factor a difference of two
perfect squares.
We are going to have this as 2x+5, the square root of this first terms is 2x.
The square root of four, is 2 and the square root of x squared is x, so we have 2x.
The square root of one is one, so we have 2x+1 times 2x-1.
That is our example finished!
We have a little bit of factoring by grouping and a little bit of a review there of factoring
a difference of squares.
I have two more examples for you:)
Now, this third example looks very similar to the first two, and it is, but it poses
a little bit of a special problem that causes my students to make some common errors with
it.
So, let's see what happens.
Again, we have four terms, we have a third degree polynomial, so hopefully this is factorable
by grouping.
Between the six and seven there is no common factors so all we have the ability to factor
out an x squared.
So, we take that out.
6 x cubed divided by x squared is equal to 6x because there were three x's and I took
two x's out.
Negative 7x^2 divided by x^2 just gives us negative seven.
Ok, now here is the tricky part with this problem.
When you factor by grouping, you must show that you are factoring out a common monomial
from the first two terms and you must show that you are factoring out a common monomial
from the last two terms.
Well, I already have a perfect match, 6x-7 and 6x-7.
Well, then what?!
This is where the mistake generally happens with these problems.
The first example I could factor out a three, the second example I could factor out a negative
one, this example I must show with factoring by grouping that I am factoring out something
out of the last two terms...otherwise i am not going to know how to finish it...
I cannot just write a set of parenthesis and do 6x-7, so I need to divide something out
of these two terms that does not change the way that they look or change their values.
They have no common factors anyway, so what can I divide out of these numbers that is
going to keep them the same?
I can divide by one.
So, we are going to show that we are dividing out a positive one and that is for academic
purposes, I need to see that plus one so that I can finish my factoring by grouping process.
Six divided by one is just still 6x, and negative seven divided by one is just going to be negative
seven.
It appears that nothing has changed, but see I have that plus sign separating my two terms
that I want to again factor out a common factor in our two terms of 6x-7.
So, if I divide out 6x-7 from the first term I am left with just x squared, and if I divide
out 6x-7 from the second term I am just left with the positive one.
This is done!
I have finished my factoring by grouping and again I just showed this to you because when
the students see that match a lot of times they don't know how to continue.
Now this question here, I have got many many variables in the same algebraic expression.
But, it is four terms and generally when I am given four terms and I am asked to factor,
that is a factor by grouping process.
This would be no equation that I could solve even if I set it equal to zero.
This is just a problem where we want to practice that factor by grouping process again.
So, these those first two terms ab with a leading coefficient of one so there is no
numerical value that you can factor out, but these two terms both have a factor of a.
We are going to divide that a out.
ab divided by a is just going to be b, and minus 2ac... 2ac divided by a means that I
am taking that a and dividing it out...so we are left with 2c.
Ok, now can I take something again out of these last two terms to make this show up
again.
Well, these two terms both have a common factor of three, so we are going to factor out that
positive three.
3b divided by 3 is equal to b.
Negative six divided by positive three is negative 2c.
Voila, I have a matching factor in each of these two terms.
We again are going to take that out.
We are going to factor out that b-2c, and we have a plus three.
That is the end of my video.
So, I am Mr. Tarrou.
BAM!!!
Go Do Your Homework:)