>> Welcome to the Cypress College Math Review on Factoring Trinomials Advanced.
In objective 1, you will learn how to factor quadratic type expressions using the AC method
when the leading coefficient is equal to 1.
In this first example, we have y to the 6th minus 14y to the 3rd plus 24.
To use the AC method, we need this to be a quadratic expression.
So we can do a temporary substitution where we let x equal y to the 3rd,
then x squared would be y to the 3rd squared which is y to the 6th.
And now we can make these substitutions in the original expression,
and it becomes x squared minus 14x plus 24.
Now here, we see the leading coefficient is 1, we know that we are going
to make the grid here to use the AC method.
And in the top here, we take the leading coefficient times the constant term which is 24.
We put the B term right here, negative 14, and because the AC term is 24
and the B term is negative 14 we know both factors of 24 will have to be negative.
So we start walking through the factors.
Negative 1 times negative 24 would add to be negative 25,
which is not the negative 14 we need, but right away we see
that negative 2 times negative 12 gives us the positive 24 but adds to be negative 14.
And because the leading coefficient is 1 here, these immediately give us our factors,
so we would get x minus 2 times x minus 12.
Now, remember that this was a temporary substitution
because the original expression was in terms of the y variable.
So we have to go back to that and wherever we see an x, we are going to put y cubed.
So this becomes y cubed minus 2 times y cubed minus 12.
Now we always need to FOIL out to check our factoring.
First times first gives us y to the 6th.
The outer inner term, we have a negative 12y cubed and a negative 2y cubed.
Those combined to give us negative 14y cubed.
And then last times last gives us positive 24, so we have a correct factorization.
In this example, we have to factor the quantity a minus 1
to the 10th plus 8 times the quantity a minus 1 to the 5th plus 15.
Now this trinomial can be turned into a quadratic type trinomial
and the way you recognize that is because the highest degree is always double
of the next highest degree.
When that happens, this can be a quadratic type.
So for example, we would let x equal the quantity a minus 1 to the 5th power.
And then the x squared term would be that quantity a minus 1 to the 5th squared
which is equal to a minus 1 to the 10th which is exactly what our highest degree is here.
Now if we make these substitutions,
then this trinomial becomes x squared plus 8 times x plus 15.
And now this looks like a very simple thing to factor.
So using the AC method, we make the X grid.
Our leading coefficient times the constant term is 15.
The B term is 8.
All the terms are positive so we know both factors will be positive.
And we know 1 and 15 would add to be 16, so that's not it because we need 8.
And we see quickly here that the factors need to be 3
and 5 so that they will multiply to be 15 but add to be 8.
And because the leading coefficient is a 1,
these two give us the factorization immediately, x plus 3 times x plus 5.
And remember, this is a temporary substitution that we made.
We have to go back and put it in the same form as the original expression.
Wherever we see an x, we're going to replace it with a minus 1 to the 5th power.
So in this first factor, then we would get a minus 1 to the 5th plus 3.
And the second factor would become quantity a minus 1 to the 5th plus 5.
Now, we always want to FOIL to check our answer.
First times first will give us a minus 1 to the 10th.
For the outer inner term, we're going to get 5 times a minus 1
to the 5th plus 3 times a minus 1 to the 5th and that gives us our 8 times a minus 1 to the 5th.
And then last times last will give us 15 so we have a correct factorization.
In this example, we will factor h to the 8th power minus 4 times h to the 4th,
k to the 3rd minus 12k to the 6th.
This is another one that we can turn into a quadratic type expression.
Notice that this exponent of 8 is double the exponent of 4 for the h variable.
And then the k variable, the exponent of 6, is double the exponent of 3.
So for this problem, we are going to end up having two variables that we substitute.
We could let a equal h to the 4th so that a squared would then be h
to the 4th squared which is h to the 8th.
We will also let b equal k to the 3rd so that b squared is equal to k
to the 3rd squared which is k to the 6th.
Now, when we do these substitutions, we replace h to the 8th with a squared minus 4 times h
to the 4th gets replaced with a, k to the 3rd gets replaced with b minus 12,
and then k to the 6th will get replaced with b squared.
We see the leading coefficient is 1 so we can use the short form of the AC factoring.
The leading coefficient times the last coefficient is negative 12.
The middle coefficient is negative 4.
So we need two numbers that multiply to negative 12 but add to be negative 4.
We could try 1 and negative 12 but that adds to negative 11.
And next, 2 and negative 6 multiplied to be negative 12 but add to be negative 4.
Because the leading coefficient is 1, these immediately give us the factors,
but we have to be careful here because we have two variables going on.
So this means the factorization will be a plus 2 times b. Because remember,
we have this b variable going on.
And then the second factor will be a minus 6 times b. Now we have to substitute back
to the variables that were in the original expression.
Wherever we see an a, we're going to replace that with h to the 4th.
Wherever we see a b, we're going to replace that with k to the 3rd.
The second factor then will become h to the 4th minus 6k to the 3rd.
Now we're ready to FOIL to check our answer.
First times first will give us h to the 8th.
The outer term is negative 6 h to the 4th k cubed.
The inner term is positive 2 h to the 4th k cubed.
So, negative 6 plus a positive 2 gives us that negative 4 h
to the 4th k cubed that we need in the middle term.
And then last times last gives us our negative 12k to the 6th.
So we have a correct factorization.
It's time to check your understanding.
So pause the video to try this one on your own, then restart the video
when you're ready to check your answer.
In this example, we have quantity x plus y
to the 4th power plus 35 times the quantity x plus y squared minus 36.
You see right away that this can be turned into a quadratic type trinomial
because the highest degree is double the next highest degree.
So we can do a substitution where we let a equal x plus y squared,
then a squared would be the quantity x plus y squared,
squared which would turn it into x plus y to the 4th power.
When we do these substitutions, x plus y to the 4th power, gets replaced with a squared.
We have plus 35 times the x plus y squared gets replaced with a and then minus 36.
This is a simple quadratic trinomial that we can factor with the AC method
where the leading coefficient is 1.
We put the constant term here of negative 36, the B term, positive 35,
and we get right away the factorization is going to be a minus 1 and a positive 36
because they multiply to be negative 36 but add to be 35.
Because the leading coefficient is a 1, these numbers immediately give us the factorization
and we get a minus 1 times a plus 36.
We have to substitute back to the original variables.
Everywhere we see an a, we are going to replace it with an x plus y quantity squared.
So in this first factor, we are going to get x plus y quantity squared minus 1,
and then the second factor we are going to get x plus y quantity squared plus 36.
Now let's use FOIL to check our answer.
First times first gives us quantity x plus y to the 4th,
the outer term is a positive 36 times the quantity,
the inner term is a minus 1 times the quantity so the positive 36
and the minus 1 give us the positive 35 times that quantity of x plus y squared.
And then last times last we get negative 36.
So we have a correct factorization.
However, on this problem, we're not quite done yet.
Did any of you notice that this first factor here is the difference of two squares?
Now, this didn't come up on any of our previous examples.
So if you didn't notice that, don't feel bad, but this is an additional step
because this is not quite factored all the way.
This first one is going to factor as x plus y plus 1 times x plus y minus 1 because this is
like having a squared minus b squared and we get a plus b,
a minus b. The second factor cannot be factored anymore, so we bring it right
down as x plus y quantity squared plus 36.
Now, we have a complete factorization.
And if you saw this one and did that factoring, great job.
If you didn't see it, that's OK.
Just a reminder that when you end up with the difference of two squares,
you do have to factor it further.
Below are some practice problems you can try on your own.
You may either pause the video to work on them now or write them down to work on them later.
After a few seconds, the answers will be revealed.
In objective 2, you will learn how to factor quadratic type expressions using the AC method
when the leading coefficient does not equal 1.
In this first example, we have 12x to the 6th plus 8x to the 3rd minus 7.
And you can see that it's going to be a quadratic type expression
because the highest exponent 6 is double the middle exponent of 3.
Now we can do that temporary substitution where we let a equal x to the 3rd,
then we know a squared would be x to the 3rd squared which is x to the 6th.
So we can rewrite this expression as 12a squared plus 8a minus 7.
Let's make our X grid.
We know the AC term, 12 times negative 7, is negative 84.
The B term is positive 8 because the factors have to multiply
to be a negative but add to be a positive.
We know that the larger magnitude factor will be the one that we make positive.
So let's walk through the factors of negative 84 and see which ones work.
If we do negative 1 times 84, we know that adds to 83 which is not 8, negative 2 times 42 adds
to be 40 which is not 8, negative 3 times 28 adds to 25 which doesn't work,
negative 4 times 21 adds to be 17, again, doesn't work, but we see when we get
to negative 6 times a positive 14, those two multiply together to be negative 84
but they do add to be positive 8.
Now when the leading coefficient does not equal 1, these two numbers tell us how to break
up that middle term into minus 6a plus14a.
We know the leading term comes right down 12a squared, the third term comes right
down minus 7, and now we're ready to factor two by two.
From the first two terms, we can factor out 6a and we would have left 2a minus 1.
Let's check that.
Six a times 2a is 12a squared, 6a times a negative 1 is negative 6a.
Whatever math symbol is here, in this case a plus sign, comes right down.
And from these two, we can factor out a 7 and we have left 2a minus 1.
Once again we check, 7 times 2a is 14a, 7 times negative 1 is negative 7.
Now from these two terms, we can factor out 2a minus 1 and we have left 6a plus 7.
But remember, this all was a temporary substitution and we have to convert back
to the variables from the original expression.
Everywhere we see an a we will instead put x cubed,
so this becomes 2x cubed minus 1 times 6x cubed plus 7,
and we are ready to FOIL to check our answer.
First times first we have 2x cubed times 6x cubed is 12x to the 6th.
When we combine the outer and inner terms,
we'll get 14x cubed minus 6x cubed gives us the 8x cubed.
And then last times last gives us a negative 7, so we have a correct factorization.
In this example we have to factor 3 times the quantity y plus 2
to the 4th power plus 13 times the quantity y plus 2 to the 2nd power minus 10.
Once again, we recognize this as a quadratic type expression, because 4 is double of 2.
We're going to use the technique where we do a temporary substitution.
So let's let a equal y plus 2 quantity squared.
Then a squared would be that same quantity y plus 2 quantity squared,
squared which is the quantity y plus 2 to the 4th.
Now we can do those substitutions
and temporarily the expression will become 3a squared plus 13a minus 10.
This is now a simple AC method type factoring problem.
So we make our X grid.
The leading coefficient times the constant term,
3 times negative 10 is negative 30, the B term is positive 13.
They have to multiply to be negative but add to be positive so we know we're going
to have a negative times a positive and we need the larger magnitude factor to be positive.
So negative 1 times 30 adds to be 29, which is not 13,
but already negative 2 times a positive 15 multiplies
to be our negative 30 but adds to be positive 13.
These numbers when the leading coefficient is not 1, tell us how to break
up the B term into negative 2a plus 15a.
We know the leading term comes right down 3a squared, the constant term comes straight
down minus 10, and now we are ready to factor two by two.
From these first two, the only thing we can factor out is a and we would have 3a minus 2.
Always check a times 3a is 3a squared, a times a negative 2 is negative 2a.
Whatever math symbol is here, in this case a plus, we bring it down.
From these two, we can factor out a 5.
And we would have 3a minus 2.
We always check, 5 times 3a is 15a, 5 times negative 2 is negative 10.
Now from these two terms, we can factor out a 3a minus 2 and we have left a plus 5.
Now remember, this all was a temporary substitution.
So we have to convert back to the variables from the original expression
and everywhere we see an a we are going to substitute it
with the quantity y plus 2 squared.
So this first factor becomes 3 times the quantity y plus 2 squared minus 2.
And the second factor becomes quantity y plus 2 squared plus 5.
We are ready to FOIL to check our answer.
First times first gives us 3y plus 2 to the 4th.
Combining the outer and inner terms,
we get 15 times the quantity y plus 2 squared minus 2 times the quantity y plus 2 squared,
which is 13 times the quantity y plus 2 squared, and then last times last gives us the minus 10.
So we have a correct factorization.
In this example, we have to factor 8x to the 2/3 minus 26x to the 1/3 plus 15.
Once again, we recognize that this is also a quadratic type expression
because 2/3 is double of 1/3.
Using the technique where we do the temporary substitution, let's let a equal x to the 1/3,
then a squared would be x to the 1/3 quantity squared.
And remember, a power to a power you multiply.
So this becomes x to the 2/3.
Now we're ready to do the substitution and the expression becomes 8a squared minus 26a plus 15.
Using the AC method, we make the X grid.
The leading coefficient times the constant term 8 times 15 is 120 and the B term is negative 26.
Our two factors have to multiply to be a positive but add to be a negative,
so we know that they're both going to be negative.
So let's walk through the factors of 120 until we find the ones that work.
Negative 1 times 120 would add to be negative 121, that doesn't work.
Negative 2 times negative 60 adds to be negative 62 which does not work.
Negative 3 times negative 40 adds to be negative 43, getting close but still doesn't work.
Negative 4 times 30 adds to be negative 34.
Negative 5 times negative 24 adds to be negative 29, we're getting really close.
And we can see that negative 6 times negative 20 multiplied to be 120 add to be negative 26.
Now, if you can see these factors right away, then of course you just want to write them down
and you don't need to go through this whole list in order to get there.
However, I've found that students tend to do better when they go
through an organized list starting from 1 and working their way up then they don't end
up missing any factors that would work.
Now because the leading coefficient is 8 and not 1, these two numbers tell us how to break
up the middle term into negative 6a minus 20a, the leading term comes right down 8a squared,
and the constant term comes right down plus 15.
Now we're ready to factor two by two.
So from these first two, we can factor out a 2a and we would have left 4a minus 3.
Let's check it, 2a times 4a is 8a squared, 2a times a negative 3, negative 6a.
Whatever the math symbol is here, in this case a minus sign, we bring that down.
And then from these two, we can factor out a 5.
And when we do that division, we would have left 4a minus 3.
Let's check it.
Negative 5 times 4a, negative 20a.
Negative 5 times negative 3, positive 15.
Now the greatest common factor of these two terms is 4a minus 3
and we have left over 2a minus 5.
But remember, we did a temporary substitution so we will have to go back
to the variables in the original expression.
Everywhere we see an a, we will put instead x to the 1/3 power.
So this becomes 4 times x to the 1/3 minus 3 times the quantity 2 times x to the 1/3 minus 5.
And we're ready to FOIL to check our answer.
First times first gives us 8x to the 2/3 because x to the 1/3 times x
to the 1/3 means you add those together, 1/3 plus 1/3 is 2/3.
The outer inner terms become minus 20x to the 1/3 minus 6x to the 1/3 which is minus 26x
to the 1/3, and then a negative 3 times a negative 5 is a positive 15.
So we have a correct factorization.
It's time to check your understanding.
So pause the video to try this one your own, then restart the video
when you're ready to check your answer.
In this example, we have to factor 20x
to the 4th plus 36x squared y to the 8th plus 9y to the 16th.
Now this can be turned into a quadratic type expression because we notice that the exponent
of 4 is twice the exponent of 2,
and the exponent of the y variable 16 is twice the exponent
of the middle y variable which is 8.
So we are going to do a double substitution here.
So let's let a equal x squared so that a squared would be x squared,
squared which of course is x to the 4th.
We are also going to let b equal y to the 8th so that b squared would then be y
to the 8th squared, which is y to the 16th.
Now let's come over to the original expression and make those substitutions.
Then this would become 20 times a squared plus 36 times a times b plus 9 times b squared.
Using the AC method, we make the X grid,
and the leading coefficient times the last coefficient, 20 times 9, is 180.
The B term is a positive 36.
So we know that both factors will be positive because when we multiply and we get a positive,
when we add them we get a positive.
So let's work our way through the factors of 180 and find the ones that add to be 36.
We know 1 and 180 add to be 181, which doesn't work.
We know 2 times 90 adds to be 92, doesn't work.
Three times 60 adds to be 63, does not match that 36.
Four times 45 gives us 49, still doesn't work.
Five times 36 adds up to be 41, getting closer.
And then we see that 6 times 30 will multiply to be 180 but add to be 36.
Because the leading coefficient is not 1, these two numbers tell us how to break
up the middle term into 6ab plus 30ab.
And of course we're adding that 6.
The leading term comes straight down as 20a squared,
and the last term come straight down as plus 9b squared.
Now we're ready to factor two by two.
From these first two we can factor out a 2a.
And when we do the division, 20a squared divided by 2a is 10a plus 6ab divided by 2a is 3b.
Always check it.
Two a times 10a, 20a squared.
Two a times 3b, 6ab.
Whatever math symbol is here, which in this case is a plus, we bring it straight down.
Now from these two, we can factor out a 3b.
If we do the division, 30ab divided by 3b is going
to be 10a plus 9b squared divided 3b will be 3b.
Let's check it, 3b times 10a is 30ab, 3b times 3b is 9b squared.
Now from these two terms, the greatest common factor is 10a plus 3b,
and we have left 2a plus 3b.
And this was our temporary substitution.
So we have to return to the variables in the original expression.
Everywhere we see an a, we will trade it for an x squared,
so this first factor becomes 10x squared plus, everywhere we see a b, we are going to trade it
for y to the 8th so this second term in here becomes 3y to the 8th.
In the second factor here, we've got 2 times x squared which substitutes for the a,
plus 3 times y to the 8th which substitutes for the b,
and we're ready to FOIL to check our answer.
First times first we see we will get 20x to the 4th.
When we combine the outer term with the inner term, we get 30x squared y
to the 8th plus 6x squared y to the 8th which add to be 36x squared y to the 8th.
And then last times last we get 9y to the 16th.
So we have a correct factorization.
Below are some practice problems you can try on your own.
You may either pause the video to work on them now or write them down to work on them later.
After a few seconds, the answers will be revealed.
In factoring trinomials advanced, objective 3, you will learn how
to factor quadratic type expressions using trial and error.
In this example, we have 4x to the 1/2 minus 13x to the 1/4 minus 12.
We recognize this as a quadratic type expression because the highest degree 1/2 is double 1/4.
So we will do that temporary substitution where we let a equal x to the 1/4,
a squared will then equal x to the 1/4 squared which is x to the 2/4 which is x to the 1/2.
Now let's make those substitutions
and our expression becomes 4 times a squared minus 13 times a minus 12.
And now for this, we're going to factor using trial and error.
So let's look at the factors of 4.
It could be 1 times 4 or 2 times 2.
For the factors of 12, we have 1 times 12, 2 times 6, or 3 times 4.
We have to pick one to start.
So maybe to get 4a squared, maybe we'll try doing the 2 times 2.
So we'll try 2a times 2a.
And for 12, we have three choices.
Maybe we would try 3 times 4.
Now we know one of them is going to be positive and one is going to be negative
because of this negative 12 here.
Before we decide which one should be negative, let's take a look at the cross products.
We get 6a here.
We get 8a here.
Now because the middle term is negative, we know we're always going
to want the larger cross product to be negative, which would make this 4 negative.
However, when we add these up, we get a negative 2a where we needed negative 13a.
So maybe we will try trading out the 3 and the 4 for a 2 and a 6.
So we'll say 2a times 2a and we'll put in a 2 and a 6.
The first cross product is 4a, the second cross product is 12a,
and we always want the bigger one to be negative since the middle term is negative.
That would make that a negative 6.
This is a negative 12.
When we add these two we get negative 8a.
That's not the negative 13a that we need.
So maybe instead of using 2 times 2, we'll try doing 1 times 4.
And if we do a times 4a, maybe we would try the 3 and the 4 here.
This first cross product is 12a.
The second cross product is 4a.
If we make the bigger one negative, we would get negative 8a, which doesn't work.
We could try reversing these.
If we do a and 4a and switch the 4 and 3, this first cross product is 16a,
the second cross product is 3a, and we want the larger one to be negative.
Now when we add these two together, we get negative 13a
which is exactly what we were looking for here.
So this now tells us that our factors are going to be a minus 4 times 4a plus 3.
Now remember, we did this temporary substitution here, so we need to go back to the variables
from the original expression and everywhere that we see a,
we are going to substitute with x to the 1/4.
So this becomes x to the 1/4 minus 4 times 4.
And this a gets substituted with x to the 1/4 again plus 3.
It's to FOIL to check our answer.
First times first we get x to the 4th times 4x to the 4th which is 4x to the 1/2.
Next we look at the outer and inner terms.
The outer is 3x to the 4th, the inner is negative 16x to the 4th.
Those two add to be negative 13x to the 4th.
And then last times last we get a negative 12.
So we have a correct factorization.
However, we're not finished factoring with this one because we noticed
that this first factor is the difference of two squares.
We know that when we have a squared minus b squared,
that's going to factor into a plus b times a minus b.
Now if a squared is x to the 1/4, then that means a will be x to the 1/8.
So this first factor will factor into x to the 1/8 plus 2 times x to the 1/8 minus 2.
The second factor cannot be factored anymore so we just bring it down 4x to the 1/4 plus 3.
And now, we have a complete factorization.
In this example, we have to factor 6 times the quantity x plus y
to the 6th plus 29 times the quantity x plus y to the 3rd plus 35.
We verify that it's a quadratic type expression because the exponent
of 6 is double the exponent of 3.
We will do that temporary substitution where we let a equal x plus y to the 3rd,
then a squared would be quantity x plus y to the 3rd that quantity squared
which is quantity x plus y to the 6th.
Now let's go ahead and do these substitutions
and the expression becomes 6 times a squared plus 29 times a plus 35.
We are electing to use trial and error in this objective.
So we take a look at the factors of 6 and we know we could use 1 times 6 or 2 times 3.
Next we look at the factors of 35 and that could be 1 times 35 or it could be 5 times 7.
Very often we'll start out by trying the factors that are closest together.
So let's do that and we would have 2a times 3a.
For 35, we could use 5 times 7.
And now it's time to check the cross products.
We know that all the terms are going to be positive
because the constant term is positive and the middle term is positive.
So if we look at this first cross product, we get 15a.
If we look at the second cross product, we get 14a.
And when we add those two together, we have 29a and that matches our middle term.
Now this means we can do the factorization as 2a plus 5 times 3a plus 7.
Now remember, this was a temporary substitution to make things easier to see.
And now, we have to return our variables to those in the original expression.
So everywhere we see an a, we will put instead quantity x plus y to the 3rd power.
So this first factor is going to become 2 times the quantity x plus y to the 3rd power plus 5.
And the second factor is going to become 3 times the quantity x plus y to the 3rd power plus 7.
Now we're ready to FOIL to check our answer.
First time first gives us 6 times the quantity x plus y to the 6th.
Next we do the outer and inner terms.
We have 14 times the quantity x plus y cubed plus 15 times the quantity x plus y cubed
which combine to give us the 29 times the quantity x plus y cubed.
And then last times last we get 35.
So this is a correct factorization.
In this example, we will factor 15x to the 14th plus 44x to the 7th minus 20.
We verify that this is a quadratic type expression because the highest exponent
of 14 is double, the next highest exponent which is 7.
We will use the technique where we do the temporary substitution and let a equal x
to the 7th, then a squared would be x to the 7th squared which is x to the 14th.
Now when we do the substitutions, this expression turns
into 15a squared plus 44a minus 20.
We are electing to use the trial and error method.
So what we need consider, the factors of 15 which are 1 times 15 or 3 times 5,
and the factors of 20 which are 1 times 20, 2 times 10, or 4 times 5.
Very often we elect to start out by using the factors that are closer together.
So let's try 3a times 5a and 4 times 5.
Because our constant term is negative,
we know that one of the cross products will negative and one will be positive.
Because the middle term is positive, we know that we will make sure that the largest
of the two cross products is the one that's positive and the smaller one is negative,
so that when we add them together we end up with a positive number.
The first cross product here we get 20 times a, the second cross product here is 15a.
We would make the smaller one negative which makes this 5 negative.
Now when we add these up, we get 5a which does not match our 44a
that we need, so this one didn't work.
We could try switching the 4 and the 5.
So maybe we will keep the 3a and 5a the same and we'll put the 5 here and the 4 here.
The first cross product is 25a.
The second cross product is 12a.
Make the smaller one negative.
Let's add those up and we get 13a which still is not the 44a.
Maybe we'll do 2 times 10 to get the 20.
So we'll leave the 3a and the 5a alone and we'll try 2 times 10.
These cross products give us 10a and 30a.
Make the smaller one negative.
Add those two up and we get 20a, which does not match the 44a.
Let's try switching these two around, 3a times 5a, put the 10 here and the 2 here.
This first cross product is 50a.
The second cross product is 6a.
Make the smaller one be negative.
When we add these two up, we get 44a which is exactly the middle term that we needed.
So that tells us that we can create our factors
to be quantity 3a plus 10 times the quantity 5a minus 2.
Now this was a temporary substitution that we did.
We will need to return to the variables of the original expression,
so everywhere that we see an a we will be replacing that with an x to the 7th.
So this becomes in the first factor, we will have 3 times x to the 7th plus 10.
And then the second factor, we will have 5 times x to the 7th minus 2.
Let's FOIL to check our answer.
First times first gives us 15x to the 14th.
The outer and inner terms will combine.
We have negative 6x to the 7th plus 50x to the 7th which adds to 44x to the 7th.
Then last times last is negative 20.
So we have a correct factorization.
It's time to check your understanding.
So pause the video to try this one on your own, then restart the video
when you are ready to check your answer.
In this example we have to factor 12x to the 4/3 minus 28x to the 2/3 plus 15.
We verify that this is a quadratic type expression because the exponent
of 4/3 is double the exponent of 2/3.
We will do a temporary substitution where we let a equal x to the 2/3,
then a squared would be x to the 2/3 squared.
Remember, when you take a power to a power, you multiply.
So this becomes x to the 4/3.
And now we can do these substitutions.
The original expression becomes 12a squared minus 28a plus 15.
Using trial and error, we need the factors of 12 which are 1 times 12, 2 times 6,
and 3 times 4 as well as the factors of 15 which are 1 times 15 or 3 times 5.
It's common to start off using the factors that are closer together.
So we can try 3a times 4a and 3 times 5 for the constant term.
Now because our constant term is positive but our middle term is negative,
we know that both factors will have to be negative.
And let's check the cross products.
This first one gives us negative 12a, the second one gives us negative 15a,
and those add to be negative 27a but that doesn't match our middle term of negative 28a.
Maybe we would try switching these two.
So we have 3a and 4a again but we'll try switching the negative 5 and the negative 3.
The first cross product is negative 20a, the second cross product is negative 9a which add
to be negative 29a, again, not quite what we need.
Maybe we'll try switching up.
Instead of 3a and 4a, maybe we'll try 2a and 6a.
So if we do those and we'll try the negative 3 and the negative 5 here,
the first cross product is negative 18a, the second cross product is negative 10a,
and those two add to be negative 28a which is exactly what we need.
So that tells us we can do our factorization here.
The first factor is going to be 2a minus 3 and the second factor is going to be 6a minus 5.
Now again, this was a temporary substitution and we need to return
to the variables in the original expression.
Everywhere we see an a, we will replace it with an x to the 2/3.
So this becomes 2x to the 2/3 minus 3 times 6x to the 2/3 minus 5.
Now we are ready to FOIL to check our answer.
First times first gives us 12x to the 4/3.
Combining the outer and inner terms, we have a negative 10x to the 2/3 plus a negative 18x
to the 2/3, which gives us our negative 28x to the 2/3.
And then last times last we get a positive 15.
So this is a correct factorization.
Below are some practice problems you can try on your own.
You may either pause the video to work on them now or write them down to work on them later.
After a few seconds, the answers will be revealed.
In objective 4, you will learn how
to factor quadratic type expressions with the greatest common factor.
In this example, we have to factor 24x to the 7th y squared minus 20x
to the 4th y to the 4th plus 4xy to the 6th.
Looking at these three terms, we see that the greatest common factor is 4xy squared.
So we will factor that out from all three terms, 4xy squared.
And what we have left, just do the division, 24x to the 7th y squared divided
by 4xy squared is 6x to the 6th minus, if we divide 20x to the 4th y to the 4th
by 4xy squared, we get 5x cubed y squared.
And if we divide 4xy to the 6th by 4xy squared, we get y to the 4th.
Now in this trinomial, we notice that we have a quadratic type expression because the exponent
of 6 is double the exponent of 3 and the exponent of 4
and the y variable doubles the exponent of 2 in the y variable.
So we will do that temporary substitution where we let a equal x cubed,
which means a squared would be x cubed squared which is x to the 6th.
We also need a second substitution where we let b equal y squared,
then b squared will be y squared, squared which is y to the 4th.
Next we will substitute these temporary variables into the original expression.
So let's bring down that 4xy squared.
We're just going to leave that alone.
We're going to do the temporary substitution here on this trinomial.
So this becomes 6a squared minus 5ab plus b squared.
Let's go ahead and practice trial and error method.
So what we'll need to do is come up with the factors of 6 which are either 1 times 6
or 2 times 3, and the factors of 1 here which will only be 1 times 1.
So, it's often customary to try the two factors that are closest together.
So let's try 2a times 3a.
And then of course the only way to get b squared is going
to be b times b. Now the third term is positive, which tells us that these two will have
to have the same sign, because the middle term is negative we know
that both of these will be negative.
And now let's check the cross products, 3a times a negative b is negative 3ab,
and the second cross product, 2a times negative b is negative 2ab.
When we add those up, we get negative 5ab which is exactly what we needed in the middle term.
So let's continue to carry down the 4xy squared.
And now we have our factorization of this already.
The first factor is going to be 2a minus b and the second factor is going
to be 3a minus b. Now remember, this was a temporary substitution that we did.
So we will need to return to the variables from the original expression
that 4xy squared is already in the original variables.
But now, everywhere that we see an a, we will be replacing that with x cubed
and everywhere we see a b, we will be replacing that with y squared.
So the factorization becomes 2 times x cubed minus y squared times the quantity 3 times x
cubed minus y squared.
Now we are ready to FOIL to check our answer.
And we'll be checking with this trinomial up here.
First times first gives us 6x to the 6th.
The outer and inner terms combine to be negative 2x cubed y squared minus 3x cubed y squared
which is the negative 5x cubed y squared.
And then last times last gives us a plus y to the 4th.
So we have a correct factorization.
In this example, we have to factor 30y times the quantity x minus 4
to the 6th plus 27y times the quantity x minus 4 to the 3rd minus 27y.
We can see that these three terms do have a greatest common factor,
which is equal to 3 times y. So we're going to factor that out right from the beginning,
3y and then we will have the quantity.
So let's do the division here.
This first term divided by 3y becomes 10 times the quantity x minus 4
to the 6th plus the second term divided by 3y becomes 9 times the quantity x minus 4
to the 3rd minus the third term divided by 3y becomes 9.
Now we recognize this is a quadratic type expression because the exponent
of 6 is double the exponent of 3.
So we'll use the technique where we let a equal the quantity x minus 4 to the 3rd power,
then a squared would be quantity x minus 4
to the 3rd power squared which is x minus 4 to the 6th.
And now, we can do those substitutions into this trinomial.
So first, we bring down the 3y, and then the trinomial becomes 10a squared plus 9a minus 9.
And once again, let's practice using trial and error, so we will need the factors of 10
which are 1 times 10 or 2 times 5.
We need the factors of 9 which are 1 times 9 or 3 times 3.
Let's start off using the two factors that are closer together.
And we have 2a times 5a.
We'll try these two, 3 times 3 because those are closer together.
Now because the constant term is negative,
we know that one of these will be negative and one will be positive.
The middle term is positive.
So when we check our cross products,
we will want to make the larger magnitude cross product be the positive
and the smaller one be the negative.
This first cross product is 15a, and the second cross product is 6a.
Now remember, we want the larger one to be positive and we'll take the smaller one
and change it to a negative, which means we have to make that 3 negative.
Now let's add these two up and we get 9a
which is exactly what we needed for our middle term over here.
So now we know we have the factors, let's go ahead and carry this down here.
We've got the 3y.
And now, this trinomial will factor as 2a plus 3 times the quantity 5a minus 3.
Remember that this all was a temporary substitution.
We will need to return to the variables of the original expression.
So this becomes 3y which is already in the original variables.
Everywhere we see the a variable, we are going to trade it
for the quantity x minus 4 to the 3rd power.
So this become 2 times the quantity x minus 4
to the 3rd plus 3 times 5 times the quantity x minus 4 to the 3rd minus 3.
Now let's FOIL to check our answer.
First times first gives us 10 times the quantity x minus 4 to the 6th which we see here.
Combining the outer and inner terms,
we get a minus 6 times the quantity plus 15 times the quantity,
which gives us 9 times the quantity x minus 4 to the 3rd.
And then last times last gives us that minus 9.
So we have a correct factorization.
In this example, we have to factor 20x to the 3/2 minus 31x to the 5/4 minus 9x.
We notice right away that these three terms have a greatest common factor
of x. So let's factor that out.
If we take 20x to the 3/2 and divide it by x, we get 20x to the 1/2 minus,
if we divide 31x to the 5/4 by x, we're going to get 31x
to the 1/4 minus, and then 9x divided by x is 9.
We notice right away that this is a quadratic type expression because the exponent
of 1/2 is double the exponent of 1/4.
So let's do that temporary substitution where we let a equal x to the 1/4,
then a squared will be x to the 1/4 squared which is x to the 2/4 which is x to the 1/2.
We continue to bring down this x. And if we do the substitutions,
our trinomial becomes 20a squared minus 31a minus 9.
Let's go ahead and practice factoring with trial and error.
The factors of 20 would be 1 times 20, 2 times 10, or 4 times 5.
The factors of 9 would be 1 times 9 or 3 times 3.
Let's start with the two factors that are closest together.
And try 4a times 5a.
And if we start with 3 times 3 to make the 9.
Now we have to be careful about our positive and negative signs because we need a negative
for the constant term, which means one of these will be negative and one will be positive.
We need a negative for the middle terms.
So that tells us that when we do our cross products, we'll want to make the larger
of the two cross products the one that's negative.
So let's see what we have here.
The first cross product becomes 15a, the second one is 12a.
If we make the larger one negative, then that means we have to make this 3 negative.
Now when we add these two together, we get negative 3a
which definitely does not match the negative 31a that we need.
And we can see even if we switched the signs of the 3s,
then all that would do is make this a positive 3a.
So that's not going to work.
So we either have to change the 4 and the 5 or we have to change the 3s.
You don't really know which one is going to be, but maybe we'll try leaving the 4 and 5 the same
but put a 1 and a 9 for the constant term.
So we've got 4a times 5a, and we will put a 1 and 9.
Now let's check the cross products.
This first one would be 5a.
The second one would be 36a.
We want the larger to be negative which mean the 9 has to be negative.
Now let's check our addition here, 5a minus 36a is negative 31a
which is exactly what we needed here for the middle term.
So if we continue here, this x comes straight down, and now we can factor this trinomial
into these two factors, we've got 4a plus 1 times 5a minus 9.
Now remember, this was a temporary substitution that we did, so we will need to return
to the variables from the original expression and this becomes x. Now everywhere we see an a
in these two factors, we will have to replace it with x to the 1/4.
So this first factor becomes 4x to the 1/4 plus 1.
The second factor becomes 5x to the 1/4 minus 9.
And now let's FOIL to check our answer.
First times first is 20x to the 1/2.
Combining the outer term and the inner term, we get negative 36x to the 4th plus 5x to the 4th,
which is negative 31x to the 4th.
And then last times last is negative 9.
So we have a correct factorization.
It's time to check your understanding.
So pause the video to try this one on your own, then restart the video
when you are ready to check your answer.
In this example, we have to factor 12x to the 9th minus 46x to the 5th plus 30x.
We see that these three terms have a greatest common factor of 2x.
So let's factor that out first.
If we take the first term and divide it by 2x,
we will have 6x to the 8th minus the second term divided by 2x is going to be 23x
to the 4th plus the third term divided 2x will be 15.
Now we see that this is a quadratic type expression because the exponent
of 8 is double the exponent of 4.
So let's do that temporary substitution where we let a equal x to the 4th,
then a squared would be x to the 4th squared which is x to the 8th.
Here we're going to bring down that factor 2x.
And then in the trinomial, we will substitute and get 6a squared minus 23a plus 15.
Let's go ahead and factor by trial and error.
The factors of 6 are 1 times 6 or 2 times 3.
The factors of 15 are 1 times 15 or 3 times 5.
As this customary, we'll start off using the factors closest together,
so we have 2a times 3a.
And for the 15, we'll use 3 times 5.
Now let's think about the positive and negative signs.
We see the constant term is positive,
which means our two factors have to have the same sign.
The middle term is negative, so that means they are both going to be negative.
And let's check the cross products.
The first cross product is a negative 9a, the second cross product is a negative 10a,
and those add up to be negative 19a which does not match the negative 23a that we need.
Well, maybe we'll try switching these two around.
So keep the 2a and 3a but switch the negative 5 with the negative 3.
The first cross product is negative 15a, the second cross product is negative 6a,
and those two add to be negative 21a which also does not match the negative 23a.
So maybe we will try 1 times 6, so we have a times 6a.
And we'll try the negative 3 times the negative 5.
The first cross product is negative 18a, the second cross product is negative 5a,
and those two add to be negative 23a which matches what we need here in the middle term.
So let's continue.
We bring down the 2x and then we factor this trinomial into a minus 3 times 6a minus 5.
Remember that this was a temporary substitution, so we will need to return
to the variables of the original expression.
We bring down that 2x which is already in the original variables.
Everywhere we see an a, we are going to replace it with x to the 4th,
so this factorization becomes x to the 4th minus 3 times the quantity 6x to the 4th minus 5.
Let's FOIL to check our answer.
First times first is 6x to the 8th.
Combining the outer and inner terms, we get negative 5x to the 4th minus 18x to the 4th
which is negative 23x to the 4th.
And then last times last is a positive 15, so this is a correct factorization.
Below are some practice problems you can try on your own.
You may either pause the video to work on them now or write them down to work on them later.
After a few seconds, the answers will be revealed.