WEBVTT
00:00:02.330 --> 00:00:04.100
Multiplying binomials.
00:00:04.500 --> 00:00:11.640
In this video, we will learn how to multiply two binomials using the FOIL method and the area method.
00:00:12.560 --> 00:00:16.310
Letβs start by reminding ourselves of what a binomial is.
00:00:16.650 --> 00:00:21.440
A binomial is an expression that contains the sum or difference of two terms.
00:00:21.660 --> 00:00:25.560
So, for example, three π₯ squared plus five π₯ is a binomial.
00:00:25.860 --> 00:00:28.190
And so is π¦ minus seven.
00:00:28.760 --> 00:00:33.730
So if we have two binomials and we want to multiply them, what would that look like.
00:00:33.990 --> 00:00:42.890
For example, if we have the binomial π₯ plus two and we multiply it with binomial π₯ plus three, we can write this with a multiplication symbol in the middle.
00:00:43.200 --> 00:00:50.620
Notice that we have parentheses around each binomial so that we know that we multiply all of each one by all of the other one.
00:00:51.020 --> 00:00:56.990
We most commonly see this type of problem without the multiplication symbol between the two binomials.
00:00:57.510 --> 00:01:06.940
We may also see that multiplication of two binomials refer to as expanding, or expanding the product, or using the distributive property.
00:01:07.870 --> 00:01:11.890
Now, letβs talk about how to actually multiply our two binomials.
00:01:12.230 --> 00:01:14.530
There are a number of different methods for this.
00:01:14.790 --> 00:01:17.330
But here, weβre going to use two different ones.
00:01:17.470 --> 00:01:20.040
The first method is the FOIL method.
00:01:20.240 --> 00:01:28.900
Here, the acronym FOIL represents First, Outer, Inner, and Last and refers to the position of the terms in the binomial.
00:01:29.390 --> 00:01:32.240
The second method is the area or grid method.
00:01:32.510 --> 00:01:37.190
And this method of multiplication is commonly seen for multiplying integers.
00:01:37.950 --> 00:01:44.830
So letβs take a look at an example of multiplying two binomials and see how we would apply both of these methods.
00:01:47.250 --> 00:01:51.490
Expand the product two π₯ plus one, 3π₯ minus two.
00:01:52.760 --> 00:01:58.110
In this question, the word product indicates that weβll be multiplying our two binomials.
00:01:58.410 --> 00:02:03.310
Weβre going to use two different methods to demonstrate the expansion of these binomials.
00:02:03.430 --> 00:02:05.230
The first is the FOIL method.
00:02:05.260 --> 00:02:07.830
And the second is the area or grid method.
00:02:08.160 --> 00:02:16.230
So starting with the FOIL method, this means that we take our first, then the outer, then the inner, and then the last terms and multiply.
00:02:16.550 --> 00:02:21.260
Looking at our binomials, the first terms are the two π₯ and the three π₯.
00:02:21.490 --> 00:02:25.000
So we multiply those, writing two π₯ times three π₯.
00:02:25.390 --> 00:02:29.330
The outer terms will be our two π₯ and our negative two.
00:02:29.530 --> 00:02:32.460
So we add two π₯ times negative two.
00:02:32.760 --> 00:02:36.350
The inner terms will be our plus one and our three π₯.
00:02:36.820 --> 00:02:39.760
Therefore, we add our one times our three π₯.
00:02:40.260 --> 00:02:44.620
And finally, the last two terms will be the plus one and the negative two.
00:02:45.030 --> 00:02:48.010
So we add plus one times negative two.
00:02:48.440 --> 00:02:51.210
We may not always need to write this line of workings.
00:02:51.460 --> 00:02:54.130
But it can be helpful to illustrate our method.
00:02:54.810 --> 00:02:56.670
And now, we work out the products.
00:02:57.470 --> 00:03:01.580
Two π₯ times three π₯ is six π₯ squared.
00:03:02.250 --> 00:03:11.890
Two π₯ times negative two is negative or minus four π₯ plus one times three π₯ is equivalent to plus three π₯.
00:03:12.300 --> 00:03:16.540
And finally, plus one times negative two is negative two.
00:03:17.080 --> 00:03:22.970
The next thing to do at this stage of the question is to simplify by collecting any like terms.
00:03:23.480 --> 00:03:25.900
Here, we have two terms in π₯.
00:03:26.320 --> 00:03:35.790
So we can write this as six π₯ squared minus π₯ minus two, because our negative four π₯ plus three π₯ is equivalent to a negative π₯.
00:03:36.490 --> 00:03:40.310
Letβs see how we would get the same answer using the grid method.
00:03:41.020 --> 00:03:46.640
To set up our grid, we put one binomial horizontally and one binomial vertically.
00:03:47.040 --> 00:03:51.010
The first binomial will give us two π₯ and one horizontally.
00:03:51.230 --> 00:03:56.830
And the second binomial of three π₯ minus two can be split into the terms three π₯ and negative two.
00:03:57.480 --> 00:04:03.980
We then calculate the product of each row and column, starting with three π₯ times two π₯.
00:04:04.010 --> 00:04:06.290
This will give us six π₯ squared.
00:04:06.570 --> 00:04:10.220
Next, three π₯ times one is three π₯.
00:04:10.660 --> 00:04:15.630
In our second row, negative two times two π₯ is negative four π₯.
00:04:15.960 --> 00:04:19.340
And finally, negative two times one is negative two.
00:04:19.870 --> 00:04:25.820
We then simplify by adding all four terms and seeing if we can collect any like terms.
00:04:26.200 --> 00:04:28.490
We start with six π₯ squared.
00:04:28.860 --> 00:04:37.520
We then have a plus three π₯ and a negative four π₯, which simplifies to negative π₯ and then our final term of negative two.
00:04:38.370 --> 00:04:45.190
So using either of our methods would give us the answer of six π₯ squared minus π₯ minus two.
00:04:47.500 --> 00:04:53.030
In the next example, weβll see a problem where the binomials include a number of different variable terms.
00:04:53.360 --> 00:04:59.700
However, we can use the same process and the same methods to get an answer for the expansion of these two binomials.
00:05:02.280 --> 00:05:08.020
Use the distributive property to fully expand two π₯ plus π¦, π₯π¦ minus two π§.
00:05:08.020 --> 00:05:19.120
In this question, the phrase βto use the distributive propertyβ means that we can multiply the sum by multiplying each addend separately and then adding the products.
00:05:19.700 --> 00:05:23.150
We could use a number of methods to expand the brackets.
00:05:23.230 --> 00:05:25.680
Here, weβre going to look at two of those.
00:05:26.000 --> 00:05:29.820
The first method is the FOIL method, which is an acronym.
00:05:29.910 --> 00:05:35.240
We can recall that it stands for the First, the Outer, the Inner, and the Last terms of our binomials.
00:05:35.620 --> 00:05:43.560
So with our two binomials set, we start by multiplying the first two terms, which is two π₯ times π₯π¦.
00:05:43.820 --> 00:05:47.790
Next, our outer terms will be two π₯ times negative two π§.
00:05:47.790 --> 00:05:54.770
Itβs always worth taking extra care when we have the variable π§, because we donβt want to confuse it with the digit two.
00:05:55.500 --> 00:05:59.760
Next, our inner terms will be π¦ multiplied by π₯π¦.
00:06:00.130 --> 00:06:05.280
And finally, we add on the product of our last terms, which is π¦ times negative two π§.
00:06:06.060 --> 00:06:08.420
We now need to simplify our products.
00:06:08.830 --> 00:06:11.330
The coefficient of our first term is two.
00:06:11.620 --> 00:06:14.600
π₯ multiplied by π₯ will be π₯ squared.
00:06:14.860 --> 00:06:16.020
And we have our π¦.
00:06:16.020 --> 00:06:19.390
So the first term simplifies to two π₯ squared π¦.
00:06:19.790 --> 00:06:26.520
The coefficient of our second term is found by multiplying plus two by negative two, giving us negative four.
00:06:26.770 --> 00:06:28.700
And we then have π₯ times π§.
00:06:28.700 --> 00:06:34.770
Our third term simplified will be π₯π¦ squared since we have π¦ times π¦.
00:06:35.090 --> 00:06:38.410
And our final term will be negative two π¦π§.
00:06:38.410 --> 00:06:45.900
At this stage, we always check if we can simplify our answer by collecting any like terms.
00:06:46.350 --> 00:06:49.940
And as there are none here, this would be our final answer.
00:06:50.300 --> 00:06:56.330
As an alternative method, we could use the area or grid method to expand our binomials.
00:06:56.940 --> 00:07:03.740
To set up our grid, we split up our binomials into their component terms with one on the row and one on the column.
00:07:03.890 --> 00:07:06.360
And it doesnβt matter which way round we put them.
00:07:06.700 --> 00:07:11.820
To fill in each cell in our grid, we multiply the row value by the column value.
00:07:12.150 --> 00:07:19.510
So taking our first cell, we have π₯π¦ times two π₯, which would be two π₯ squared π¦.
00:07:19.880 --> 00:07:25.680
For our next cell, we multiply π₯π¦ by π¦ to give us π₯π¦ squared.
00:07:25.900 --> 00:07:31.910
On our next row then, we have negative two π§ times two π₯, which would give us negative four π₯π§.
00:07:31.910 --> 00:07:36.450
And our final grid value would be negative two π¦π§.
00:07:37.250 --> 00:07:51.110
To find our answer from this grid, we add our four products, giving us two π₯ squared π¦ minus four π₯π§ plus π₯π¦ squared minus two π¦π§, which is the same as we achieved using the FOIL method.
00:07:52.510 --> 00:07:58.750
Letβs now look at a question where we subtract the expansion of two binomials from another term.
00:08:01.760 --> 00:08:07.190
Expand and simplify seven minus three minus π¦, π¦ plus two.
00:08:08.180 --> 00:08:13.320
In this question, we can see we have two binomials which are multiplied together.
00:08:13.520 --> 00:08:18.420
Weβre going to expand these binomials first and then subtract the answer from seven.
00:08:18.770 --> 00:08:22.340
Weβre going to use the grid method to multiply these binomials.
00:08:22.550 --> 00:08:25.890
And notice the negative in front of them isnβt included.
00:08:26.280 --> 00:08:31.250
Setting up our grid, we have three minus π¦ as our first binomial.
00:08:31.460 --> 00:08:35.290
Our second binomial can be split into the terms π¦ and two.
00:08:35.660 --> 00:08:38.390
And we can write that with or without the plus sign.
00:08:38.800 --> 00:08:42.090
It doesnβt matter which way round do we put our binomials.
00:08:42.750 --> 00:08:48.170
Filling in our grid, we start with π¦ times three, which is three π¦.
00:08:48.550 --> 00:08:53.760
Next, we have π¦ times negative π¦, which will give us negative π¦ squared.
00:08:54.110 --> 00:08:57.380
On the next row, we have two times three, which is six.
00:08:57.670 --> 00:09:03.820
And the final term is calculated by two times negative π¦, which is negative two π¦.
00:09:04.290 --> 00:09:09.330
To take our answer from the grid then, we add together the four products weβve just calculated.
00:09:09.650 --> 00:09:14.090
Starting with our largest exponent of π¦, we have negative π¦ squared.
00:09:14.510 --> 00:09:19.150
Next, we notice that we have two terms in π¦, which we can collect together.
00:09:19.310 --> 00:09:23.390
So three π¦ plus negative two π¦ will give us plus π¦.
00:09:23.750 --> 00:09:27.840
And then, we add on our final term from the grid which is plus six.
00:09:28.340 --> 00:09:40.740
So to answer the question then of seven minus three minus π¦, π¦ plus two, we replace what weβve worked out in our expansion, giving us seven minus minus π¦ squared plus π¦ plus six.
00:09:40.950 --> 00:09:46.580
Itβs important to include all of these in parentheses since we want to subtract all of these terms.
00:09:47.050 --> 00:09:52.420
It can be helpful to write the next line where we distribute the negative across all the terms.
00:09:52.790 --> 00:09:58.270
So we have seven minus minus π¦ squared, which is plus π¦ squared.
00:09:59.000 --> 00:10:02.320
A negative times plus π¦ will give us negative π¦.
00:10:02.570 --> 00:10:06.470
And finally, a negative times plus six will give us negative six.
00:10:07.020 --> 00:10:10.580
To simplify then, we check if there are any like terms.
00:10:10.860 --> 00:10:20.170
We can see that we have a seven and a negative six, which is equivalent to one, giving us a final answer of π¦ squared minus π¦ plus one.
00:10:20.600 --> 00:10:22.930
We could write these terms in any order.
00:10:23.200 --> 00:10:27.430
But by convention, we usually write them in decreasing exponent values.
00:10:27.820 --> 00:10:32.500
Here, we have our π¦ squared first, then our π¦ term, and then our constant.
00:10:34.080 --> 00:10:38.520
In our next example, weβll look at binomials that have higher exponent values.
00:10:38.800 --> 00:10:42.430
And weβre going to use the rules of exponents to help us solve the problem.
00:10:45.080 --> 00:10:53.760
Find π΄π΅ given π΄ equals five π₯ cubed minus three π₯ and π΅ equals negative six π₯ squared plus three π₯.
00:10:55.070 --> 00:11:00.050
To find π΄π΅ in this question, that means we need to multiply π΄ and π΅.
00:11:00.540 --> 00:11:09.310
So π΄π΅ can be found by multiplying all of five π₯ cubed minus three π₯ by all of negative six π₯ squared plus three π₯.
00:11:09.810 --> 00:11:16.910
Since π΄ and π΅ are both binomials, we can use a method for multiplying or expanding two binomials.
00:11:17.470 --> 00:11:28.170
So setting up a multiplication or an area grid, we can split up our binomials into their separate terms, remembering to include the negatives wherever they appear.
00:11:28.920 --> 00:11:33.820
To find the value in each cell, we take the product of the row and the column value.
00:11:34.340 --> 00:11:41.560
The coefficient of the term in the first cell can be found by multiplying negative six and five, which is negative 30.
00:11:41.840 --> 00:11:45.760
For our π₯ term, we multiply π₯ squared by π₯ cubed.
00:11:46.000 --> 00:11:50.520
And here, we may need to pause and remind ourselves of one of the rules of exponents.
00:11:50.630 --> 00:11:57.120
And that is that π₯ to the power of π times π₯ to the power of π is equal to π₯ to the power of π plus π.
00:11:57.320 --> 00:12:02.640
So our calculation π₯ squared times π₯ cubed is equal to π₯ to the power of five.
00:12:02.860 --> 00:12:07.920
Therefore, the first term in our grid is negative 30π₯ to the power of five.
00:12:08.430 --> 00:12:20.110
In the next cell of our grid, we have the coefficient, which will be equal to negative six times negative three, which will be 18 since two negative values multiplied will always give a positive value.
00:12:20.670 --> 00:12:29.090
For the π₯-value then, we have π₯ squared times π₯, which is equivalent to π₯ squared times π₯ to the power of one, giving us π₯ cubed.
00:12:29.370 --> 00:12:32.450
And so the whole term will be 18π₯ cubed.
00:12:33.000 --> 00:12:39.800
In the next row, we have three π₯ times five π₯ cubed, which will give us 15π₯ to the fourth power.
00:12:40.120 --> 00:12:45.230
The final term in our grid is calculated by three π₯ times negative three π₯.
00:12:45.450 --> 00:12:48.160
And thatβs equal to negative nine π₯ squared.
00:12:48.740 --> 00:13:01.770
To find the answer using the grid then, we add our four products together, giving us negative 30π₯ to the fifth power plus 15π₯ to the fourth power plus 18π₯ cubed minus nine π₯ squared.
00:13:02.160 --> 00:13:05.980
As there are no like terms, we canβt simplify this any further.
00:13:06.130 --> 00:13:08.810
So this is our final answer for π΄π΅.
00:13:11.130 --> 00:13:21.850
Before we summarize what weβve learned in this video, it may be useful to note that multiplying two binomials is part of the method we use for multiplying three or more binomials.
00:13:22.210 --> 00:13:24.650
Itβs outside the scope of this video.
00:13:24.650 --> 00:13:41.210
But as a rough guide, if we were multiplying three binomials, for example, π₯ plus three times π₯ plus seven times two π₯ minus five, we would expand two of the binomials, for example, to give us π₯ squared plus 10π₯ plus 21 and then multiply that by two π₯ minus five.
00:13:41.470 --> 00:13:45.250
Here, we could use the grid method to find our final answer for this.
00:13:46.110 --> 00:13:55.960
So in summary then, in this video, we learned how to multiply two binomials using two methods, the FOIL method and the area or grid method.
00:13:56.510 --> 00:14:00.040
We need to be careful when multiplying negative values.
00:14:00.650 --> 00:14:05.490
And finally, we may need to use an important exponent rule when multiplying binomials.
00:14:05.760 --> 00:14:12.590
That is that π₯ to the power of π times π₯ to the power of π is equal to π₯ to the power of π plus π.