Multiplying Monomials
Lesson 31
In lesson 31, your students will be multiplying monomials again in the warm-up. They will be finding these products to review because they will be multiplying polynomials in the lesson. In our solve problem, the length of a rectangle is represented by the binomial 2x-4. The width of a rectangle is represented by the binomial 3x+2. What is the area of the rectangle? To study our problem, or “S” the problem, we will underline the question. What is the area of the rectangle? We will also ask ourselves, what is this problem asking me to find? This problem is asking me to find the area of the rectangle. Problem one, the quantity of x+1 times the quantity of x+2. We will represent our x+1 vertically with one long yellow rectangle, to represent the x and 1 small yellow square to represent the +1. We will then represent our x+2 with one long yellow rectangle to represent the x and 2 small yellow squares to represent the 2. Because we are finding the area, we are going to multiply our width times our length. Our x times x gives us an x2. Our x times 1 gives us and x. And our x times 1 gives us another x. We also have to multiply our 1 times x, which is x. 1 by 1 is 1. And 1 times 1 is also 1. So we have created our rectangle, and if you look at the product, our answer is, x2, put our like terms together, which is 3x are 2. To represent this pictorially, we represent the x+1 vertically, we will use 1 long yellow rectangle and 1 small yellow square. x+2 will be one long yellow rectangle and 2 small yellow squares. We have a width of x and a length of x, so we have x times x, or x2, x(1)=x, x(1)=x; we also have 1 (x) which is x; 1 times 1 is 1; 1 times 1 is 1. So our answer is the rectangle inside which is made up of 1x2…1-2-3 x’s…and 2 ones, or x2+3x+2. We can also find our answer algebraically using the box method. We will represent the x+1 vertically, and the x+2 horizontally… x(x)=x2; x(2)=+2x; x(+1)=+1x or just x; and +1(+2)=+2. You look at your diagonal, they are like terms… so we have 1x2+3x (because x+2x is 3x) +2.
In problem one, we have 2x2, which we will represent with 2 large yellow squares, plus 4x, which we will represent with 4 long yellow rectangles. Your students can arrange them into a rectangle. And once they have created their rectangle, we are going to identify the length and the width. If you notice we have x(x), so we need to use our long yellow rectangles so that we can make our x2; we also have x(x) to make our second large yellow square, so we will use another long yellow rectangle. Then we have x, with a product of x, so we need something times x to give us x, so that will be a 1 because 1(x)=x; and 1(x) is also x; and we need another 1 here. So we have now found our width and our length. Our first binomial is (x+2) and our second is (2x). In problem 1, we will put our 2x2+4x inside the box. And we have to ask ourselves, what is our greatest common factor. What term will divide evenly into both terms? 2x will divide evenly into both terms; and 2x2 divided by 2x is x; +4x divided by 2x is +2. So we have an answer of 2x (x+2), which is the same as (x+2) 2x. Because the commutative property of multiplication says that the order in which we multiply does not matter. In problem 2, we will represent our 3x2 with 3 large yellow squares; we will represent our 3x with three long yellow rectangles. We can see this is one way to make your rectangle. Line them up in their positions. We have an x2, so we need an x (x) to create our x2. With another 2x2 , we will need another x for each of them; and here we have an x(x), so we will use a 1 because 1(x)=x. We have now found the width and the length of our rectangle. So we see that our answer is (x+1) times 3x. With our commutative property we also know that is the same as 3x (x+1). To factor problem 2, we will put our 3x2 inside the box, +3x. And we have to ask ourselves, what is our greatest common factor? What will divide into each of the terms evenly? Our greatest common factor is 3x. 3x2 divided by 3x…the 3’s cancel out and x2 divided by x, is x…our +3x divided by 3x; 3x divided by 3x is 1. So from here we see that we have 3x(x+1).
In problem 2, we have the product of two binomials. We have (x+2) (x+1). We will represent our x+2 vertically because it is the first binomial. And then we will represent x+1 horizontally. When we have a width of x and a length of x, x(x)= x2; when we have a length of x and a width of 1, 1(x)= x; here we have a one unit times and x, so we have 1; and then we have a 1(1) and 1(1) =1; 1(x)=x; 1(1)=1. We have found the area of a rectangle with a width of x+2 and a length of x+1. When we remove these, we can see that our answer…we need to combine our like terms, and our answer is x2+3x+2. We can draw a pictorial of this problem, if we represent our x+2, we will have one long yellow rectangle and two small squares; one long yellow rectangle and one small square. x(x)=x2; x(1)=x; 1(x)=x; 1(1)=1; 1(x)=x; and 1(1)=1. Leaving us with the product of x2+3x+2. If we are going to multiply the two binomials using the box method, we will represent the x+2 vertically and represent the x+1 horizontally. x(x)=x2; x(+1)=+x; +2(x)=+2x; +1(+2)=+2. We combine our like terms which are on the diagonal, and we have an answer of x2+3x+2. When we compare our answer of problem 1 to problem 2, we see that they have the same answer. That’s because the only difference between problem 1 and problem 2 are the 2 binomials are switched. This has x+1 first, and here it is second; this has x+2 second and here it is first. Because of the commutative property, no matter which order you multiply in, you will still get the same answer. This holds true for binomials as well as just numbers.
When we multiply the binomials (x+3) (x+2), we will represent our x+3 vertically with one long yellow and three small squares, which are yellow. We also have x+2, so we will represent our x with one long yellow rectangle and our +2 with 2 small yellow squares. We are finding our area, so we are saying x(x)=x2; x(1)=x; and x(1)=x. Then we move to our next ones units, 1(x)=x; 1(1)=1; and 1(1)=1; 1(x)=x; 1(1)=1; and 1(1)=1; and we also have one more 1(x)=x; 1(1)=1; and 1(1)=1. We have created the area of the rectangle, and if we want to find our answer, we will combine our like terms of our product. We have an x2, we have 5x, and we have 1-2-3-4-5-6 ones units. This equals x2+5x+6. To show how to multiply problem 1 using the box method, we will represent our first x+3 vertically and our second binomial x+2 horizontally. x(x)=x2; x(+2)=+2x; +3(x)=+3x; and a positive times a positive is a positive, 3(2)=6. Our diagonal has the like terms, so our answer is x2+5x+6. In problem 2, we will represent our first binomial with one long yellow rectangle, and because it is x-1, we will represent the -1 with a red small square. We have 2x+4, we will represent this horizontally—we will represent our 2x and our +4 with 4 small yellow squares. x(x)=x2; yellow times yellow is yellow; and x(x)=x2; yellow times yellow is yellow and x(1)=x; x(1)=x; x(1)=x; and x(1)=x. Here we have red; however, and red times yellow is red; so a 1(x) gives us a red small x; red times yellow is red; red times yellow, -1(1)=-1, -1, -1, and -1. We have found the area created by the two binomials, and when we take the original ones away we can find our answer. We see that we have 2x2; however, we have two different colors in our x’s so in order to find our final answer, we have to create zero pairs. One yellow, one red- zero pair; one yellow, on red-zero pair. So we are left with 2x2+2x… either + -4 or -4. In problem 2, we will use the box method to multiply the two binomials. We will represent our x-1 vertically. And our 2x+4 horizontally. 2x(x)=2x2; x(+4)=+4x; 2x(-1)=-2x; and -1(4)=-4. Once again our diagonals have like terms. So we have 2x2… -2x+4x=+2x-4.
In problem 3, we are going to multiply x+3, which can be represented by a yellow long rectangle and 3 small yellow squares times x-2, which will be represented by a yellow long rectangle and because it is -2, we will use 2 red small squares. x(x), a positive times a positive, is x2; x(-1)= -x; x(-1)=-x; 1(x)=x; 1(-1)=-1; 1(-1)=-1; 1(x)=+x; and 1(-1)=-1, -1; 1(x)=x; -1…1(-1)=-1. So we have found our area with a binomial width of x+3 and a length of x-2. And we remove our original binomials, we see that we have created and x2; we have zero pairs with our x’s, so we have to remove one yellow, one red- zero pair; and one yellow, one red- zero pair, leaving us with one yellow x or +x; and 1-2-3-4-5-6 reds, or -6. To solve problem 3 using the box method, we will represent our x+3 vertically, and our x-2 horizontally. x(x)=x2; x(-2)=-2x; x(+3)=+3x; and a positive times a negative is a -6. We combine our like terms in the middle, and we are left with x2+x (because 3x-2x=x) -6.
In problem 4, we will represent our first binomial vertically with one long yellow rectangle and three small red squares. We will represent our second binomial horizontally with one long yellow rectangle and one small yellow square. x(x)=x2; x(1)=x; negative times a positive, -1(x)=-x; -1(1)=-1; -1(x)=-x; -1(1)=-1; -1(x)=-x; -1(1)=-1. We have found the area, so we will take away our original length and width. And our answer, we must combine like terms, we have 1x2; we have a zero pair in our x’s, one yellow-one red, which leaves us with 2 red long rectangles, or -2x; and three ones units which are red, so -3. To represent Problem 4 using the box method, we will represent the x-3 vertically and the x+1 horizontally. x(x)=x2; x(+1)=+x; x(-3)=-3x; and -3(+1)=-3. When we look at our diagonal, they have like terms. So we have x2-2x (because -3+1=-2) -3.
Lesson 31, Part 7
We can multiply the two binomials, (2x-1) (x+2). We will represent the 2x-1 vertically with 2 long yellow rectangles and one red square. We will represent the x+2 with one long yellow rectangle and two yellow squares because it is a positive 2 or +2. x(x)=x2; x(1)=x; and x(1)=x; x(x), we’re finding the area is x2; x(1)=x; and x(1)=x; -1(x)=-x; -1(1)=-1; -1(1)=-1. We can remove our 2x-1 and our x+2. We see that we have 2x2; we have two colors in our x’s, and we have -2. x2+x2= 2x2; one yellow-one red, is a zero pair; when we remove it we have 1-2-3 x’s, or 3x and two small red squares is -2. In problem 5 we will use the box method to multiply the two binomials, (2x-1) (x+2). 2x(x)=2x2; 2x(+2)=4x; -1(x)=-x; -1(+2)=-2. When we combine our diagonal, they are like terms. So we have, 2x2, -x+4x=3x; -2. In problem 6 we will represent our first binomial with two long yellows and three small yellow squares. We will represent our second binomial with one long yellow and one small red square. Yellow times yellow, x(x)…we have an+x2 but yellow times -1, x(-1)=-x …x times x gives us a positive x squared. X times negative 1 is negative x. 1 times x is positive x. but x times -1 is -1. 1 times x is positive x, 1 times -1 is another -1, 1 times x is positive x, 1 times negative 1 is a negative 1. We have created the area and now we can find out the answer by combining our like terms. We have 2 x squared’s, one yellow one red is a zero pair and one yellow one red is a zero pair, so we are left with 1 x. and three reds is -3. In problem six we can solve this problem using the box method. We will represent our 2x +3 vertically and our x-1 horizontally. 2x times x is 2x squared. 2x times -1 is -2x, positive 3 times x is positive 3x , and a positive 3 times a negative one is negative 3. We combine our like terms on the diagonal, we have our 2x squared plus 3x minus 2x is a plus x minus 3.
We are also going to review the FOIL method in this lesson. The foil method F O I L stands for first, outer, inner, and last. This is another way to multiply binomials and still get the four terms that we need. Multiplying your first terms, you will have 5x times 2x which is 10x squared. To multiply your outer terms you have 5x times 1 or a positive 1. So plus 5x , I stands for inner so we will multiply the two inner terms, negative 3 times 2 x is -6x and the last is -3 times positive 1 which is -3. You still must combine your two like terms from your outer and your inner so your final answer is 10x squared minus x minus 3.
We can use FOIL to solve problem 2, we will multiply our first two terms, j times j which is j squared, we can multiply our outer terms j times 3 which is j times 3 or 3j. We multiply our inner terms positive 1 times j or just j, and our last terms positive 3 times positive 1 is positive 3. We can combine our inner and our outer because they are like terms, so our final answer, is j squared plus 4j plus 3.
To use the FOIL method to solve problem 3, we will multiply our first two terms, 2n times n, is 2n squared, multiply our first term times our outer term, 2n times 6 will give us 12n. Our inner term -1 times n is –n, and our last two terms -1 times positive 6 is -6. We can combine our outer and our inner term and our final answer is 2n squared, plus 11 n, minus 6.
We have already “S” the problem, so we know this problem is asking us to find, the area of the rectangle. In “O” organize the facts, we have to read the problem and organize the facts. The length of a rectangle is represented by the binomial 2x-4. FACT. The width of a rectangle is represented by the binomial 3x+2. FACT. Both of these facts are necessary so we are going to list them. In “L” we are going to line up our plan. What is the area of the rectangle? Because we are finding area, we are going to use multiplication. To write in words what your plan of action will be, we will multiply the length by the width. In “V” verify your plan with action, we are going to estimate our answer. Because our length and width are binomials, we know our answer should be a polynomial. We can carry out our plan by multiplying the two binomials using the box method or using the FOIL method. If we use the box method, we will represent our first binomial for the width of the box, as 3x+2. And our length as 2x-4. 3x(2x)=6x2; 3x(-4)=-12x; 2x(2)=+4x; +2(-4)=-8. We see that our diagonals are like terms. So our answer is 6x2-8x-8. We can also use the FOIL method: 2x(3x)=6x2, for our first two terms; our outer terms is 2x(2)=4x; our inner terms are -4(3x)=-12x; and our two last terms are -4(+2)=-8. When we combine our like terms we have the same as 6x2-8x-8. In “E” we will examine our results. Does your answer make sense? We were finding the area of a rectangle with a width and a length of binomials, so yes it does make sense to have a polynomial as your answer. Is your answer reasonable? We estimated that our answer would be a polynomial, so yes our answer is reasonable. Is your answer accurate? If your students use the box method, they could use FOIL to check their answer--- or if your students used FOIL, they could use the box method to check their answer. Some students may still want to draw a pictorial or use the manipulatives to find their answer. Write your answer as a complete sentence. The area of the rectangle is 6x2-8x-8. To close the lesson we will look at the essential questions. Question 1, when multiplying two binomials, how do you end up with four terms? There are 2 terms by 2 terms which gives you a product of 4 terms. Our second question is, when can monomials be added or subtracted? Monomials can be added or subtracted when they are like terms.