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Multiplying Polynomials Lesson 32 In lesson 32, our warm-up has students multiplying binomials. They are going to review multiplying binomials because they will be multiplying polynomials in this lesson. Our solve problem, the length of a rectangle is represented by the trinomial 3x2+2x-4. The width of the rectangle is represented by the trinomial x2+5x-1. What is the area of the rectangle? When we study the problem, or “S” the problem, we will first underline the question. What is the area of the rectangle? We will then answer the question, what is this problem asking me to find? This problem is asking me to find the area of the rectangle. We are going to use the box method to multiply polynomials. We will represent our first term, x+2, vertically. And we will represent our second term horizontally. The box method will help keep your students organized so they will not loose any terms in this process. x(x2)=x3; x(+3x)=3x2; and x(+1)=+x; +2(x2)=+2x2; +2(+3x)=+6x; +2(+1)=+2. From here to get our final answer, we have to combine like terms. Our diagonals show us that there are like terms. We only have one x3; we have 2x2+3x2=5x2; we have 6x+x=7x; +2. This is our final answer. In problem two, we will multiply the binomial times the trinomial. We will represent our first binomial vertically. And our trinomial horizontally. We will use the box method to keep us organized. 2x(x2)=2x3; 2x(-x)=-2x2; 2x(+4)=+8x; +1(x2)=+x2; +1(x)=-x;+1(+4)=+4. We see that once again we have like terms on the diagonal. So, we have 2x3; +x2-2x2=-x2; -x+8x=+7x; +4. In problem 1, we are going to multiply two trinomials. We will represent the first trinomial vertically. And the second trinomial horizontally. Because we have 3 terms times 3 terms, we will end up with 9 terms after we multiply. So the box method will help us stay organized. x2(x2)=x4; we will then multiply our x2(3x), so you get +3x3; x2(+2)=+2x2; +2x(x2)= +2x3; +2x(+3x)=+6x2; +2x(+2)=+4x; +1(x2)= +x2; +1(+3x)=+3x; and +1(+2)=+2. From here we are going to combine our like terms, which once again are our diagonals. We only have one x4; +2x3+3x3=5x3; +x2+6x2+2x2=9x2; +3x+4x= 7x; +2. To multiply our first trinomial, we will represent our trinomial vertically; and the second trinomial horizontally. 3x2(x2)=3x4; 3x2(+2x)=+6x3; 3x2(-1)=-3x2; -x(x2)=-x3; -x(+2x)=-2x2; -x(-1), negative times a negative… is a +x. +4(x2)=+4x2; +4(+2x)=+8x; +4(-1)=-4. Our diagonals are our like terms. So we have 3x4; -x3+6x3=5x3; +4x2-2x2 -3x2=-x2; +8x+x=9x; -4. In problem 3, we are going to multiply a trinomial by a binomial. We will represent our trinomial vertically and our binomial horizontally. x2(2x)=2x3; x2(-3)=-3x2; +5x(+2x)=+10x2; +5x(-3)=-15x; +4(+2x)=+8x; +4(-3)=-12. To find our final answer, we are going to combine our like terms. 2x3; +10x23x2=+7x2; +8x-15x=-7x; -12. In problem 4, we have two products added together. So we have 2x (x+1) and this is added to or plus, 3x(x2-5x+1). 2x(x)=2x2; 2x(+1)=2x; 3x(x2)=3x3; 3x(-5x)=-15x2; 3x(+1)=3x. We have a 2x2+2x+3x3-15x2+3x. Our largest number of x’s is 3x3 so we can write that first. We then have two x2, one is a +2 and one is a -15, so we can combine those two and get -13x2. And then we have a +2x and +3x, which we can combine to get 5x. We can use the distributive property to multiply polynomials. We would have to multiply the x (x2)…to get x3; the x (-x)….to get –x2; and the x(-2)…to get -2x. We would then have to multiply the +1(x2)…to get +x2; +1(-x)…-x; and +1(-2)…to get -2. We can combine like terms. We will begin with our x3. There one x2 plus another x2, Negative and positive, so they actually cancel and leave us with 0x2. -2xx=-3x; -2. To complete problem 2, we will use the distributive property. Simplify the first product. 4x(x)=4x2; 4x(-2)=-8x. We will add this to the product (x+2) (x+3). We can use FOIL to multiply these two binomials. x (x)=x2 ; our outer, x (-3)= -3x ; our inner, +2(x) = 2x ; and our last +2 (-3)=-6. We now need to combine like terms. We have one 4x2 and another x2 , which gives us 5x2; we have a -8x, -3x, +2x which gives us -9x; and we have a -6 for a minus 6. To complete our solve problem, we have already “S” the problem and know that this problem is asking us to find the area of the rectangle. In “O” organize the facts, we must identify our facts. The length of a rectangle is represented by the trinomial 3x2+2x-4. This is a fact. The width of the rectangle is represented by the trinomial x2+5x-1. This is also a fact. Both of these facts are necessary so we will list them. In “L” lining up our plan, we must choose an operation and because we are finding the area, we are going to use multiplication. We will write our plan of action in words. We will multiply the length times the width. In “V” verify your plan with action, we are going to estimate. Our estimate is going to be a polynomial because our length is a polynomial and so is our width; so the product will also be a polynomial. We can carry out our plan. I’m going to multiply the two trinomials using the box method. Some students may choose to use the distributive property in order to multiply these polynomials. Because I have a trinomial times a trinomial, I need three rows and three columns. (3x2+2x-4) and (x2+5x-1). 3x2(x2)=4x4; 3x2(5x)=+15x2; 3x2(-1)=-3x2; +2x(x2)=2x3; +2x(+5x)=10x2; +2x(-1)=-2x; -4(x2)=-4x2; -4(+5x)=-20x; -4(-1)=+4. When we look at our diagonals, they are like terms. So we have 3x4; +2x3+15x3=17x3; -4x2+10x2-3x2=3x2; -20x-2x= -22x; +4. In “E” we will examine our results. Does your answer make sense? Our question asked, what is the area of the rectangle? Because our length and width were polynomials, it does make sense to have a polynomial for an answer. Is your answer reasonable? Our estimate was a polynomial, so yes our answer is reasonable. And is your answer accurate? You could have your students check their work using a different method. We will now write our answer as a complete sentence. The area of the rectangle is 3x4+17x3+3x2-22x+4. To close the lesson, we will look at the essential questions. Number 1, when multiplying polynomials, is it possible to determine the number of terms you should have before combining like terms? YES, you can multiply the number of terms in each factor to determine the total number of terms in the product before combining like terms. In other words, if you have a trinomial times a trinomial, there are 3 terms in the first factor and 3 terms in the second factor…so you should have 9 terms before combining like terms. Question 2, is it possible to have more than two like terms when multiplying polynomials? YES, depending on the types of polynomials you are multiplying, you could have more than two like terms.