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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
rectangle? Because our length and width were polynomials, it does make sense to