Factoring Special Cases
Lesson 35
The warm up for lesson 35 has the students review their roots. This is an important
skill because today we will be looking at the difference of squares. And students
must know their perfect squares. The area of the rectangle is represented by the
trinomial x2-18x+81. What could be the length of the rectangle? To “S” this problem,
we will first underline the question, what could be the length of the rectangle? This
problem is asking me to find the possible lengths of the rectangle. If we want to
factor x2-4. We will represent it with our algebra tiles. We will have one large x2 which
is yellow because it is positive and 4 red small squares because the four is a -4, or
negative 4. We will begin by placing our x2 in the top corner. And in this case, we can
make a square with our 4, however we have no x’s…when we don’t have any x’s, we
can create our possibility by adding zero pairs. We must keep our colors together, so
we can set our two yellows here and our two reds here. Here, we have x(x); we know
that we have an x, and because it is yellow, we have a positive one and another
positive one; however here, we have a positive x but because all of these are red, we
will have 2 reds. Our answers are going to be x+2; x-2. We can check our answer by
using the FOIL method or the box method. x+2 and x-2. x(x)=x2; -2x; +2x; -4. We
notice that our diagonals or like terms cancel, leaving us with x2-4. We want to factor
the polynomial x2-9. We have 1-2-3-4-5-6-7-8-9. We place our x2 in the corner and
then we are going to create a three by three square. We create a three by three
square because we know that our number of x’s are going to have to be equal
because we are going to have to add them as zero pairs. We add 1 zero pairs, and 2
zero pairs, and 3 zero pairs. So you can put your three positive x’s here or your three
positive x’s here, either way is correct. The book has the positives down here and the
negatives here. We know that we have x(x); and if we want to find what we multiplied
the x by to get another positive x, it’s going to be +1—we have three of those, so, we’ll
have three ones; and we have three negative x’s, so we will have three negative x at
the top. So we see that our answer is (x+3)(x=3). We can check our answer using
the box method. x(x)=x2; -3x; +3x; -9. Look at your diagonals, they cancel out leaving
you with x2-9. So our answer is correct.
Our third problem, we have 4x2 and we have a negative 1 or a -1. If we want to find a
way to make a square, we will start by using our x squares. And then we move to our
-1. We will have to add zero pairs. And we can’t mix up colors, so we will have to
have two yellows in one spot and two red in another spot. It doesn’t matter because
the commutative property allows us to mix those two binomials. I’m just modeling
the way we have it in the teacher’s guide. Here we have a width of x and a width of x.
We know that we have another positive x and another positive x. x(x) gives us x2; x(1)
times what? gives us 1, we know this has to be a positive 1. x times what? that
gives us a -1, we know this has to be a negative one. So my two factors are 2x+1 and
2x-1. From here I can check these using the FOIL method or the box
method…2x(2x)=4x2; -2x; +2x; -1. In each of these examples my like terms are
canceling, leaving me with 4x2-1. In problem 1, we are going to factor, x2+4x+4. We
have an x2, we have a 4x, and we have a plus 4. So, we are going to put x2 in the top
corner; we are going to move to our ones unit; now I’m going to make a square out of
them, a two by two square. Then that allows me to fill in my spaces with my x’s.
Because they are all positive, our width is x+2, and x+2. This factors into (x+2) and
(x+2). We can check our answer using the box method…x2+2x+2x+4. Our like terms
are on the diagonal, so it does equal x2+4x+4. Our answer checks. We are going to
factor x2+6x+9. We will start by placing our x square in the corner and then move to
those ones units. We have 9, so let’s try 3 times 3. If we do 3 times 3, we can fill in
our spaces. Everything is positive. We see that our factors are x+3; and x+3. We
can check this answer using our box method or using FOIL. x2+3x+3x+9. Our
diagonal has like terms, so it equals x2+6x+9. We are correct.
In problem 3, we are going to start with our x2, we have 4 negative x’s, and a positive
4. Here we can create a 2 by 2. We can create our square using our –x’s. We have
to be very careful because you have negatives involved. We know that x(x) gives us
our x2. We have to ask ourselves, what do we multiply this x by to get a negative x, so
it’s a -1; but the same thing has happened here, what do we multiply our x by to get a
negative 1x over here; and then we must check, a negative times a negative does
give us a positive. So our two factors are x-2 and x-2. We can check using the box
method or FOIL. If we use FOIL, our first two terms are x (x) or x2; and x (-2)= -2x,
those are our outer terms; our inner terms are -2(x) or -2x; and our last terms are -2(2) or +4; x2-4x+4, is where we started so our answer checks. If we wanted to factor
x2=49 using the box method and splitting our middle term, we have to complete our
puzzle. Our standard form is ax2+bx+c. Our x2 or the coefficient of our x2, which is a,
is a 1. And our c is a -49. In this case, we don’t have a b term, so b is equal to 0. 1
(-49)= -49; what are the two factors of -49 that add to give you zero? -7 (7)=-49; and
-7+7= 0. If we know that -7 + 7 =0, then we know that -7x+7x+0x or 0. So we can
split that -0x middle term into
-7x+7x. We have x2-7x+7x-49. Our greatest
common factor of the first row is x. Our greatest common factor of the second row is
+7. Our greatest common factor of the first column is x. Our greatest common factor
of the second column is -7. So our two factors are (x+7) (x=7). Your students can
always check their answers using the FOIL method or the box method.
In problem 2, we are going to factor the trinomial algebraically, using the box method.
We have to solve our puzzle. Our a is 1 and our c is 25, so they have to multiply to be
25. And our b is 10. We need two numbers that multiply to be 25 and add to be 10.
These two numbers are 5 and 5. If 5+5=10, then 5x+5x+=10. We can split our
middle term into 5x+5x; x2+5x+5x+25. Our greatest common factor of the first row
is x. Our greatest common factor of the second row is +5. Our greatest common
factor of the first column is x. Our greatest common factor of the second column is
+5. Our two factors are (x+5) and (x+5). Your students can check their answers
using the FOIL method or using the box method. In problem 3, we are going to factor
our trinomial algebraically. We must first solve our puzzle. Our a=1; our c=+144. So
a(b)=144. Our b= -24, so that equals -24. What two numbers multiplied to be +144
that add to be -24? -12. Because -12 + -12 = -24, -12x + -12x = -24x. We can split
this middle term to -12x + -12x. x2-12x-12x+ 144. Our greatest common factor of
our first row is x. Our greatest common factor of our second row is 12. Our greatest
common factor of our first column is x. Our greatest common factor of the second
column is 12. We have to look at our signs here, because we see that we have a -
12x and x and +12. We know that 12 times 12 is going to be a positive 144, and -12
times -12 is also going to be a positive 144. So our two binomials will be (x-12) and
(x-12). If your students did not catch themselves the first time…if they had written
the answer as x+12 and x+12, when they check their answer using the FOIL method
or the box method, they should automatically see that their middle term would not 24x and they will go back to the box method to see how they could change their
answer.
We have already “S” this problem and so know this problem is asking us to find the
possible lengths of the rectangle. In “O” we are going to organize our facts. The
area of the rectangle is represented by the trinomial x2-18x+81. This is our only fact
and it is important so we will write it down. In lining up a plan, we know that our
operation will be to factor or divide. If we write in words what our plan of action will
be, we are going to factor the area of the rectangle. In “V” verify your action, we will
estimate that our answer will be a polynomial. We have the trinomial x2-18x+81.
And in order to factor, I am going to use the box method. So, I must solve my
puzzle…something times something equals a times c; something plus something
equals b. My a=1; my c=81, so these have to equal 81; and my b has to equal -18. 9 (-9)= 81; and -9+ -9= -18. If we know that -9 + -9 = -18, then -9x + -9x = -18x. I
can split my middle term into x2-9x-9x+81. My greatest common factor of the first
row is x. My greatest common factor of the second row is 9. My greatest common
factor of the first column is x. My greatest common factor of my second column is 9;
however we know that if we have 9 (9)=81, and -9 here—we are going to have to
have x-9 and x-9 as our two factors. In “E” we examine our results. Does your
answer make sense? We were looking for possible lengths of a rectangle where the
area of the rectangle was given as a polynomial, so yes- our answer makes sense. Is
your answer reasonable? Our estimate said that our answer would be a polynomial,
so yes our answer is reasonable. Is your answer accurate? We can check our
answer using the foil method or the box method. If we use foil, our first two terms
are x(x)=x2; Our outer terms are x (-9)= -9x; our inner terms are -9 (x)=-9x; and our
last terms are -9 (-9)=+81. x2-18x+81. So our answer checks. We will now write our
answer as a complete sentence. The length of the rectangle is (x-9). We will close
our lesson by looking at the essential questions, Number one, can you factor all
binomials? No, not all binomials can be factored. Binomials can be factored if they
have a greatest common factor or if they are the difference of squares. Number 2,
can you use grouping to factor all polynomials with four terms? No, you can only use
grouping to factor four terms, if after you group them, and take out the greatest
common factor, they leave a common binomial.