Download There`s a Problem in a box

There’s a Problem in a box
Name:_____________________
We have looked at factoring trinomials (3 term polynomials) using a box to help
find the two factors. Now we will use the box to look at factoring polynomials
with 4-terms.
Example problem: x3 + 3x2 + 2x + 6
x3
3x2
2x
6
Answer:
Look how easy it is to fill in this box! 4 terms fit nicely into 4 boxes! Now all you
need to do is find the common stuff!
Practice makes Perfect problems:
1) 7x3 – 14x2 – x + 2
2) 25x3 + 5x2 + 30x + 6
3) 6x3 – 16x2 + 21x – 56
4) 28x3 + 16x2 – 21x – 12
Let’s do some without the box. Some people like to think outside the box, so
here is a way to factor the 4-terms without the box.
5) 7x3 – 14x2 – x + 2
6) 25x3 + 5x2 + 30x + 6
7) 6x3 – 16x2 + 21x – 56
8) 28x3 + 16x2 – 21x – 12
Last chapter review stuff:
9) (x2 + 3x – 8) + (4x – 2)
11) (-3x2 + 2) (4x – 5)
10) (2x – 9) – (3x2 + 8x – 11)
12) (x – 1) (x + 7)
Trinomials
Box or short cut, you decide
Name:_____________________
Today will be a bit tougher because we will not have all the information to put
into the box at the start of the problem. We are going to do some multiplying,
looking at the middle term, and use our knowledge of factors to help create the
box. There is also a short cut if you get really good at these problems.
Example problem: x2 – 3x – 10
Answer:
left
x2
-10
with
1) Start by placing the first term in the upper
box.
2) The third term goes into the lower right box
3) The two remaining boxes will be filled in
two terms that we have to find.
Multiply the numbers in the boxes: (1)(-10) = -10
Find the number of the middle term: -3
Find two numbers that multiply to the first answer (-10) and add to the second
answer (-3): (-5)(2) works
Put the numbers in the two empty boxes (with the variable x) and find common
stuff.
x2
-5x
2x
-10
Note: It doesn’t matter where the -5x and 2x are
placed in the diagonal
Now factor to solve:
The short cut is just to find what multiplies to the last number and adds to the
middle number. The two numbers will be the answer to the factoring problem.
Example problem: x2 – 3x – 10 (again)
Perfect Practice (both methods to start with):
1) x2 – x – 2
2) x2 + 7x + 6
3) x2 – 15x + 56
What happens with this special problem?
5) x2 + 5x – 12
4) x2 – 1x – 12
Trinomials part2
Name:_____________________
These aren’t the nice trinomials you are used to playing with. They hit and bite
so be careful how you deal with them!
Notice that all these problems are different from yesterday because there is a
number in front of the x2 term that is not “1”. Luckily, we still follow the same
method as yesterday. I am sorry to report that we won’t be able to use a nice
short cut for these factoring problems.
Example problem: 3x2 – 8x + 4
Answer:
1) Start by placing the first term in the upper left box.
3x2
2) The third term goes into the lower right box
3) The two remaining boxes will be filled in with two
4
terms
that we have to find.
Multiply the numbers in the boxes: (3)(4) = 12
Find the number of the middle term: -8
Find two numbers that multiply to the first answer (12) and add to the second
answer (-8): Hey, (-6)(-2) works
Put the numbers in the two empty boxes (with the variable x) and find common
stuff.
3x2
-6x
-2x
4
Note: It doesn’t matter where the -6x and -2x are
placed in the diagonal
Now factor to solve:
Start your boxing training here (you may want to hit your teacher at some point
due to all of the box method going on these past 2 chapters):
1) 3x2 – 2x – 5
2) 2x2 + 11x + 5
3) 5x2 – 19x + 12
4) 6x2 + 5x – 6
What happens when you get a really big number as your multiplied number? I
suggest you try the square root of the number before giving up and working on a
crab boat in Alaska. You may make better money, but this is a lot safer.
5) 16x2 + 40x + 25
6) 100x2 + 180x + 81
Don’t forget that you may face some that you can’t factor. Those are known as
our “Prime” answers (Just like no solution for equation problems).
Ex: 3x2 – 4x + 11
This factoring stuff is really
Name:______________________
making me mad! I have to go to my box now!
I have now made the problems a bit more challenging. You can
thank me later over the dinner table during Thanksgiving Break!
Most of the problems now have a common factor in them that
needs to be dealt with first. Once we find the GCF we can use
our nice box method to factor again! Just watch out for your
final answer because we have to see that common factor again.
It is just like an old flame that won’t go away!
Trinomial example:
polynomial example:
4-term
8x3 – 34x2 + 8x
105x2 – 90x
63x4 + 54x3 –
(A) What is common?
common?
(B) Now what does the problem look like?
NWDTPLL?
(C) Factor with the box
with the box
(A) What is
(B)
(C) Factor
(D) The final answer with everything
TFAWE
Here are some problems that you can do with friends or
family…..
(D)
1) 30x3 + 105x2 – 10x – 35
3x2 – 21x
2) 12x4 – 84x3 +
3) -36x2 – 36x + 27
28x
4) -10x3 + 74x2 –
Some more review problems to keep you in mathematical shape!
5) (6x + 1)(4x-3)
8)(4x + 9)
7) Add and subtract: No boxes needed here!
(-2x2 + 3x – 11) + (x2 – 19) – (4x2 – 8x – 13)
6) (7x2 –
Binomial factorization
Name:______________________
You can’t be serious. More factoring!
I know your mind is just about to explode with all of the
factoring we have done. At least this time there won’t be a
box to deal with (you still can use it if you want). In fact,
there is a nice short cut method to deal with problems that
just have two parts. All we need to have is:
(1) A Binomial (a polynomial with two terms)
(2) Perfect square terms
*numbers are perfect squares
*variables have even powers
(3) A minus sign in between
Let’s look at two examples:
1) 25x2 – 81
Here we have two terms, perfect squares, and a minus sign! How
lucky are we! To get the answer, just split the perfect square
numbers up into their square roots, take half of the exponent,
and use a “+” and “-“
at the end.
Answer:
(5x + 9) (5x – 9)
2) 12x11 – 27x5
Oh no! I’m freaking out since there are not perfect square
numbers or even exponents! What are we to do? Never fear,
just look for something common to help! There is a common 3x5
in both parts, so lets take it out!
3x5 (4x6 – 9) Few. I now have my nice perfect squares (hey,
aren’t squares shaped like a box! It does show up again!).
Answer: 3x5 (2x3 + 3) (2x3 – 3) **Notice it doesn’t matter if
you use the + first. It could also be 3x5 (2x3 – 3)(2x3 + 3)
If we take another look at the first example 25x2 – 81 we can
really make it into a trinomial with the middle term 0x. It
will then look like:
25x2 + 0x – 81
It is now a trinomial that we can factor using the box method
again.
Multiplies (-2025)
Adds to (0)
25x2
-81
This is a big #, so use the square root of 2025 (pos answer)
because you can’t square root the negative number in this case.
Now just factor like normal:
Armed with this new method to defeat the evil polynomials that
roam the Earth and beyond, you may now go forth and try to
factor some other problems.
Here are some nice ones without a GCF
1) 36x8 – 1
2) 49x2 – 100
3) 121 – 169x2
4) 16x10 – 225
Now some ones that do have something in common (besides
frustrating you)
5) 98x2 – 200
7) 81x4 – 900x2
6) 343x2 – 7x4
8) 400 – 36x2
These old friends decided to stop by so you could visit with
them again.
9) (x – 4)(3x – 7)
– 15)
10) (3x + 2) - (11x
Factor for problems 11-14
11) 28x3 + 63x2 – 32x – 72
12) 3x2 + 1x – 10
13) -14x5 – 37x4 – 5x3
+ 4x
14) -2x4 + 4x3 – 2x2
Binomials, Trinomials,
And 4-term polynomials
Oh my!
Name:________________________
To help with the decision of how to factor, we are going to make a
decision chart for factorization.
Short Cuts:
Box method:
Problems to try:
1)
x2 – 4x + 3
3)
30
x2 – 49
4)
x3 – 6x2 + 5x –
5)
4x2 + 28x + 40
6)
-3 + 192x12
Solving Equations by Factoring
2)
5x2 – 9x – 2
Name__________________
Don’t panic now that there is an equal sign in our problems.
We can get through the extra steps, but you must remain calm
and keep hands and feet in the car at all times! Sometimes the
problems will be half done for you, so you must have been on
the “good” list this year.
Problems that have already been factored will usually have ( )
and the = 0 part. If this is the case, then the hard part of
the problem has already been taken care of.
Example:
3x(x – 4)(2x + 3) = 0
Just take the individual parts and set them equal to zero.
Then solve for the variable.
3x = o
2x + 3 = 0
x – 4 = 0
Now you try some factored problems:
1) x(x-7) = 0
3) (2z + 1)2 = 0
mathematically**
2)
4b(b + 4) = 0
**Don’t forget what squaring means
The hard part comes when the problems aren’t factored and we
have to do the dirty work. The steps to solve these problems
are:
1) Set the problem equal to 0!
Move parts around by adding
or
subtracting terms. Try to make the first term positive and
listed in
descending order of powers.
2) Follow the factoring guidelines.
3) Set the individual factors equal to zero and solve the
equations.
Examples (a binomial and trinomial)
Binomial:
= -12x
75x = 12x3
Try these problems.
GCF).
4) y2 = 3y
Trinomial:
3x2 – 63
Most DO NOT have a GCF (only 4 have a
5)
c2 - 9 = 0
6) h2 + 14 = -9h
7)
8) j2 - 64 = 0
= 0
9)
10)
11)
5z2 - 5z = 150
4v2 + 3v = 10
12)
- 6 = -4x2
2x3 + 30x2 = -88x
-f - 6 = -f2
r3 - 2r2 - 15r
13) 23x
14)
70 = 0
t2 = 5t
15) p2 - 19p +
16)
rectangle with length x.
x2 - 2x represents the area of a
Find x if the area is 35.
Challenge Problems
Name:______________________
Here are some problems that will challenge your knowledge of
factoring. Be careful because a lot of them can be solved
further than you are used to going. Some can’t be factored or
solved.
What do you do when you can’t factor or solve?
Ex: x2 – 9x + 15
shortcut of what
This is a trinomial that can use the
multiplies to 15 and adds up to -9.
The
only problem is
that there aren’t two numbers that do
this.
Since
there isn’t an = in the problem the
answer is “prime”.
Ex: x2 – 9x + 15 = 0
is “No Sol.”
Now that there is an = 0 part, the answer
This is because we can’t find the
answer to x that will
make the equation true.
Challenge Problems:
1)
x4 – 10x2 + 9= 0
2)
x2 – 324
3) x2 + 2x – 6 = 0
5)
3x4 + 9x2
7) 5x9 – 1280x
4) 3x2 – 2x – 5
6)
3x2 + 7x + 2 = 0
8) x2 + 49
9)
– 4
x2 + 8x + 16 = 0
10)
x4 – x2 + 4x
11) x2 – 21x – 72 = 0
2 = 0
12)
7x2 – 9x +