Binomial Theory Using Polynomial Expansion

If we expand:
 (x+a)(x+b)=x²+(a+b)x+ab
(x+a)(x+b)(x+c)=x³+(a+b+c)x²+(ab+ac+bc)x+abc
(x+a)(x+b)(x+c)(x+d)=x⁴+(a+b+c+d)x³+(ab+ac+ad+bc+bd+cd)x²+(abc+abd +cd+bcd)x+abcd

Lets say that we want to have 2 numbers or a binomial we would change the expressions above to:
(x+1)², (x+1)³, and (x+1)⁴
They equal:
(x+1)²=x²+(1+1)x+1X1=x²+2x+1
(x+1)³=x³+(1+1+1)x²+(1X1+1X1+1X1)x+1X1X1=x³+3x²+3x=1
When when we expand a binomial, the number preceding each x is just the number of combinations of a,b,c variables preceding our x in our polynomial expansion.

(x+1)⁴=x⁴+4x³+6x²+4x+1
(x+1)ⁿ=xⁿ+[nC(n-1)]xⁿ⁻¹+[nC(n-2)]xⁿ⁻²+[nC(n-3)]xⁿ⁻³...........

Now lets lets see what happens if we use a variable like y instead of 1.
(x+y)²=x²+(y+y)x+y.y=x²+2yx+y²
(x+1)³=x³+(y+y+y)x²+(y.y+y.y+y.y)x+y.y.y=x³+3yx²+3y²x+y³
(x+y)⁴=x⁴+4yx³+6x²y²+4xy³+y⁴

Our expansion becomes
(x+y)ⁿ=xⁿ+[nC(n-1)]xⁿ⁻¹y+[nC(n-2)]xⁿ⁻²y²+[nC(n-3)]xⁿ⁻³y³...........yⁿ

Lets say we have (x+y)⁶=x⁶+6x⁵y+15x⁴y²+20x³y³+15x²y⁴+6xy⁵+y⁶

If we had a number like 3 instead of y, we would substitute 3 where we have y and it would look like this:

(x+3)⁶=x⁶+6x⁵(3)+15x⁴(9)+20x³(27)+15x²(81)+6x(243)+y⁶(729)=x⁶+18x⁵+135x⁴+540x³+ 1215x²+1458x+729y⁶

As you can see there is a lot of calculation involved in expanding binomial series.  There is a way that you can just plug a number in from a table. This table is called Pascal's Triangle.  We will do post on this next.  Many people skip binomial expansion and just use Pascal's Triangle.

Fast Multiplication Using Foiling

Many people have used Foiling to distribute quadratic equations.  It actually can be very helpful in multiplying very fast.

When we foil we multiply the first terms together, then we multiply the outside, then the inside, and finally the last terms.  So lets see this in action.

11X15  Not so easy right?

Lets rewrite it:

(10 +1)(10+ 5)

First term is 10X10=100
Outside: 5x10=50
Inside: 1x10=10
Last: 5x1=5
So 100+50+10+5=165

Still doesn't feel to easy?  There are too many numbers so lets modify our foil and lets make three spaces.

100  + 60    +   5     

First       O+I       Last

Okay a little easier

How about these numbers?:

1x1=1

5x1+1x1=6
5x1=5

Quite a bit simpler

Okay line them all up like:        1

                                                   +   6

                                                   +       5

It's 165.

Lets try 13X15:

1x1=                                        1

5x1+3x1=                                     8

3x5 = ?????? what do we do now?

Not to worry.  It is easy.  I promise.

1x1=                                        1

5x1+3x1=                                     8

3x5 =                                             1  5

So our answer is 195.  Since our last multiplier was 2 digits, we just carrier it into the next column which is shared by 8.

How about multiplying a number like (63)(72)?

6x7=                                        4 2

2x6+3x7=                                   33  

3x2 =                                                6

The answer is 4536.  Now that is easy to do on paper but to do it in your head is probably still a little difficult.  Let's break this a part but we are going to start with the middle.  2x6=12 and 3x7=21 they add up to 33.  33 is easy to remember.  We have 6 x 7 is 42 with a carry from 33 of 3 which is 45 and 2 times 3 is 6 so we have 4536.

Okay lets do something even more difficult:  49x83.

Middle is 3x4 +9x8 =12+72=84 remember this number.
Our first number is 8x4=32.
Our last number is 3 times nine 27.
We have a carry from the last digit of 2 and add this to 4=6.
We have a carry of 8 and add that to 32 =40.
So our number is 4067.

Okay lets try:

14x31
Do you visualize and see 434?  If you didn't, I will post more worked process.  It takes a bit of traininng your mind to do these.  After you practice, you will be able to do these really well.

Try these numbers and write down the answer after you figured these out and check with a calculator.
67x85

49x72

88x55

31x96

I will do a post with more worked practice but if you do thes

Foiling Quadratic Functions

(X+A)(X+B) is the multiplication of two factors in a quadratic equation.  We also call it distribution.  A very easy way to solve these equation is with a term called Foiling.  Lets break it down.

First + Outside + Inside + Last

Example:

Expand (X+5)(X+3)

First (X)(X)=X² + Outside (X)(3)=3X +  Inside (5)(X)=5X + Last (5)(3)=15=X² + 8X + 15

Ah ha.  Now you know what foil means.

Lets try expanding (X-4)(X+3)

First (X)(X)=X² + Outside (X)(3)=3X +  Inside (-4)(X)=-4X + Last (-4)(3)=-12=X² -X -12

Something a little harder

Expand (-3X+7)(2X-4)

First (-3X)(2X)=-6X² + Outside (-3X)(-4)=-12X +  Inside (7)(2X)=14X + Last (7)(-24)=-28=

=-6X² +2X -28

Whohoo! Now try a few yourself:

1) (X+5)(X+2)

2) (X-3)(X+5)
3) (X+9)(X+1)

4) (3X+2)(X-7)
5) (6x-4)(3X-2)

Answers (1) x²+7x +10 (2) x²+2x-15 (3) x²+10x+9 (4) 3x²-19x-14 (5) 18x²+8

Learning to multiply 3's 6's 7's and 9's

3's 6's 7's and 9's are hard numbers to multiply.   There are complete tutoring agencies teaching math because of 3's 6's 7's and 9's

The series does not flow as easily as others.

3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75 78 81 84 87 90

The numbers that are high lighted are the numbers on the 10 x 10 multiplication table that gives students the most problems.

I have my students figure out the series the series to 6 9 18 21 and 27 after they learn to multiply 2 4 5 8 10.  6 and 9 are pretty simple by this time so to complete the threes  18 21 and 27 need to be memorized. 3 x 6 =18 3 x 7 =21 and 3 x 9=27




If the student has learned the 2's 4's 5's 8's and 10's the table will be as follows after the student learns the 3's:

1 2 3 4 5 6 7 8 9 10 1 1 2 3 4 5 6 7 8 9 10 2 2 4 6 8 10 12 14 16 18 20 3 3 6 9 12 15 18 21 24 27 30 4 4 8 12 16 20 24 28 32 36 40 5 5 10 15 20 25 30 35 40 45 50 6 6 12 18 24 30 48 60 7 7 14 21 28 35 56 70 8 8 16 24 32 40 48 56 64 72 80 9 9 18 27 36 45 72 90 10 10 20 30 40 50 60 70 80 90 100

The next series is 6 and 36 42 and 54 are hard numbers for students.  6 x 6=36  6 x 7 = 42 6 x 9 = 54

	1	2	3	4	5	6	7	8	9	10
1	1	2	3	4	5	6	7	8	9	10
2	2	4	6	8	10	12	14	16	18	20
3	3	6	9	12	15	18	21	24	27	30
4	4	8	12	16	20	24	28	32	36	40
5	5	10	15	20	25	30	35	40	45	50
6	6	12	18	24	30	36	42	48	54	60
7	7	14	21	28	35	42		56		70
8	8	16	24	32	40	48	56	64	72	80
9	9	18	27	36	45	54		72		90
10	10	20	30	40	50	60	70	80	90	100

The next series is 9 and 7.  Memorize 9 x 7 =63 and 9 x 9= 81.  The 10 x 10 multiplication table is done.

Do plenty of exercises and you will have this constantly in your memory.

Fast Math Squaring Numbers That End in 5

Squaring numbers that end in 5 are very easy and can be done very fast.

The fast math tricks have actually many different ways to solve but I will present 3.

1) Addition method:

Lets start with a number 55.
Subtract 5 from the original number: 55-5=50.

We will call this number our base number.

Square the base number: 50²=2500

Multiply the base number by 10: 10x50=500

Add the square of the base number to the base number x 10 and add 25: 2500+500+25=3025

If we wanted to find out what 115² equals, we would do the following:
115-5=110
110²=12100

110x10=1100
115²=12,100+1100+25=13,225

2) Subtraction method:

Lets start with a number 55 again.
Add 5 from the original number: 55+5=60.

We will call this number our base number.

Square the base number: 60²=3600

Multiply the base number by 10: 10x60=600

Subtract the base number x 10 from the base number squared and add 25: 3600-600+25=3025

Pretty cool so far right?

Lets use this methodology to find 95²

95+5=100
100²=10,000

100x10=1000
95²=10,000-1000+25=9025

3) Multiplication method

Lets start with a number 55 again.

Add 5 from the original number: 55+5=60.

We will call this number our base number 1.

Subtract 5 from the original number: 55-5=50

We will call this base 2.

Multiply base 1 to base 2 and add 25: 50x60+25=3025

Wow!  That is really amazing!

Lets try 205²
205-5=200

205+5=210
200X210=42000
42000+25=42025

In later post, I will post how these apply to squares of other numbers.

If you practice and memorize 2 or 3 digit number multiples, you can multiply really fast. I will post some practice on doing these with each unit number digit and you will get really fast and impressive and you will do some difficult problems quickly on standardized tests like the SAT.