Completing the Square
For Leading Coefficients that aren't 1

We looked at completing the square for equations who's leading coefficients are 1.  If the leading factor is not 1 we factor the number out of the leading coefficient and the second term so that the leading coefficient is one and then we proceed to complete the square.

The technique is very easy.

Lets look at a quadratic equation in the standard form.

Y=AX²+BX+C 
We factor A out of both the A and B term.
Y=A(X²+[B/A]X)+C 
Notice how we don't factor the A term out of C

To write this equation in vertex form, we take 1/2 of the B/A value and write our equation like this:

Y=(X+[1/2][A/B])²+k

It looks kind of messy but lets use an example.
Y=2X²+4X-3

Our equation now looks like this:

Y=2(X²+[4/2]X)-3 =2(X²+2X)-3
Y=2(X+1)²-3-(2)(1)²=2(X+1)²-5
Vertex is at X=-1 Y=-5 or (-1,-5)

So watch what we did with the k term.  
Lets expand 2(X+1)²

2X²+4X+2So we gained a 2 in the expansion of our modified equation so we have to also subtract a 2.
So our form becomes A(X-h)²+k
k is calculated by taking C-A(h)²

So lets do another example:
Y=-3X²-12X+4
Factor the -3 out of the first and second term:
-3(X²+4X)+4
Complete the square:
-3(X+2)²+4-(-3)(2)²=-3(X+2)²+16
The vertex is at:  X=-2  Y=16 or (-2,16)

Practice:
Y=6X²+18X+2
Y=-2X²+5X+3
Y=-7X²-21X+14

Answers posted in next post:

Completing the Square

Completing the square is a technique used to take a quadratic equation and put it in a form called the vertex form.  This is an important technique that does not stop just in algebra.  We will revisit this in the calculus post for integration techniques.

The technique is very easy.

Lets look at a quadratic equation in the standard form.

Y=X²+BX+C  

To write this equation in vertex form, we take 1/2 of the B value and write our equation like this:

Y=(X+1/2B)²+k

Both of these equations are equivalent so
We expand (X+1/2B)², we have (X+1/2B)²=X²+BX+1/4B²

We know that (X+1/2B)²-k will expand to: X²+BX+1/4B²+k=X²+BX+C 

So C=1/4B² + k or if we want to find k we do the following:
k=C-1/4B²

Not too bad right?  We are going to now do an example.

Y=X²+6X+8
B=6
1/2B=3

C=8

(X+1/2B)²=(X+3)²

k=8-3²=-1

So our equation becomes:

Y=(X+3)²-1
Vertex form is (X-h)²+k
The vertex is at x=-3 and y=-1 or (-3,-1)

Lets look at:

Y=X²-4X-3

B=-4

1/2B=-2

C=-3

(X+1/2B)²=(X-4)²
k=-3-(-2)²=-3-4=-7

So our Vertex form becomes:

Y=(X-2)²-7
Our vertex is at: (-2,-7)

Note that:   Even though our 1/2B term was negative we still ended up subtracting it from the C term.

Homework:

Factor the following into the vertex form by completing the square:

Y=X²+12X-7
Y=X²-6X+6

Y=X²+11X-1
Y=X²-14X-31
Y=X²-9x+4

The solutions are posted here.

In future posts, we will be looking at completing the square on equations with multipliers on the leading coefficient, a synthetic way of completing the square, and an explanation of why completing the square works. 

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.

Polynomial Expansion 

(x+a)(x+b)=x²+(a+b)x+ab
Is the first and many times the last way we learn to expand polynomial equations.  The equation on the left side of the equal sign is known as the factored form and the equation on the right side is known as the general form.  This is very simple.  Third, forth, fifth, and nth order polynomials are very easy to understand also even though most people are not taught how to do it.

Lets look at the expansion of a third order polynomial:

(x+a)(x+b)(x+c)=x³+(a+b+c)x²+(ab+ac+bc)x+abc

And lets look at a forth order expansion:

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

How do we arrive at these expansion solutions?  Lets multiply through and see.

(x+a)(x+b)(x+c)

Lets multiply (x+a)(x+b)=x²+(a+b)x+ab

Lets multiply  (x+a)(x+b)(x+c)=(x+c)(x+a)(x+b)=(x+c)[x²+(a+b)x+ab]

Multiplying x =x³+(a+b)x²+abx
Multiplying c=cx²+c(a+b)x+abc=cx²+(ca+cb)x+abc=

(x+c)[x²+(a+b)x+ab]=x³+(a+b)x²+abx+cx²+(ca+cb)x+abc
Combine terms:  x³+(a+b)x²+cx²+(ca+cb)x+abx+abc=x³+(a+b+c)x²+(ab+ac+bc)x+abc

This is a little tedious but there is a pattern:

• The first term is going to be an exponent of the order of the polynomial.
• The second term is going to be the addition of all the roots times xⁿ⁻¹
• The last term will be the multiplication of all the roots.  This is consistent among all polynomials.
• The middle terms are just permutations of the roots.

So lets look at (x+a)(x+b)(x+c)(x+d)(x+e)(x+f)=

• First lets look at the second term:  That will be (a+b+c+d+e+f)x⁵
• The last term is: abcdef
• The third term is ab+ac+ad+ae+af+bc+bd+be+bf+cd+ce+cf+de+df+ef
• The forth term is abc+abd+abe+abf+acd+ace+acf+ade+adf+aef+bcd+bce+bcf+bde+bdf+bef+cde+cdf +cef+def
• The fifth term: abcd+abce+abcf+acde+acdf+adef+bcde+bcdf+bdef+cdef
• The sixth term is abcde+abcdf+acdef

So (x+a)(x+b)(x+c)(x+d)(x+e)(x+f)=x⁶+(a+b+c+d+e+f)x⁵+(ab+ac+ad+ae+af+bc+bd+be+bf+cd+ce+cf+de+df+ef)x⁴+(abc+abd+abe+abf+acd+ace+acf+ade+adf+aef+bcd+bce+bcf+bde+bdf+bef+cde+cdf

+cef+def )x³+(abcd+abce+abcf+acde+acdf+adef+bcde+bcdf+bdef+cdef)x²+(abcde+abcdf+acdef)x + abcdef

That is pretty long.  The main idea is that is the second term and the last term are always very easy to ascertain.  While we have formulas for the quadratic equation, we do not have similar equations for finding roots of higher order polynomials but there are many things we can use to solve this equation like multiple similar roots, even and odd roots as well as other roots.   We will look at those in the subsequent posts; We will also look at expansion of the equations with the leading or first terms being numbers other than 1;   We will also use these equations to explain the binomial expansion theory; We will also look at algorithms for solving and expanding polynomials.

Lots of good things to follow.

¹  ²  ³  ⁴  ⁵  ⁷  ⁸  ⁹

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  Binary

What do we mean by binary?  All digits have 2 numbers 1 and 0.

This is not initially intuitive but it is not too bad after we compare it to a system we commonly use which is based on tens.  This system is called the decimal system.

When we count in tens we do it as follows.
1 2 3 4 5 6 7 8 9 and after we get to the end of are units digit we carry a 1 to our next digit and get 10.
So as we go further from 10 it becomes 11 12 13 14 15 16 17 18 19 and we get to the end of our units digit and we carry a 1 into our tens digit and add it to the 1 already there and we have 20.

You may be asking why we are doing this.  We already know this and it seems natural but our new number system is similar but because we only have 2 numbers we go like this.
1 10 11 100 101 110 111 1000 1001...............
Okay this is becoming more clear.  As you can see our 2 numbers are 1 and 0 but every time we go from 1 to the next number we carry into the next digit. 

Lets go a little further in our counting:
1001 1010 1011 1100 1101 1110 1111 10000 10001 10010 10011 10100 10101 10110 10111 11000 11001 11010 11011 11100 11101 11110 11111 100000

Okay now you got it.  Practice counting by looking away from the numbers above and check once you are done.  You will get it pretty quick.

Now how would we convert binary to our decimal number system:

Lets look at a number. 11101 Now lets break it a part and we have 1 in the 5th place and 1 in the 4th place and 1 in the 3rd place 0 in the 2nd place 1 in the unit or ones place.

So this is more clear we will count along with the decimal equivalent so we can see what is happening.
1 2    3   4     5     6     7     8       9       10     11      12     13     14     15    16      17        18       19
1 10 11 100 101 110 111 1000 1001 1010 1011  1100 1101 1110 1111 10000 10001 10010 10011
 20       21        22      23       24       25       26       27       28      29     
10100 10101 10110 10111 11000 11001 11010 11011 11100 11101

Now that doesn't seem too easy does it?  There is an easier way but this demonstrates how the number systems correspond.
First any digit that is 0 will represent 0 for that digit.
The first digit is equal to 1 so 1=1 and 0=0
The second digit equal to 2 so 10=2 and 00=0
The third digit is equal to 2² =4 so 100=4
The forth digit is equal to 2³=8 so 1000=8
The fifth digit is equal to 2⁴ =16 so 10000=16

So in our 11101=10000+1000+100+1=16+8+4+1=29

So each digit equals 2⁽ⁿ⁻¹⁾ .  If we have a number, 1001001=2⁽⁷⁻¹⁾ +2⁽⁴⁻¹⁾+1=  2⁶ +2³+1=64+8+1=73

Try some numbers and use a binary to decimal calculator to test your answer.

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.

30 x 30 Multiplication Table

It is very common for students to learn to multiply with a 10 x 10 multiplication table.  After this many students just use a calculator.  Most students will practice on speed and I have had students in calculus that still cant multiply simple numbers without a calculator.  I am going to present ways to find patterns and also how to progress from a 10 x 10 multiplication table to a 30 x 30.  Understanding a 30 x 30 table will help on standardized tests as well as help students later in their math career.  

Cut and paste and do whatever you want with this table but feel free to use it as you see fit.

	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22	23	24	25	26	27	28	29	30
1	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22	23	24	25	26	27	28	29	30
2	2	4	6	8	10	12	14	16	18	20	22	24	26	28	30	32	34	36	38	40	42	44	46	48	50	52	54	56	58	60
3	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
4	4	8	12	16	20	24	28	32	36	40	44	48	52	56	60	64	68	72	76	80	84	88	92	96	100	104	108	112	116	120
5	5	10	15	20	25	30	35	40	45	50	55	60	65	70	75	80	85	90	95	100	105	110	215	120	125	130	135	140	145	150
6	6	12	18	24	30	36	42	48	54	60	66	72	78	84	90	96	102	108	114	120	126	132	238	144	150	156	162	168	174	180
7	7	14	21	28	35	42	49	56	63	70	77	84	91	98	105	112	119	126	133	140	147	154	261	168	175	182	189	196	203	210
8	8	16	24	32	40	48	56	64	72	80	88	96	104	112	120	128	136	144	152	160	168	176	285	192	200	208	216	224	232	240
9	9	18	27	36	45	54	63	72	81	90	99	108	117	126	135	144	153	162	171	180	189	208	308	216	225	234	243	252	261	270
10	10	20	30	40	50	60	70	80	90	100	110	120	130	140	150	160	170	180	190	200	210	220	230	240	250	260	270	280	290	300
11	11	22	33	44	55	66	77	88	99	110	121	132	143	154	165	176	187	198	209	220	231	242	253	264	275	286	297	308	319	330
12	12	24	36	48	60	72	84	96	108	120	132	144	156	168	180	192	204	216	228	240	252	264	276	288	300	312	324	336	348	360
13	13	26	39	52	65	78	91	104	117	130	143	156	169	182	195	208	221	234	247	260	273	286	299	312	325	338	351	364	377	390
14	14	28	42	56	70	84	98	112	126	140	154	168	182	196	210	224	238	252	266	280	294	308	322	336	350	364	378	392	406	420
15	15	30	45	60	75	90	105	120	135	150	165	180	195	210	225	240	255	270	285	300	315	330	345	360	375	390	405	420	435	450
16	16	32	48	64	80	96	112	128	144	160	176	192	208	224	240	256	272	288	304	320	336	252	368	384	400	416	432	448	464	480
17	17	34	51	68	85	102	119	136	153	170	187	204	221	238	255	272	289	306	323	340	357	274	391	408	425	442	459	476	493	510
18	18	36	54	72	90	108	126	144	162	180	198	216	234	252	270	288	306	324	342	360	378	296	414	432	450	468	486	504	522	540
19	19	38	57	76	95	114	133	152	171	190	209	228	247	266	285	304	323	342	361	380	399	318	437	456	475	492	513	532	551	570
20	20	40	60	80	100	120	140	160	180	200	220	240	260	280	300	320	340	360	380	400	420	440	460	480	500	518	540	560	580	600
21	21	42	63	84	105	126	147	168	189	210	231	252	273	294	315	336	357	378	399	420	441	462	483	504	525	542	567	588	609	630
22	22	44	66	88	110	132	154	176	208	220	242	264	286	308	330	252	274	296	318	440	462	484	506	528	550	572	594	616	638	660
23	23	46	69	92	215	238	261	285	308	230	253	276	299	322	345	368	391	414	437	460	483	506	529	552	575	595	618	641	667	690
24	24	48	72	96	120	144	168	192	216	240	264	288	312	336	360	384	408	432	456	480	504	528	552	576	600	624	648	672	696	720
25	25	50	75	100	125	150	175	200	225	250	275	300	325	350	375	400	425	450	475	500	525	550	575	600	625	650	618	700	725	750
26	26	52	78	104	130	156	182	208	234	260	286	312	338	364	390	416	442	468	494	520	546	572	595	624	650	676	702	728	754	780
27	27	54	81	108	135	162	189	216	243	270	297	324	351	378	405	432	459	486	513	540	567	594	618	648	675	702	729	756	783	810
28	28	56	84	112	140	168	196	224	252	280	308	336	364	392	420	448	476	504	532	560	588	616	641	672	700	728	756	784	812	840
29	29	58	87	116	145	174	203	232	261	290	319	348	377	406	435	464	493	522	551	580	609	638	667	696	725	754	783	812	841	870
30	30	60	90	120	150	180	210	240	270	300	330	360	390	420	450	480	510	540	570	600	630	660	690	720	750	780	810	840	870	900

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.