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. 

Simple Factoring of Quadratics

How do you factor quadratics?

We are going to learn a simple method for factoring quadratics when the leading multiplier is 1 and the quadratic has non-complex integer solutions that are positive.
If you are at this point you have learned how to FOIL if not visit the link.

Lets expand (x+8)(x+21)
(x+8)(x+21)=x²+29x+168

Now how would we factor something like this if we did not know what the factors are.
If we expand a quadratic, it looks like this:  (x+A)(x+B)=x²+(A+B)x+AB
The second term or the number before x is the addition of 2 roots or A+B
The last term is the multiplication of 2 roots or AB.

First If we factor 168 into all of its roots, we will find all potential factors:

168 factors into these pairs:     1         168
                                                 2         84
                                                 3         56
                                                 4         42
                                                 6         28
                                                 8         21
                                                 12       14
                                                 24         7

Now we see which 2 pairs add up to 29.  The only 2 that add up to 29 are 8 and 21.

Lets do a couple of simpler problems where we do not already know the factors:

Example 1:      x²+12x+32

Lets factor 32:                                                     32
                                                                          2    16
                                                                                2    8
                                                                                     2   4
                                                                                         2   2
                                                      or
                                                    1         32
                                                    2         16
                                                    4          8

So
1 and 32 add up to 33
2 and 16 add up to 18
4 and 8 add up to 12 so this is our solution set so it looks like this factored:
x²+12x+32=(x+4)(x+8)

One more example:
x²+10x+21

 Lets factor 21:                                                 1        21
                                                                         3         7

We see the only 2 number that add up to 10 are 3 and 7 so:
x²+10x+21=(x+3)(x+7)

In a future post we will do numbers with negative roots.

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

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