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. 

Solution Set for Basic Completion of the Square

Y=X²+12X-7

B=12
1/2B=6
(X+1/2B)²=(X+6)²

k=-7-6²=-43

Vertex form

Y=(X+6)²-43

Vertex is at: X=-6 and Y=-43

Y=X²-6X+6

B=-6
1/2B=-3
(X+1/2B)²=(X-3)²

k=6-(-3)²=-3

Vertex form

Y=(X-3)²-3

Vertex is at: X=3 and Y=-3

Y=X²+11X-1

B=11
1/2B=11.5
(X+1/2B)²=(X+11.5)²

k=-1-(11.5)²=-1-132.25=-133.25

Vertex form

Y=(X+11.5)²-133.25

Vertex is at: X=-11.5 and Y=-133.25

Y=X²-14X-31

B=-14
1/2B=-7
(X+1/2B)²=(X-7)²

k=-31-(-7)²=-31-49=-80

Vertex form

Y=(X-7)²-80

Vertex is at: X=7 and Y=-80

Y=X²-9x+4

B=-9
1/2B=-4.5
(X+1/2B)²=(X-4.5)²

k=4-(-4.5)²=4-20.25=-16.25

Vertex form

Y=(X-4.5)²-16.25

Vertex is at: X=4.5 and Y=-16.25

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.

Pascals Triangle

Here is a Pascals triangle for up to a 20th order polynomial.  The second number in the series is the order of the polynomial.

                                                                             1

                                                                                                                    1    1

                                                                                                                  1   2   1

                                                                                                                1   3   3   1

                                                                                                            1  4    6    4   1

                                                                                                       1   5    10  10   5   1

                                                                                                     1   6   15   20   15   6   1

                                                                                                  1   7   21   35   35   21   7   1

                                                                                               1   8   28   56   70   56   28   8   1

                                                                                        1   9   36   84   126   126   84   36   9   1     

                                                                               1   10   45   120   210   252   210   120   45   10   1

                                                                          1   11   55   165   330   462   462   330   165   55   11   1  

                                                                      1   12   66   220   495   792   924   792   495   220   66   12   1

                                                                  1   13   78   286   715   1287   1716   1716   1287   715   286   78   13   1

                                                        1   14   91   364   1001   2002   3003   3432   3003   2002   1001   364   91   14   1

                                                 1   15   105   455   1365   3003   5005   6435   6435   5005   3003   1365   455   105   15   1

                                      1   16   120   560   1820   4368   8008   11440   12870   11440   8008   4368   1820   560   120   16   1

                               1   17   136   680   2380   6188   12376   19448   24310   24310   19448   12376   6188   2380   680   136   17   1

                         1   18   153   816   3060   8568 18564   31824   43758   48620   43758   31824   18564   8568   3060   816   153   18   1  
                1   19   171   969   3876   11628   27132   50388   75582   92378   92378   75582   50388   27132   11628   3876   969   171   19   1
1   20   190   1140   4845   15504   38760   77520   125970   167960   184756   167960   125970   77520   38760   15504   4845   1140   190   20   1  

The Derivative of x ^x

^{When taking the derivative of} x ^{x it often looks very hard but this is one of the functions you can always understand if you can understand how it is derived.}

^y=x x

^{The trick here is to take the ln of both sides which allows us to use the logarithmic power rule.}

lny=lnx ^x

^{Power rule of logs:}

lny=xlnx

Take the derivative of both sides and on the right side we will use the quotient rule and on the left side we will use implicit differentiation and the chain rule.

y '(1/y)=lnx +x(1/x)=lnx+1

y '=y(lnx+1)

We know y=x ^x  

                          ^so

^{y '=}x ^x(lnx+1)

Pascal's Triangle

1

1 1      1

2 1   2    1

3 1   3    3    1

4 1    4   6    4    1

5 1   5   10   10   5   1

6 1   6  15   20   15   6    1

7  1  7  21  35  35   21     7   1

On the last post we looked at binomial expansion.  It involved calculating combinations as well as carrying large numbers and was not very easy.  With Pascal's triangle we, there is no need to calculate combinations because they are given on the table.

So lets look at the table.  The numbers in red are the order of polynomial ad the number go in front of the variable from left to right.  To show this lets say that we wanted to see what (x+1)⁴:

This equation is 4th order.  So we use: 1       4       6        4         1

This becomes:   (x+1)⁴=1 x⁴   +  4 x³   +   6 x²   +    4x    +     1= x⁴+4x³+x²+4x+1

That's pretty nifty isn't it?

If we wanted (x+y)⁴=x⁴+4x³y+x²y²+4xy³+y⁴

 So you might ask if you need to memorize this table of does this table only go down to 7th order equations.  The answer is no to both questions.  You can construct this table and it goes down to as large a order as you want.

How do we construct it.

Start with                                                                                                  1

Put 1 on outside of the 1                                                                        1    1

Put one on the outside of the 1 and add the middle numbers:             1    2    1      (2=1+1)

Repeat                                                                                                1    3    3   1    (3=1+1)
Repeat                                                                                             1    4     6   4   1   (4=3+1) (6=3+3)

So if you are observing, you will see the number in the triangle is between two numbers and it is the sum of those 2 numbers.

The order of the equation starts at 1  at the two 1's ie                               1   1    1

The numbers descend from 1 from there:                                                2   1   2    1

                                                                                                              3    1    3   3     1

                                                                                                         ...................................

So just to show how easy it is if we had (x+1)⁸=x⁸+8x⁷+28x⁶+56x⁵+70x⁴ +56x³+28x²+8x+1

Homework:  Construct the 8 the through 10 order of pascals triangle.  I will be posting a large table soon so check back. 

                                                                                                            

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