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 va ..read more

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+kBoth of these equations are equivalent so We expand (X+1/2B)², we have (X+1/2B)²=X²+BX+1/4B²We know that (X+1/2B ..read more

Solution Set for Basic Completion of the SquareY=X²+12X-7B=12 1/2B=6 (X+1/2B)²=(X+6)²k=-7-6²=-43Vertex formY=(X+6)²-43Vertex is at: X=-6 and Y=-43 Y=X²-6X+6B=-6 1/2B=-3 (X+1/2B)²=(X-3)²k=6-(-3)²=-3Vertex formY=(X-3)²-3Vertex is at: X=3 and Y=-3 Y=X²+11X-1B=11 1/2B=11.5 (X+1/2B)²=(X+11.5)²k=-1-(11.5)²=-1-132.25=-133.25Vertex formY=(X+11.5)²-133.25Vertex is at: X=-11.5 and Y=-133.25 Y=X²-14X-31B=-14 1/2B=-7 (X+1/2B)²=(X-7)²k=-31-(-7)²=-31-49=-80Vertex formY=(X-7)²-80Vertex is at: X=7 and Y=-80 Y=X²-9x+4 B=-9 1/2B=-4.5 (X+1/2B)²=(X-4.5)²k=4-(-4.5)²=4-20.25=-16.25Vertex formY=(X-4.5)²-16.25Ve ..read more

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 ..read more

Pascals TriangleHere 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 ..read more

The Derivative of x xWhen 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                            soy '=x x(lnx+1 ..read more

Pascal's Triangle 11 1      12 1   2    13 1   3    3    14 1    4   6    4    15 1   5   10   10   5   16 1   6  15   20   15   6    17  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 t ..read more

Binomial Theory Using Polynomial ExpansionIf 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+4x ..read more

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 ..read more

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 ..read more