FORMULAS

Below are the algebraic laws of addition and multiplication:

Commutative Laws

a+b=b+a

a*b=b*a

Distributive Law

a*(b+c)=a*b+a*c

How do you multiply polynomials?

Take (a*x²+b*x+c)*(x+d) = (a*x²+b*x+c)*x+(a*x²+b*x+c)*d = a*x³+b*x²+c*x+a*d*x²+b*d*x+c*d= a*x³+(b+a*d)*x²+(c+b*d)*x+c*d

Some algebraic formulas

1. a²-b²=(a+b)*(a-b)

2. (a+b)²=a²+2*a*b+b²

3. (a-b)²=a²-2*a*b+b²

4. A^(mn)= (A^m)^n

5. A^(m+n)= (A^m)*A^n

6. A^(-m) = (1/A^m)

Quadratic Equation

Here we will be deriving the general solution to equations of the form Ax²+Bx+C= O where A,B, and C are real number (positive or negative non-zero) coefficients.

A less abstract example equation is 3x²+5x+10=O

This equation is of the second order (x^2) in x.

In order to solve for x, we need an expression that is first order in x, (x^1), or x= constant. If like a surgeon doing an operation we could somehow take the square root of both sides of Ax²+Bx= -C, we might obtain the desired first order equation in x. Trying to take into account the Bx term would seem to be a problem. Suppose, however, we create a perfect square on the left hand side of the equation in X^2;

Ax²+Bx+C=O

X²+Bx/A= -C/A  (Dividing by A and subtracting C/A from both sides)

X²+Bx/A+B²/4A²= B²/4A²-C/A   (complete the square)

Verify that X²+Bx/A+B²/4A²= (x+B/2A)²

Note that we have a perfect square on the left and a constant on the right.

(X+ B/2A)² = B²/4A² – C/A.

Performing the square root:

x+B/2A = ∓√(B²/4A² – C/A)        x= -B/2A ­­∓ ­√(B²/4A² – C/A)

OR

x= -B/2A ∓ √(B²-4AC)/2A                 

                         (Note: there are two roots)

Note that the quantity under the radical must be > or = to 0 in order for two real roots to exist.

If B²- 4AC is less than O, we would be attempting to take the square root of a negative number, which  is known as an imaginary number.

These conclusions come from the fact that there are no negative real numbers that have real number roots.

Example: 

√-4= ?     2² =  (-2)² = 4, Never -4

let i=(-1)^(1/2) a unit imaginary number. Then (-4)^(1/2)=2i

New derivation of quadratic formula:

Ax²+Bx+C= y

 A,B,C real and neq zero 

D= B²-4AC

Interested in the case where equation has complex roots and X is taken to be complex.

Let X= P+IQ     Y= R+IS  I=(-1^(1/2))

Y= A(P+IQ)² + B(P+IQ) + C = R+IS

      A(P²+ 2IPQ -Q²) + BP+IBQ+C = R+IS

      A(P²-Q²)+ BP+C= R   and    2IAPQ+ IBQ = IS

      AP²+BP+C – AQ²= R   and   Q(2AP+B) = S

Let S= O. Y is real. X is complex. P,Q,R,S are real. 

Therefore Q not equal zero since x complex then 2AP+B= O and P therefore equals real part of X which always equals -B/2A when X complex and Y real.

P= -B/2A

A(-B/2A)²+B(-B/2A)+C- AQ²=R = Y    (S=O) 

AB²/4A²-B²/2A+C-AQ²=R

B²/4A-2B²/4A+C-AQ²=R

B²-4AC+4A²Q²= -4AR

D/4A+AQ²=-R

Y=R= -AQ²-D/4A

Lets find the roots of the quadratic for X complex; – AQ²-D/4A=O    AQ²=-D/4A   Q=±√(-D)/2A Since Q real, D<O.

Therefore X=-B/2A±I√(-D)/2A, and X=-B/2A+-(D)^(1/2)/2A  and we have proved the quadratic formula for the case D<O.

The Factorable Cubic Equation

A  cubic equation is in general of the form:

1.) Ax³+Bx²+Cx+D= O.

To be factorable and have integer real roots, the equation must be expressible as

2.) (x-p)(x-q)(x-r) = O. The solutions or roots of the equation are p,q,r since plugging in x=p, x=q, and x=r will make the expression zero. p-p =0, q-q =0, r-r =0 and any real number times zero is zero.

Lets work with #2 up above:

(x-p)(x-q)(x-r) =0

(x²-px-qx+pq)(x-r) =0

(x²-(p+q)x+pq)(x-r) =0

x³- (p+q)x²+ pqx – rx²+r(p+q)x -pqr =0

x³ – (p+q+r)x²+ (pq+rp+rq)x- pqr =0

This to x³+Bx²/A+CX/A+D/A =0 and we get

A’= 1

B’=B/A=- (p+q+r)

C’= C/A=(pq+rp+rq)

D’=D/A=- pqr

This result leads to a simple method for solving such cubics.

x³ – (sum of the roots) x² + (sum of product pair of roots) x- (product of the roots) = 0

EXAMPLE: Solve the equation x³ – 9x²+ 23x-15= 0

We begin by factoring 15; 15= 1*3*5, -1*-3*5, -1*3*-5. The factors of 15 are 1,3,5,15 and the suitable factors -1,-3, 5 or -1,-3,-5 or 1,-3,-5 and 15.

What 3 factors 15 add up to 9? 1+3+5= 9.

What of the product? 1*3*5=15

We have the solution!

(x-1)(x-3)(x-5) =0

(x²-4x+3)(x-5) =0

x³ – 4x² + 3x – 5x² + 20x-15 = 0

X^3-9x^2+23X-15=0

SOLVING CUBICS GIVEN ONE REAL ROOT

All cubic equation graphs cross the x axis at least once; in other words, they all have at least one real root. Take y=Ax^3+Bx^2+Cx+D=0. Assume p is the real root. y=(x-p)(fx^2+gx+h).We can divide x-p into y and we get y/(x-p)=(Ax^2+(B+Ap)x+(C+p(B+Ap)).We have the roots. Example:3x^3-5x^2+10x-24=0. A=f=3.B=-5.C=10.D=-24.We are given p=2. g=B+Ap=-5+3*2=1. h=C+pg=10+2*1=12=-D/p. We can compute the roots using the the quadratic formula:y=(fx^2+gx+h)(x-p)=(3x^2+x+12)(x-2).Therefore x=-1/6+-((1-4*3*12)^(1/2))/6=-1/6+-I((143)^1/2)/6. D=g^2-4fh >0 implies real roots; D<0 implies imaginary roots. Consider y=A(x-p)(x^2+gx/A+h/A). f=A. g=B+Ap h=C+pg. Example: y=x^3-2x^2-23x+60=0.A=1. B=-2. C=-23. We are given p=3.g=B+Ap=1. f=1.h=C+pg=-20.D=g^2-4fh=1+4*1*20=81>0 and we have real roots: x=(-1+-(1-4*(-20))^1/2)/2=(-1+-9)/2=4,-5,and 3.

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