Polynomial and/or Polynomial Functions and Equations
      Definition of a polynomial or polynomial function
      Division of polynomials
      Factoring polynomials and solving polynomial equations by factoring
         Solving quadratic and cubic equations by factoring, examples
Definition of a polynomial or polynomial function
A polynomial and/or polynomial function in the variable x is an expression of the general (or standard) form 
f (x) = anxn + an-1xn-1 + . . .  + a1x + a0
consisting of n + 1 terms each of which is a product of a real coefficient ai and the variable x raised to a non-negative integral power.
If the leading coefficient of a polynomial an is not 0 then, the degree of the polynomial is n. 
The constant term a0 is the y-intercept of the polynomial.
Division of polynomials
Dividing two polynomials,  p (x) and q (x) obtained is quotient Q (x) and remainder R (x), that we write as
    p (x)/q (x) = Q (x) + R (x)/q (x  or    p (x) = Q (x) q (x) + R (x).
Two polynomials are divisible if  R (x) = 0. 
Divide the highest degree term of dividend by the highest degree term of the divisor to get the first term of the quotient.
Take the first term of the quotient and multiply it by every term of divisor. Write this result below the dividend, making sure you line up all the terms with the terms of the dividend that has the same degree. 
Subtract the result from the dividend, i.e., reverse all the signs of the terms of the result and add like terms.
Repeat the process of long division until the degree of the new obtained dividend is less than the degree of the divisor.
Examples:
 
 
 
 
 
 
Note, since each second line should be subtracted, the sign of each term is reversed.
                       b 3x4 + x2 + 5) ( x2 - x - 1)  
 
or
 
   
like     17   5 = 3 + 2/5 
  -15 
      2
Factoring polynomials and solving polynomial equations by factoring
A polynomial and/or polynomial function with real coefficients can be expressed as a product of its leading coefficient an and n linear factors of the form x - xi, where xi denotes its real roots and/or complex roots, 
f (x) = anxn + an-1xn-1 + . . .  + a1x + a0 = an(x - x1)(x - x2) . . . (x - xn).
By multiplying the parentheses on the right side and collecting like terms and then comparing the resulting coefficients with the coefficients of the given polynomial obtained are Vieta's formulas that show relations between coefficients and roots of a polynomial.
Thus, a quadratic or a second degree polynomial
a2x2 + a1x + a0 = a2(x - x1)(x - x2) = a2[x2 - (x1 + x2)x + x1x2],
and similarly a cubic or a third degree polynomial
  a3x3 + a2x2 + a1x + a0 = a3(x - x1)(x - x2)(x - x3)  
  = a3[x3 - (x1 + x2 + x3)x2 + (x1x2 + x1x3 + x2x3)x - x1x2x3].
Solving quadratic and cubic equations by factoring, examples
Example:   Factorize  (2/3)x2 - (2/3)x - using the above theorem.
Solution:    (2/3)x2 - (2/3)x - 4 = (2/3)(x2 - x - 6) = (2/3)[x2 - (3 + (- 2))x + 3(- 2)]
                  = (2/3)(x2 - 3x + 2x - 6) = (2/3)[x(x - 3) + 2(x - 3)] = (2/3)(x - 3)(x + 2)
 Example:  Given are leading coefficient a2 = -1and the pair of conjugate complex roots, x1 = 1 + and
                 x2 = 1 - i, of a second degree polynomial, find the polynomial using the above theorem.
Solution:  By plugging the given values into    a2x2 + a1x + a0 = a2(x - x1) (x - x2)
      a2x2 + a1x + a0 = -1[x - (1 + i)] [x - (1 - i)] = -[(x - 1) - i] [(x - 1) + i] 
                                = -[(x - 1)2 - i2] = -(x2 - 2x + 1 + 1)  = - x2 + 2x - 2
Example: The real root of the polynomial  - x3 - x2 + 4x - 6 is x1 = - 3, factorize the polynomial.
Solution: We divide given polynomial by one of its known factors, since 
a3x3 + a2x2 + a1x + a0 = a3(x - x1) (x - x2) (x - x3)
then we calculate another two roots of given cubic by solving obtained quadratic trinomial,
Finally we use the theorem to factorize given polynomial (see the previous example),
a3(x - x1)(x - x2)(x - x3) = -1(x + 3)[x - (1 + i)][x - (1 - i)] = -1(x + 3)(x2 - 2x + 2).
Notice that given cubic has one real root and the pair of the conjugate complex roots.
Odd degree polynomials must have at least one real root.
 Example:  Solve polynomial equation  x3  +  2x2 - x - 2 = 0 by factoring.
 Solution:          x2(x +  2)  - (x +  2) = 0
                                (x +  2)(x2 -  1) = 0
                       (x +  2)(x + 1)(x - 1) = 0
                                                 x +  2 = 0      =>     x1 = -2
                                                 x +  1 = 0      =>     x2 = -1
                                                  x - 1 = 0      =>     x3 = 1
The roots are,    x1 = -2,    x2 = -1   and    x3 = 1.
College algebra contents C
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