Algebraic Expressions
      Factoring polynomials
      Using a variety of methods including combinations of the above to factorize algebraic expressions
Factoring polynomials
A polynomial and/or polynomial function f (x) = anxn + an-1xn-1 + . . . + a1x + a0
in the variable x 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 conjugate complex roots, thus
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, for a quadratic or a second degree polynomial
a2x2 + a1x + a0 = a2(x - x1)(x - x2) = a2[x2 - (x1 + x2)x + x1x2],
and similarly, for 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].
Example:   Factorize  2/3x2 - 2/3x - using the above theorem.
Solution:    2/3x2 - 2/3x - 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 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, 
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)]
                                         = - (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.
Using a variety of methods including combinations of the above to factorize algebraic expressions
Examples:   a)  x2 - 2xy + y2 + 2y - 2x = (x - y)2 - 2(x - y) = (x - y)(x - y - 2),
b)  x2 - y2 + xz - yz = (x - y)(x + y) + z(x - y) = (x - y)(x + y + z),
c)  4x- 4xy  + y2  - z2 = (2x - y)2   - z2 = (2x - y - z)(2x - y + z),
d)  a- 7a + 6 = a- a - 6a + 6 = a(a2 -1) - 6(a -1) = (a -1)[a(a + 1) - 6] = (a -1)(a2 + a - 6) =
                             = (a -1)(a2 + 3a - 2a - 6) = (a -1)[a(a + 3) - 2(a + 3)] = (a -1)(a + 3)(a - 2).
Pre-calculus contents A
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