Topics in math and science.

absolute value, polynomials, factoring

Joe Tuncavage  BS Chemistry

  At this point we will define some important terms of algebra. 

  Absolute value:  the value of a number when sign is discarded; by definition the absolute value is always positive.
                                           For example, |3| = 3 and |-3| = 3

   As we can clearly see, the absolute value does  nothing to positive numbers.

  Algebraic expression: a symbol or combination of symbols which  represents a number.
                                 When the expression consists of several  parts separated by
                                 (+) and (-) signs, each of the parts, including sign, is called a term.

   Monomial: an expression consisting of one term only.

  Polynomial:  An expression consisting of more than one term. Special cases are binomial (2 terms); trinomial (3 terms).

  Factor:  each number used as a factor in multiplication and the product of any [combination] of the numbers multiplied together.

   Any factor(s) of a term is(are) the coefficient(s) of the remaining factors.

   √x² = | x |     By definition, the square root of a positive number (square roots of negative numbers are complex, and will be discussed in another section) must be positive.

   √(-5)² = 5  =  |-5|

   √(3.33)² = 3.33

   Absolute values are encountered in statistics and in many other applications, normally in situations where “sign” is not important.

  A polynomial is a function where there is more than one term:

  P(x) = a₀ + a₁x  + a₂x² + a₃x³ + …. + a_n-1x^n-1 + a_nxⁿ

             _n
         = ∑ a_ixⁱ  =  sum of all terms above
                   ⁱ⁼⁰

   (The a_i terms are called coefficients.)

If n=2, the function becomes P(x) = a₀ + a₁x + a₂x²

   (This is a quadratic function.)

If we set the expression to the right of P(x) = 0:

          a₀  +  a₁ x  +  a₂ x² =  0

  Let   a₀  =  c ;  a₁ = b;  a₂ = a       
      
       ax²  +   bx  +  c = 0  (the quadratic equation)

   We can derive the solution to this eqn:

   Dividing both sides by a:
    x²  +   (b/a)x  + c/a  = 0

   x²  +   (b/a)x  + c/a  +  b²/(4a²) =  b²/(4a²)
    x²  +   (b/a)x  +  b²/(4a²)  =   b²/(4a²) – c/a

   ( x + b/(2a) )²  =   b²/(4a²) – c/a
    √( x + b/(2a) )²   =  ± √ [b²/(4a²) – c/a]

    ( x + b/(2a) )   =   ± √ [b²/(4a²) – c/a]

     x = -b  ±√(b² – 4ac)
               ^2a

1.   Solve  x² – 6x –7 = 0

x = 6 ± √(36 - 4(1)(-7) )  =  6 ±  √(36 + 28) = 6 ± 8

                              2                                    2                         2
         X = 7, -1
                          The above equation could have been factored as follows:

                          (x – 7)( x + 1) = 0  x = 7, -1

      2.  Solve  x² – 4x – 3 = 0

           x =  - (-4) ± √( (-4)² – 4(1)(-3) ) = 4 ± √(16 + 12) =  4 ± √28
                              2(1)                             2                     2
          x = 4 ± (√4)( √7) = 2 ± √7

                           2