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Algebra II

The Modulus Function
  • No line with a modulus ever goes under the x-axis.
  • Any line that does go below the x-axis, when modulated is reflected above it

Polynomials
  • To find unknowns in a given identity:
    • Substitute suitable values of x
    •                 OR
    • Equalize the given coefficients of like powers of x.
  • Factor Theorem: If (x – t) is a factor of the function p(x), then p(t) = 0
  • Remainder Theorem: If the function f(x) is divided by (x – t), then the remainder, R = f(t)
Binomial Series
  • Expanding (1 + x)n, where |x| < 1
  • Factor Case: If the constant is not 1, pull out a factor from the brackets to make it 1, and use the general equation. Do NOT forget the indices.
  • Substitution Case: If the bracket contains more than one x term (e.g. 2 – x + x2), then make the last part u, expand and then substitute back in.
  • Finding the Limit of x in the Expansion: E.g. (1 + ax)n, the limit can be found by substituting ax between the modulus sign in |x| < 1 and altering it to have only x in the modulus.
Partial Fractions

  • Multiply through by (px + q), substitute x = -q/p and find A
  • Multiply through by (rx + s), substitute x = -s/r and find B

  • Multiply through by (px + q), substitute x = -q/p and find A
  • Multiply through by (rx + s)2, substitute x = -s/r and find C
  • Substitute any constant, e.g. x = 0, and find B.

  • Multiply through by (px + q), substitute x = -q/p and find A
  • Take A/px + q to the other side, subtract and simplify.
  • The linear equation left at the top is equal to Bx + C
  • Improper Fraction case
    • If the numerator has x to the degree of power equivalent or greater than the denominator, then another constant is present.
    • This can be found by dividing the denominator by the numerator and using the remainder.
Example
    1. Express the following in partial fractions:

      • Expand the brackets:
      • The greatest power of x is the same for the numerator and denominator, thus this is an improper case.
      • Make it into a proper fraction:
      • This is then written as:
      • Now proceed with the normal case for the fraction:
      • A(2x – 3) + B(x + 1) = 5 – 5x
      • when x = -1
      • -5A = 5 + 5
      • ∴ A = -2
      • when x = 3/2
      • 5/2B = 5 – 15/2
      • ∴ B = -1
      • Thus the partial fraction is:
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