Polynomials

(a+b)² = a² + 2ab + b²

(a+b)(c+d) = ac + ad + bc + bd

a² - b² = (a+b)(a-b) (Difference of squares)

a³ b³ = (a b)(a² ab + b²) (Sum and Difference of Cubes)

x² + (a+b)x + ab = (x + a)(x + b)

If ax² + bx + c = 0, then x = ( -b(b² - 4ac) ) / 2a (The Quadratic Formula for solving second degree equations ... see below.)

Solving Quadratic Equations

There are two common methods used to solve 2nd degree (quadratic) equations: factoring or the quadratic formula. In both cases, the equation being solved must be put into standard form (ax² + bx + c = 0) as a first preliminary step. Standard form requires that all variables and numbers be simplified and grouped onto one side of the equation -and- set equal to zero on the other side of the equation. For example, if the original equation is: x(x + 2) = 8 ... the standard form would be:

x² + 2x - 8 = 0

At this point ... the most common solution method is to attempt to factor the equation. The equation x² + 2x - 8 = 0 (from above) can be factored:

( x - 2 ) ( x + 4 ) = 0

Assuming the equation can be factored, one then cross equates each of the factors to zero ... and solves the sub-equations:

x - 2 = 0 -or- x + 4 = 0

x = 2 -or- x = -4

If the equation can NOT be solved by factoring, then the quadratic formula is used. The "a," "b," and "c" values for the equation are identified and then "plugged into" the quadratic formula.In our example (above): a=1, b=2, and c=-8.

Note One: In the quadratic formula it is assumed that the ENTIRE top of the equation is divided by the entire bottom.

Note Two: It is assumed that the "b² - 4ac" must be computed prior to taking the square root.

Note Three: If "b² - 4ac" is positive, then the equation has two real number roots (solutions). If "b² - 4ac" is zero, then the equation has only one real number root (solution). If "b² - 4ac" is negative, then the equation has NO real number roots (solutions) ... it has two complex number solutions.

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