Quadratic: Bridge Method

• 1). Remove the coefficient of x^2 and multiply it by the constant term.
  
  For example, 2x^2 - 5x -12 = 0 becomes x^2 - 5x - 24 = 0.
  
  
• 2). Find factors of the (new) constant term that sum to the coefficient of x.
  
  Continuing with the example, -8 and 3 are such numbers because -8*3 = -24 and -8 + 3 = -5.
  
  
• 3). Rewrite the quadratic in factored form. So x^2 - 5x - 24 = (x-8)(x+3).
  
  
• 4). Divide both constants by the coefficient you removed in Step 1.
  
  So (x-8)(x+3)=0 becomes (x-4)(x+3/2)=0.
  
  
• 5). Multiply the entire quadratic by the coefficient removed in Step 1.
  
  (x-4)(x+3/2)=0 becomes 2(x-4)(x+3/2)=0. You can multiply this out to confirm that this formula equals the same quadratic you started with in Step 1.
  
  
• 6). Determine what x makes each formula in parentheses zero. These are your zeroes.
  
  So x = 4 and x = -3/2 are the zeroes.
  
  

Cubic

• 1). Factor a cubic of the form x^3 - y^3 using the equation (x-y)(x^2+xy+y^2).
  
  For example, 27x^3 - 64 = (3x)^3 - 4^3 = (3x-4)(9x^2+12x+16).
  
  
• 2). Factor a cubic of the form x^3 + y^3 using the equation (x+y)(x^2-xy+y^2).
  
  
• 3). Solve a cubic of the form x^3 + px = q using "Vietta's substitution." Let x = w - p/3w. Then substituting this back into the original formula gives you w^3 - p^3/(27w^3) - q = 0. Then multiply through by w^3 and solve like any other quadratic. Once you know w^3 using regular quadratic factoring, you'll know w and therefore x as well.
  
  

Complex Roots of a Quadratic

• 1). Solve for complex roots using the quadratic formula, first by putting the quadratic equation in the form ax^2 + bx + c = 0, where a, b and c are constants.
  
  
• 2). Solve for b^2-4ac. If this value is negative, you will have two complex roots. If the value is positive, you won't have complex roots.
  
  
• 3). Take the square root of the positive value of b^2-4ac. Multiply the result by i, which stands for the square root of -1. If b^2-4ac is positive, don't multiply by i. Denote the result by the capital letter C.
  
  
• 4). Solve for (-b+C)/2a and (-b-C)/2a. These are your two roots.
  
  For example, x^2+x+1=0 has C=√(1-4)= √-3 = 3i. So the two roots are (-1+3i)/2 and (-1-3i)/2.