By Hand

• 1). Put the polynomial into the form ax^2 + bx + c, if the polynomial is second-order.
  
  In this case, a, b and c are constants, and x is a variable. (Note that if a=0, then the polynomial is first-order, that is, just a linear equation and can be solved by hand with basic arithmetic manipulations. In other words, bx + c = 0 gives bx = -c, or x = -c/b.)
  
  
• 2). Use the quadratic formula to solve for the zeros.
  
  The quadratic formula is x = [-b +/- √(b^2 -- 4ac)] / [2a], which is the solution to the equation ax^2 + bx + c = 0. (This can be proven using the method of completing squares or by inserting the solution into the formula to see that it holds true.)
  
  +/- here means "plus or minus." If b^2 -- 4ac is not equal to zero, the above equation gives two zeros. (Note that n-th order polynomials have n zeros.)
  
  
• 3). Use i to represent √(-1), if b^2 -- 4ac < 0.
  
  This will give complex numbers as the zeros of ax^2 + bx + c. Complex numbers are numbers that include the imaginary number i. In the x-y plane, the curve of ax^2 + bx + c would not touch the x-axis at any value for x. Therefore, some math books discard the answer as meaningless. However, defining the square root of a negative number as the product of i times the square root of the absolute value of the number gets around this obstacle.
  
  For example, √(-100) = √(-1) --- √100 = i --- 10 = 10i.
  
  

By Computer: Bisection Method

• 1). Determine two values x near the expected zero of the polynomial, P(x), such that the polynomial is of opposite sign at the two points.
  
  That is, the two values x1 and x2 should be such that sign(P(x1)) = - sign(P(x2)).
  
  Therefore, the two x values will bound the zero, and P(x1) and P(x2) will be above and below the x-axis, bounding the x-axis.
  
  
• 2). Calculate the midpoint between x1 and x2.
  
  In other words, define x3 = (x1 + x2) / 2.
  
  
• 3). Determine the sign of the polynomial at x3, that is, the sign of P(x3).
  
  
• 4). Discard the x-value that gives the same sign as x3.
  
  For example, if x1 and x3 give the same sign for P(x), discard x1.
  
  
• 5). Calculate the midpoint of the two remaining x values, as was done in Step 2.
  
  
• 6). Keep repeating steps 3-5 until P(x) is closer to zero than some tolerance, where x is the approximation of the root in that region.
  
  The x that pushes the absolute value of P(x) below that tolerance level, for example, 0.001, is the numerical approximation of the zero for the P(x). In other words, an x has been found such that |P(x) - 0| < tolerance level.
  
  
• 7). Then repeat the above process, starting at Step 1 again, to find all other zeros of P(x).
  
  (Note that n-th order polynomials have n zeros.)
  
  

By Computer: Newton-Raphson

• 1
  Newton-Raphson Method (Courtesy: CS Dept. of UC Davis)
  
  Solve for the derivative of the polynomial, P(x).
  
  The monomial of form c --- x^n, where c and n are constants, has derivative cn --- x^(n-1). The derivative of a polynomial is the sum of the derivatives of the monomials. The derivative of a constant by itself is zero, since a constant's graph is flat, that is, not varying with x. So, for example, the derivative of P(x) = 2x^2 + 3 is P`(x) = 4x. (Note the mark by the P indicating the derivative.)
  
  
• 2). Make a best guess as to the zero of the polynomial, just to have a starting point. Call it x1.
  
  
• 3). Solve x2 = x1 - P(x1) / P`(x1).
  
  
• 4). Repeat step 3 to produce x3 from x2, x4 from x3, and so on.
  
  
• 5). Stop the iteration (repetition of Step 3) after an x has been found that places P(x) as close to zero as desired.