How to Solve a Second-Order Polynomial

The general form for a second-order polynomial, which is also called the quadratic equation, is ax^2 + bx + c = 0, where a, b and c are constants. Solving the quadratic equation is one of the fundamental skills of algebra and is usually taught to first year algebra students. Due to the "x^2" term, a second-order polynomial will usually have two solutions, although there is a special case that would lead to only one solution.

Instructions

1. Write the second-order polynomial in the general form, ax^2 + bx + c = 0. This is especially true if multiple terms of the same order are present.
   -14x + 15 + x^2 = -5 - x^2
   -14x + 15 + 5 + x^2 + x^2 = 0
   2x^2 - 14x + 20 = 0

2. Divide each term by the coefficient of the second-order term, which is "a" in the general form. In this example, 2x^2 is the second-order term, so you should divide each term by two.
   2x^2 - 14x + 20 = 0
   x^2 - 7x + 10 = 0

3. Factor the second order polynomial by determining the factors of the first and third terms and determining which sign to use between them. In this example, the factors for the first term are x, x. The factors for the third term are 2, 5.
   x^2 - 7x + 10 = (x 2)(x 5)
   If the third term in the polynomial is positive, as it is in this case, then the sign in both factors is the same as the sign of the second term in the polynomial.
   x^2 - 7x + 10 = (x - 2)(x - 5)
   If the third term is negative, then the signs in the factors are opposite, and the greater value in the factor has the same sign as the second term in the polynomial.
   x^2 + 4x - 5 = (x + 5)(x - 1)

4. Add the constants in the factors to ensure they equal the coefficient of the second term. Use the signs determined in the previous step. If they don't equal the second term coefficient, you will need to select a different set of factors to use for the third term.
   x^2 - 7x + 10 = (x - 2)(x - 5)
   -2 - 5 = -7

5. Set both factors equal to zero and solve for x.
   x - 2 = 0; x = 2
   x - 5 = 0; x = 5
   The solution for this example is x = (2, 5)

Tips & Warnings

• If b = 0 and c is positive, then there is no middle term and the equation has no real solution. If b = 0 and c is negative, then the solution is x = (positive square root of c, negative square root of c) If the factors are the same, then there is only one solution. For example, the solution for (x - 2)(x - 2) is x = 2.

References

• Practical Algebra, A Self-Teaching Guide; Peter Selby and Steve Slavin; Wiley Publishing, 2nd ed., 1991