Polynomials, remainder and factor theorems

What you need to know

• A polynomial in x is a sum of terms of the form ax^n where n is a whole number. The degree is the highest power present, so 2x^3 - 3x^2 + 5 has degree 3 and a cubic always has three roots counted with repetition.
• The remainder theorem says that when f(x) is divided by (x - a), the remainder is f(a). You do not divide at all, you simply substitute. For a divisor like (2x - 3), set 2x - 3 = 0 to get x = 3/2 and evaluate f(3/2).
• The factor theorem is the remainder theorem with remainder zero: (x - a) is a factor of f(x) if and only if f(a) = 0. This is how you find the first root of a cubic, by testing the factors of the constant term.
• To find the first root of a cubic with integer coefficients, test only the factors of the constant term divided by the factors of the leading coefficient. For 2x^3 - 3x^2 - 11x + 6 you try 1, -1, 2, -2, 3, -3, 6, -6 and then the halves.
• Once one factor is found, divide the cubic by that factor using long division or synthetic division to get a quadratic, then factorise the quadratic normally. Never attempt to guess all three roots by inspection.
• When two conditions are given, such as one factor and one remainder, each condition gives you one equation in the unknown constants. Two unknowns need two conditions, and you then solve the pair simultaneously.
• For a quadratic ax^2 + bx + c with roots alpha and beta, alpha + beta = -b/a and alpha x beta = c/a. These let you answer questions about the roots without ever finding them.
• To build a new quadratic whose roots are related to the old ones, compute the new sum S and new product P, then write x^2 - Sx + P = 0. For example, roots 1/alpha and 1/beta give S = (alpha + beta)/(alpha beta) and P = 1/(alpha beta).
• For a cubic ax^3 + bx^2 + cx + d with roots alpha, beta, gamma: alpha + beta + gamma = -b/a, the sum of the products in pairs is c/a, and alpha beta gamma = -d/a.
• Useful identity: alpha^2 + beta^2 = (alpha + beta)^2 - 2 alpha beta. This converts any symmetric expression in the roots into the sum and product you already know.
• Partial fractions start from a proper fraction. If the degree of the numerator is not less than the degree of the denominator, divide first, then split the remaining proper fraction. A linear factor (x - a) gives A/(x - a), a repeated factor (x - a)^2 gives A/(x - a) + B/(x - a)^2, and an irreducible quadratic gives (Ax + B)/(x^2 + px + q).
• The quickest way to find partial fraction constants is the cover-up substitution: multiply through by the denominator, then substitute the value of x that kills one bracket at a time.
• For a cubic equation that will not factorise, you may be asked to show that a root lies between two values. Evaluate f at both ends and show that the signs are opposite, which means the curve must cross the x-axis between them.