Factor Polynomials
Tutorial with detailed solutions on factoring polynomials.
Factoring Polynomials Formulas1: One common factor. a x + a y = a (x + y) 2: Sevearl grouped common factor. a x + a y + b x + b y = a(x + y) + b(x + y) = (a + b ) (x + y) 3: Difference of two squares (1). x 2 - y 2 = (x + y)(x - y) 4: Difference of two squares (2). (x + y) 2 - z 2 = (x + y + z)(x + y - z) 5: Sum of two cubes. x 3 + y 3 = (x + y)(x 2 - x y + y 2) 6: Difference of two cubes. x 3 - y 3 = (x - y)(x 2 + x y + y 2) 7: Difference of fourth powers. x 4 - y 4 = (x 2 - y 2)(x 2 + y 2) = (x + y)(x - y)(x 2 + y 2) 8: Perfet square x 2 + 2xy + y 2 = (x + y) 2 9: Perfet square x 2 - 2xy + y 2 = (x - y) 2 10: Perfect cube x 3 + 3x 2y + 3xy 2 + y 3 = (x + y) 3 11: Perfect cube x 3 - 3x 2y + 3xy 2 - y 3 = (x - y) 3 Examples in Factoring Polynomials with SolutionsExample 1Factor the binomial 9 - 4x 2Solution Rewrite the given expression as the difference of two squares then apply formula 1 given above. 9 - 4x 2 = 3 2 - (2x) 2 = (3 - 2x)(3 + 2x) As a practice, multiply (3 - 2x)(3 + 2x) to obtain the given expression. Example 2Factor the trinomial 9x 2 + 3x - 2Solution To factor the above trinomial, we need to write it in the form. 9x 2 + 3x - 2 = (ax + m)(bx + n) Expand the product on the right above 9x 2 + 3x - 2 = abx 2 + x(mb + na) + mn For the polynomial on the left to be equal to the polynomial on the right we need to have equal corresponding coefficients, hence ab = 9 mb + na = 3 mn = -2 Trial values for a and b are: a = 1 and b = 9 or a = 3 and b = 3 Trial values for m and n are: m = 1 and n = -2, m = 2 and n = -1, m = -1 and n = 2 and m = -2 and n = 1. Trying various values for a, b, m and n among the list above, we arrive at: 9x 2 + 3x - 2 = (3x + 2)(3x - 1) As a practice, multiply (3x + 2)(3x - 1) and simplify to obtain the given trinomial. Example 3Factor the polynomial x 3 + 2x 2 - 16x - 32Solution Group terms that have common factors. x 3 + 2x 2 - 16x - 32 = (x 3 + 2x 2) - (16x + 32) Factor the grouped terms = x 2(x + 2) - 16(x + 2) Factor x + 2 out = (x + 2)(x 2 - 16) The term (x 2 - 16) is the difference of two squares and can be factored using formula 1 above = (x + 2)(x + 4)(x - 4) Check the above answer by expanding the obtained result.
ExercisesFactor the polynomials.1: (x + 1) 2 - 4 2: x 2 + 5x + 6 3: x 3 - 1 4: x 3 - x 2 - 25x + 25 Solutions to above exercises 1: (x + 1) 2 - 4 = (x - 1)(x + 3) 2: x 2 + 5x + 6 = (x + 2)(x + 3) 3: x 3 - 1 = (x - 1)(x 2 + x + 1) 4: x 3 - x 2 - 25x + 25 = (x - 1)(x + 5)(x - 5) More references and links to polynomial functionsMultiplicity of Zeros and Graphs Polynomials.Find Zeros of Polynomial Functions - Problems Polynomial Functions, Zeros, Factors and Intercepts |
