Polynomials and Polynomial Functions Chapter 5 Polynomials and Polynomial Functions

Chapter Sections 5.1 – Addition and Subtraction of Polynomials 5.2 – Multiplication of Polynomials 5.3 – Division of Polynomials and Synthetic Division 5.4 – Factoring a Monomial from a Polynomial and Factoring by Grouping 5.5 – Factoring Trinomials 5.6 – Special Factoring Formulas 5.7-A General Review of Factoring 5.8- Polynomial Equations Chapter 1 Outline

§ 5.5 Factoring Trinomials

Factoring Trinomials Recall that factoring is the reverse process of multiplication. Using the FOIL method, we can show that (x – 3)(x – 8) = x2 – 11x + 24. Therefore x2 – 11x + 24 = (x – 3)(x – 8). Note that this trinomial results in the product of two binomials whose first term is x and second term is a number (including its sign).

Factoring Trinomials Factoring any polynomial of the form x2 + bx + c will result in a pair of binomials: Numbers go here. x2 + bx + c = (x +?)(x +?) O (7x + 3)(2x + 4) F I L = 14x2 + 28x + 6x + 12 = 14x2 + 34x + 12

(x + one number) (x + second number) Factoring Trinomials Find two numbers (or factors) whose product is c and whose sum is b. The factors of the trinomial will be of the form (x + one number) (x + second number)

Examples a.) Factor x2 - x – 12. a = 1, b = -1, c = -12. We must find two numbers whose product is c, which is -12, and whose sum is b, which is -1. We begin by listing the factors of -12, trying to find a pair whose sum is -1. continued

Examples Factors of -12 (1)(-12) (2)(-6) (3)(-4) (4)(-3) (6)(-2) (12)(1) Sum of Factors 1 + (-12) = -11 2 + (-6) = -4 3 + (-4) = -1 4 + (-3) = 1 6 + (-2) = 4 12 + (-1) = 11 The numbers we are seeking are 3 and -4 because their product is -12 and their sum is -1. x2 - x – 12 = (x + 3)(x – 4)

Factor out a Common Factor The first step when factoring any trinomial is to determine whether all three terms have a common factor. If so, factor out the GCF. Example Factor 3x4 – 6x3 – 72x2. The factor 3x2 is common to all three terms. Factor it out first. = 3x2(x2 – 2x – 24) We find that -6 and 4 are the factors to the trinomial in the parentheses. Therefore, 3x4 – 6x3 – 72x2 = 3x2(x – 6)(x + 4)

Trial and Error Method Write all pairs of factors of the coefficient of the squared term, a. Write all pairs of factors of the constant, c. Try various combinations of these factors until the correct middle term, bx, is found.

Trial and Error Method Example: Factor 3t2 – 13t + 10. There is no factor common to all three terms. Next we determine that a is 3 and the only factors of 3 and 1 are 1 and 3. Therefore we write, 3t2 – 13t + 10 = (3t )(t ) Next, we look for the factors that give us the correct middle term, - 13t. Continued.

Sum of the Products of the Inner and Outer Terms Trial and Error Method Example continued: Possible Factors Sum of the Products of the Inner and Outer Terms (3t – 1)(t – 10) -31t (3t – 10)(t – 1) -13t (3t – 2)(t – 5) -17t (3t – 5)(t – 2) -11t Thus, 3t2 – 13t + 10 = (3t – 10)(t – 1).

Factor Trinomials of the Form ax2 + bx + c, a ≠ 1, Using Grouping To Factor Trinomials of the Form ax2 + bx + c, a ≠ 1, Using Grouping Find two numbers whose product is a · c and whose sum is b. Rewrite the middle term, bx, using the numbers found in step 1. Factor by grouping.

Factor Trinomials Using Substitution Sometimes a more complicated trinomial can be factored by substituting one variable for another. Example Factor 3z4 – 17z2 – 28. Let x = z2. Then the trinomial can be written Now substitute z2 for x.