8-4Factoring Trinomials:ax 2 + bx + cMain Ideas• Factor trinomialsof the forma x 2 + bx + c.• Solve equationsof the forma x 2 + bx + c = 0.Standard 11.0Students applybasic factoringtechniques to second- andsimple third-degreepolynomials. Thesetechniques include finding acommon factor for all termsin a polynomial, recognizingthe difference of two squares,and recognizing perfectsquares of binomials.Standard 14.0 Studentssolve a quadratic equationby factoring or completingthe square. (Key)New Vocabularyprime polynomialThe factors of 2 x 2 + 7x + 6 are the dimensions of the rectangle formedby the algebra tiles shown below. The process you use to form the rectangle is the same mental processyou can use to factor this trinomial algebraically.Factor a x 2 + bx + c For trinomials of the form x 2 + bx + c, thecoefficient of x 2 is 1. To factor trinomials of this form, you find the factorsof c with a sum of b. We can modify this approach to factor trinomials forwhich the leading coefficient is not 1.ALGEBRA LAB1. Complete the following table.Product of TwoBinomialsUse FOIL.a x 2 + mx + nx + ca x2+ bx + c m n a c(2x + 3)(x + 4) 2 x 2 + 8x + 3x + 12 2 x 2 + 11x + 12 24 24(x + 1)(3x + 5)(2x - 1)(4x + 1)(3x + 5)(4x - 2)2. How are m and n related to a and c?3. How are m and n related to b?You can use the pattern in the **Algebra** Lab and the method of factoringby grouping to factor trinomials. Consider 6 x 2 + 17x + 5. Find twonumbers, m and n, with the product of 6 · 5 or 30 and the sum of 17.The correct factors are 2 and 15.6 x 2 + 17x + 5 = 6 x 2 + mx + nx + 5 Write the pattern.= 6 x 2 + 2x + 15x + 5 m = 2 and n = 15= (6 x 2 + 2x) + (15x + 5) Group terms with common factors.= 2x(3x + 1) + 5(3x + 1) Factor the GCF from each grouping.= (3x + 1)(2x + 5) 3x + 1 is the common factor.Therefore, 6 x 2 + 17x + 5 = (3x + 1)(2x + 5).Lesson 8-4 Factoring Trinomials a x 2 + bx + c 441

EXAMPLEFactor a x 2 + bx + cFactor each trinomial.a. 7 x 2 + 29x + 4In this trinomial, a = 7, b = 29, and c = 4. You need to find two numberswith a sum of 29 and a product of 7 · 4 or 28. Make an organized list ofthe factors of 28 and look for the pair of factors with the sum of 29.Factors of 28Sum of Factors1, 28 29 The correct factors are 1 and 28.7 x 2 + 29x + 4 = 7 x 2 + mx + nx + 4 Write the pattern.b. 10 x 2 - 43x + 28= 7 x 2 + 1x + 28x + 4 m = 1 and n = 28= (7 x 2 + 1x) + (28x + 4) Group terms with common factors.= x(7x + 1) + 4(7x + 1) Factor the GCF from each grouping.= (7x + 1)(x + 4) Distributive PropertyIn this trinomial, a = 10, b = -43 and c = 28. Since b is negative, m + n isnegative. Since c is positive, mn is positive. So, both m and n are negative.Finding FactorsPut pairs in anorganized list so youdo not miss anypossible pairs offactors.FactoringCompletelyAlways check for aGCF first before tryingto factor a trinomial.List the negativefactors of 10 · 28or 280.10 x 2 - 43x + 28Factors of 280Sum of Factors-1, -280 -281-2, -140 -142-4, -70 -74-5, -56 -61-7, -40 -47-8, -35 -43= 10 x 2 + mx + nx + 28 Write the pattern.= 10 x 2 + (-8)x + (-35)x + 28 m = -8 and n = -35= (10 x 2 - 8x) + (-35x + 28) Group terms with common factors.= 2x(5x - 4) + 7(-5x + 4) Factor the GCF from each grouping.= 2x(5x - 4) + 7(-1)(5x - 4) -5x + 4 = (-1)(5x - 4)= 2x(5x - 4) + (-7)(5x - 4) 7(-1) = -7= (5x - 4)(2x - 7) Distributive Propertyc. 3 x 2 + 24x + 45Look for the pairsof factors with asum of -43.The GCF of the terms 3 x 2 , 24x, and 45 is 3. Factor this out first.3 x 2 + 24x + 45 = 3( x 2 + 8x + 15) Distributive PropertyThe correct factors are -8 and -35.Now factor x 2 + 8x + 15. Since the leading coefficient is 1, find twofactors of 15 with a sum of 8. The correct factors are 3 and 5.So, x 2 + 8x + 15 = (x + 3)(x + 5). Thus, the complete factorization of3 x 2 + 24x + 45 is 3(x + 3)(x + 5).1A. 5 x 2 + 13x + 6 1B. 6 x 2 + 22x - 8 1C. 10 y 2 - 35y + 30442 **Chapter** 8 Factoring