# a*(variable)2 + b*(variable) + c

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a*(variable)2 + b*(variable) + c
CH Factoring polynomials of the form: a*(variable)2 + b*(variable) + c Factor: 6x2 + 11x + 4 STEP 1: Is there a GCF of all terms? NO STEP 2: How many terms are there? 3 Is it of degree 2? YES * Is it in the form a*(variable)2 + b*(variable) + c? YES In this example a = 6, b=11, c = 4 The trick for these trinomials is to multiply a (the coefficient of x2) to c, the constant term, 4. ac = 6*4 = 24. Next, find a pair of factors of that number that add up to the middle term’s coefficient. Since = 11, so let’s use those factors and rewrite the middle term, 11x, as 3x + 11x. 6x x = 6x2 + 3x + 8x + 4 Now we have 4 terms, let’s factor by grouping. = 3x(2x + 1) + 4(2x + 1) common factor = (2x + 1)(3x + 4) ac 24 Factors of 24 Sum of those Factors 1, 24 1+24 = 25 2, 12 = 14 3, 8 3 + 8 = 11 4,6 4 + 6 = 10 3 8 11 b

BOX METHOD 6x2 + 11x + 4 As before, find a pair of factors of 24 (since 6* 4 = 24) that add up to the middle term’s coefficient, 11. We already figured out that 3 * 8 = 24 and = 11, so our factors to use are 3 and 8. First, make a box with the first term, 6x2 in the upper left corner and then last term term, 4 in the lower left corner. GCF’s: Then put in the factors multiplied by x in the other boxes (it doesn’t matter which ones). That is, we will put 3x and 8x in the other boxes. We then proceed to find the GCF of each row and each column of the box. If there is no common factor, just use 1. Now use these GCF’s for your factorization: 2x 1 3x 6x2 3x 4 8x 4 (3x + 4)(2x + 1) !!!

STEP 1: Is there a GCF of all 3 terms? NO
Example 5 Factor: 8y2 – 10y - 3 STEP 1: Is there a GCF of all 3 terms? NO STEP 2: How many terms are there? 3 Is it of degree 2? YES * Is it in the form a*(variable)2 + b*(variable) + c? YES In this example, a = 8, b = -10, c = -3 ac = 8*-3 = -24 b = -10 What pair of factors of -24 will add up to -10? In the previous example we only had to look at each pair once since the last term ac was positive and the middle term, b, was also positive. Now in this example, a is positive (8) and c is negative (-3) so ac= -24, which is negative. The middle term’s coefficient, b, is -10. Factors of 24 Sum of those Factors -1, 24 -1+24 = 23 1, -24 = -23 2, -12 = -10 -2, 12 = 10 -3, 8 = 5 3, -8 = -5 4, -6 = -2 -4,6 = 2 So we will split the middle term, -10y in to 2y + -12y

What was that polynomial again? 8y2 – 10y - 3
GROUPING METHOD: =8y2 + 2y + -12y - 3 =2y(4y + 1) + -3(4y+ 1) =(4y + 1)(2y – 3) BOX METHOD: GCF’s: 2y -3 Since both terms in the right column have a negative coefficient, factor out a negative number. 4y 8y2 -12y 1 2y -3 FACTORIZATION: (4y + 1)(2y - 3)

STEP 1: Is there a GCF of all 3 terms? YES. GCF=4y
Example 7 Factor: 24x2y – 76xy + 40y STEP 1: Is there a GCF of all 3 terms? YES. GCF=4y Factor out 4y from the polynomial. 4y(6x2 - 19x + 10) STEP 2: How many terms are there? 3 Is it of degree 2? YES * Is it in the form a*(variable)2 + b*(variable) + c? YES In this example, a = 6, b = -19, c = 10 ac = 6*10 = 60 b = -19 Since b is negative and ac is positive, both factors of ac must be negative in order for the product to be positive and the sum to be negative. Factors of 24 Sum of those Factors -1, -60 = -61 -2, -30 = -32 -3, -20 = -23 -4, -15 = -19 -5, -12 = -17 -6, -10 = -16 So we will split the middle term, -19x in to -4x + -15x

COMPLETE FACTORIZATION:
Let’s do the grouping method this time: 4y(6x2 - 19x + 10) Let’s just work inside the parentheses for now, but don’t forget that 4y at the end! Inside the parentheses: 6x x x = 2x(3x - 2) (3x – 2) = (3x – 2)(2x – 5) COMPLETE FACTORIZATION: 4y(3x – 2)(2x – 5)

8.4 SPECIAL FACTORING Remember these? (a + b)2 = a2 + 2ab + b2 (a – b)2 = a2 – 2ab + b2 When you see a trinomial that starts and ends with a perfect square, it’s possible the factorization could be a square of a binomial. Example: Factor: 4x2 – 20x + 25 STEP 1: Is there a GCF of all the terms? ___ STEP 2: How many terms are there? __ Is it of degree 2? ___ Are the first and last terms perfect squares? ___ 4x2 can be rewritten as (2x)2, so it is a perfect square. 25 can be rewritten as 52, so it is a perfect square. The bases of those squares are 2x and 5. The middle term of a trinomial can be factored into the square of a binomial is 2*base of the first term * base of the second term. The middle term is -20x = -2(2x)(5). So this trinomial is the square of the DIFFERENCE OF THE BASES (since the middle term is negative). = (2x – 5)2 Check: (2x – 5)2=(2x – 5)(2x – 5)= = (2x)(2x)+ (-5)(2x) + -5(2x) + (-5)(-5) = 4x x + -10x + 25 = 4x2 – 20x + 25

The Difference of Squares
Recall this one: (a – b)(a + b) = a2 – ab + ab – b2 = a2 – b2 So going the other way, a2 – b2 can be factored into (a - b)(a + b) Example: Factor 25x2 – y2 STEP 1: Is there a GCF of all terms? NO STEP 2: How many terms are there? 2 Check if this is a difference of two squares. 25x2 = (5x)2 y2 = (y)2 SO…. 25x2 – y2 = (5x – y)(5x + y) CHECK: (5x – y)(5x + y) = 25x2 - 5xy + 5xy + (-y)(y) = 25x2 – y2