a*(variable)2 + b*(variable) + c CH. 8.3 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 3 + 8 = 11, so let’s use those factors and rewrite the middle term, 11x, as 3x + 11x. 6x2 + 11x + 4 = 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 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 3 + 8 = 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 1 + -24 = -23 2, -12 2 + -12 = -10 -2, 12 -2 + 12 = 10 -3, 8 -3 + 8 = 5 3, -8 3 + -8 = -5 4, -6 4 + -6 = -2 -4,6 -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 -1 + -60 = -61 -2, -30 -2 + -30 = -32 -3, -20 -3 + -20 = -23 -4, -15 -4 + -15 = -19 -5, -12 -5 + -12 = -17 -6, -10 -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: 6x2 + -4x + -15x + 10 = 2x(3x - 2) + -5(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) = 4x2 + -10x + -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

Ch. 8.5 Solving Quadratic Equations A quadratic equation is an equation in the form: ax2 + bx + c = 0 Notice this is a trinomial that is set equal to 0. If this trinomial can be factored, we can use the “Principle of Zero Products” to solve this equation. Principle of Zero Products If the product of two factors is zero, then at least one of the factors must be zero. That is, If a*b = 0, then a=0 or b=0. Example 1: Solve 2x2 + x = 6 This does not look at first like a quadratic equation, but if we subtract 6 from both sides, we will have a zero on the right side. 2x2 + x – 6 = 0 Now factor the polynomial. Is there a GCF of all terms? NO How many terms are there? 3 Is it degree 2? YES Is it the form ax2 + bx + c? YES. a=2, b=1, c=-6 ac =2*-6 = -12 b =1 Use factors -3 and 4 since -3+4 = 1 2x2 -3x+4x – 6 = 0 x(2x-3) + 2(2x-3) = 0 (2x-3)(x+2) = 0 The factors are 2x – 3 and x+2. If 2x – 3 = 0 Then 2x = 3 x = 3/2 If x + 2 = 0 x = -2 The possible solutions are x = 3/2 and x= -2 This solution set can be written in braces, not ().

What are we being asked to find? Two consecutive positive integers. Example 2: Solve: (x – 3)(x-10) = -10 We can’t use the “Zero Product Property yet because this product = -10, not 0. We must expand it , get everything on the left hand side and zero on the right hand side, then re-factor it. Example 3: The sum of the squares of two consecutive positive odd integers is equal to 130. Find the two integer. What are we being asked to find? Two consecutive positive integers. Let n = first integer Let n + 2 = second integer. (If n=3, n+2 = 5, if n=5, n+2 = 7, etc…) Information: The sum of the squares of the two integers is 130. n2 + (n+2)2 = 130 n2 + n2 + 4n + 4 = 130 Simplify.. 2n2 + 4n + 4 = 130 Get everything on the left hand side. 2n2 + 4n - 126 = 0 Is there a GCF of all terms? ___ What is it?____ 2(n2 + 2n - 63) = 0 How many terms are inside the parentheses? __What degree? ___ What factors of -63 can be added to get the middle term, 2? 2( n + 9 )( n - 7)= 0 So the zero product property says either n+9= or n-7 = 0 for their product to be 0. n=-9 is not a possible answer because it is not positive. n=7 n+2 = 7+2 =9 7 and 9 are the two positive consecutive integers.