Warm- Up#26 (Friday, 10/30) Evaluate: (3𝑥+1) 2 Factor by grouping: 2x² – 10x + 3x – 15 =0 3. How are you suppose to behave when there is a sub teacher? Provide three specific examples.

Homework Factor Packet page 5 and pg 6

Step by Step Is there a GCF? Is it a binomial? Yes No Factor as the product of the GCF and one other factor—i.e. GCF•(the other factor). Look at the other factor and go to the next step below with it. No Go the the next step. Is it a binomial? Is it a difference of two squares? (a2-b2) Yes—Factor as (a+b)(a-b). No—It can’t be factored any more. Go to the next step.

Is it a trinomial? Yes Do you recognize it as a pattern for a perfect square trinomial? (a2+2ab+b2) or (a2-2ab+b2) Yes—Factor as (a+b)2 or (a-b)2 No—Go to next step. Use the ac and b pattern to look for factors. Can you find factors of ac that add up to b? Yes—Rewrite the equation with those factors, group, and factor. No—You can’t do anything else. If there’s no GCF, it’s a prime polynomial. No Go to the next step.

Is it a four-term polynomial? Yes Are there two sets of terms that you can group together that have a common factor? Yes—Group and factor. No—If it doesn’t have a GCF, it’s a prime polynomial. No If it doesn’t have a GCF, it’s a prime polynomial.

Factoring Polynomials by using the Diamond Method

Diamond Method for factoring ax2 + bx + c Factor out the GCF, if any For the remaining trinomial, multiply ac Look for factors of ac that sum to b Rewrite the bx term as a sum using the factors found in step 3 Factor by grouping Check by multiplying using FOIL

Ex: Factor b2 + 6b + 5 1. there is no GCF 2. the lead coefficient is 1  (1b )(1b ) 3. Look for factors of 5 1, 5 & 5, 1 (b + 1)(b + 5) or (b + 5)(b + 1) 4. outer-inner product? (b + 1)(b + 5)  5b + b = 6b or (b + 5)(b + 1)  b + 5b = 6b Either one works  b2 + 6b + 5 = (b + 1)(b + 5) 5. check: (b + 1)(b + 5) = b2 + 5b + b + 5 = b2 + 6b + 5

2. the lead coefficient is 1  (1y )(1y ) Ex: Factor y2 + 6y – 55 1. there is no GCF 2. the lead coefficient is 1  (1y )(1y ) 3. Look for factors of – 55 1, -55 & 5, - 11 & 11, - 5 & 55, - 1 (y + 1)(y – 55) or (y + 5)(y - 11) or ( y + 11)(y – 5) or (y + 55)(y – 1) 4. outer-inner product? (y + 1)(y - 55)  -55y + y = - 54y (y + 5)(y - 11)  -11y + 5y = -6y (y + 55)(y - 1)  -y + 55y = 54y (y + 11)(y - 5)  -5y + 11y = 6y  y2 + 6y - 55 = (y + 11)(y – 5) 5. check: (y + 11)(y – 5) = y2 – 5y + 11y - 55 = y2 + 6y – 55

Factor completely: 2x2 – 5x – 7 Factors of the first term: 1x & 2x Factors of the last term: -1 & 7 or 1 & -7 (2x – 7)(x + 1)

Factor Completely. 4x2 + 83x + 60 Nothing common Factors of the first term: 1 & 4 or 2 & 2 Factors of the last term: 1,6 2,30 3,20 4,15 5,12 6,10 Since each pair of factors of the last has an even number, we can not use 2 & 2 from the first term (4x + 3)(1x + 20 )