6.3 Factoring Trinomials II - Valencia College

6.3 Factoring Trinomials II - Valencia College

6.3 Factoring Trinomials II Ax2 + bx + c Factoring Trinomials Review X2 + 6x + 5 (x )(x )

Find factors of 5 that add to 6: 1*6 = 6 2*3 = 6 (x + 2)(x + 3) 1+6 = 7 2+3 = 5 Factoring Trinomials where a 1

Follow these steps: 1. Find two numbers that multiply to ac and add to b for ax2 + bx + c 2. Replace bx with the sum of the 2 factors found in step 1. ie: ax2 + bx + c becomes ax2 + mx + nx + c, where m and n are the factors found in step 1. 3. Use grouping to factor this expression into 2 binomials

2x + 5x + 2 2 Step 1: ac = 2*2 = 4 1*4 = 4 1+4 = 5 2*2 = 4 2+2 = 4 m = 1 and n = 4 Step 2: Rewrite our trinomial by expanding bx 2x2 + 1x + 4x + 2 Step 3: Group and Factor (2x2 + 1x) + (4x + 2) x(2x + 1) + 2( 2x + 1) (2x + 1) (x + 2)

2x + 5x + 2 2 Questions for thought: 1. Does it matter which order the new factors are entered into the polynomial? 2. Do the parenthesis still need to be the same? 3. Will signs continue to matter when finding m and n? 4. Does it matter how we group the terms for factoring?

3z + z 2 2 Step 1: ac = 3*-2 = -6 -1*6 = -6 -2*3 = -6 -1+6 = 5 -2+3 = 1 1* -6 = -6 1+-6 = -5 2* -3 = -6 2+-3 = -1 m = -2 and n = 3

Step 2: Rewrite our trinomial by expanding bx 3z2 + 3z 2z 2 Step 3: Group and Factor (3z2 + 3z) + (-2z - 2) 3z(z + 1) - 2( z + 1) (z + 1) (3z - 2) 3z + z 2 2 Step 1: ac = 3*2 = 6 -1*6 = -6 -2*3 = -6

-1+7 = 6 -2+3 = 1 1* -6 = -6 2* -3 = -6 1+-7 = -6 2+-3 = -1 m = -2 and n = 3 Step 2: Rewrite our trinomial by expanding bx 3z2 + 3z 2z 2 Notice that I changed the order of m and n between step 1 and step 2. Why do you

think I did this? Do you have to change the order to get the correct answer? 3z + z 2 2 What are the 3 steps for solving this quadratic equation? Step 1: Multiply a*c. Find the factors that multiply to ac and add to b Step 2: Expand bx to equal mx + nx Step 3: Group and Factor

4x 22x + 30x 3 2 Step 0: Factor out the GCF: 2x 2x(2x2 11x + 15) Step 1: a*c = 30

-1*-30 = 30 -2*-15 = 30 -3*-10 = 30 -5*-6 = 30 -1+-30 = -31 -2+-15 = -17 -3+-10 = -13 -5+-6 = -11 4x 22x + 30x 3

2 Step 0: Factor out the GCF: 2x 2x(2x2 11x + 15) Step 1: a*c = 30 -1*-30 = 30 -2*-15 = 30 -3*-10 = 30 -5*-6 = 30 -1+-30 = -31

-2+-15 = -17 -3+-10 = -13 -5+-6 = -11 4x 22x + 30x 3 2 Step 2: Expand bx to equal mx + nx -11x = -5x + -6x

2x(2x2 5x 6x + 15) Step 3: Group and Factor 2x((2x2 5x )( 6x + 15)) 2x(x(2x 5) -3(2x 5)) 2x(2x 5) (x 3) 4x 22x + 30x 3 2

Step 2: Expand bx to equal mx + nx -11x = -5x + -6x 2x(2x2 5x 6x + 15) Step 3: Group and Factor 2x((2x2 5x )( 6x + 15)) 2x(x(2x 5) -3(2x 5)) Note: The Parenthesis are the Same 2x(2x 5) (x 3) Practice

1. 3x2 + 5x + 2 2. 6x2 + 7x 3 3. 6 + 4y2 11y Practice 1. 3x2 + 5x + 2 (3x + 2)(x + 1) 2. 6x2 + 7x 3 (3x 1)(2x + 3) 3. 6 + 4y2 11y (4y 3)(y 2)

Review What is Step 0? When do you need to include this step? When will your factors both be negative? When will you have one negative and

one positive factor? How do you check your answers? ??? Questions ??? 6.4 Special Types of Factoring 1. Differnce of Squares 2. Perfect Square Trinomials (Sum and Difference of Cubes is not included) Difference of Squares

Think back to Chapter 5. What happened when we multiplied a sum and difference? (a b)(a + b) = a2 b2 So, the reverse is also true. a2 b2 = (a b)(a + b) x 25 2

Notice that we do not have a bx term. This means that we only have the F and L in foil; therefore, none of the procedures from 6.1, 6.2, or 6.3 will work. We need to use a2 b2 = (a b)(a + b) where a = x and b = 5

X2 25 = (x 5)(x + 5) x 36 2 We need to use a2 b2 = (a b)(a + b) where a = x and b = 6 X2 36 = (x 6)(x + 6)

Practice 4x2 9 100 16t2 49y2 64z2 Practice

4x2 9 a = 2x, b = 3 (2x 3) (2x + 3) 100 16t2 a = 10, b = 4t (10 4t) (10 + 4t) 49y2 64z2 a = 7y, b = 8z (7y 8z) (7y + 8z) Perfect Square Trinomials

Think back to Chapter 5. What happened when we squared a binomial? (a + b)2 = a2 + 2ab + b2 (a b)2 = a2 2ab + b2 So, the reverse is also true. a2 + 2ab + b2 = (a + b)2 a2 2ab + b2 = (a b)2 x2 + 10x + 25

This can be worked 2 different ways The first way is the simplest, but depends on whether you recognize the equation as a perfect square trinomial. a2 + 2ab + b2 = (a + b)2 Where a = x and b = 5 x2 + 10x + 25 = (x + 5)2 x2 + 10x + 25

This can be worked 2 different ways The second way is to use the method we learned in 6.2 x2 + 10x + 25 5*5 = 25 and 5+5 = 10 (x + 5) (x + 5) or (x + 5)2 4x2 - 4x + 1

This can be worked 2 different ways The first way is the simplest, but depends on whether you recognize the equation as a perfect square trinomial. a2 + 2ab + b2 = (a + b)2 Where a = 2x and b = 1 4x2 - 4x + 1 = (2x 1)2

4x2 - 4x + 1 This time we need to use the 6.3 method 4*1 = 4 -2 * -2 = 4 and -2 + -2 = -4 (4x2 2x) ( 2x + 1) 2x(2x 1) 1(2x 1) (2x 1) (2x 1) or (2x 1)2 Practice

x2 4xy + 4y2 9a2 60a + 100 25y2 + 20yz + 4z2 Practice x2 4xy + 4y2 a = x, b = 2y (x 2y)2

9a2 60a + 100 a = 3a, b = 10 (3a 10) 25y2 + 20yz + 4z2 a = 5y, b = 2z (5y + 2z) Review

What methods can you use to factor a Difference of Squares? What methods can you use to factor a Perfect Square Trinomial? What clues should you look for to identify a Difference of Squares? What clues should you look for to identify a Perfect Square Trinomial? ??? Questions ???

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