Name:________________________ Date:______________ 1 Chapter 6 Factoring Polynomials Lesson 1 Standard Factoring Monomials Example 1 Example 2 Example 3 California Algebra I Standard 11.0 Find a common factor to all of the terms in a polynomial. is the reverse of multiplying polynomials. To factor an expression means to write an equivalent expression that is a product of two or more expressions. To factor a monomial – find two monomials whose products equal the first. Factoring out a common factor. 6a + 12b + 18c = 61a + 62b + 63c) = 6(a +2b+3c) 5x + 15 = 51x+53 = 5(x + 3) 3ab2 + 6a = 3ab2 + 3a(2) = 3a(b2 + 2) 16a2b2+ 20a2 = 4a2(4b2) + 4a2(5) = 4a2(4b2 +5)

Name:________________________ Date:______________ 2 Chapter 6 Lesson 1 Factoring Polynomials Factoring cont. Example 4 Example 5 Try These Factoring out a common factor. 16a 2 b a 2 = 4a 2 (4b 2 ) + 4a 2 (5) = 4a 2 (4b 2 +5) 15x 5 –12x 4 +27x 3 -3x 2 = 3x 2 5x 3 -3x 2 4x 2 +3x 2 9x-3x 2 1 = 3x 2 (5x 3 -4x 2 +9x-1) Factor 1)x 2 + 3x=_______2)a 2 b + 2ab =________ =_______ =__________ 3) 3x 6 – 5x 3 + 2x 2 = __________________ =__________________ 4) 9x 4 -15x 3 + 3x 2 =__________________ =__________________ 5) 2p 3 q 2 +p 2 q + pq =__________________ =___________________________

Name:________________________ Date:______________ 3 Chapter 6 Lesson 2 Factoring Differences of two Squares Standard Try These Algebra I 11.0 – Recognize and factor the difference of two square. For a binomial to be a difference of two squares two conditions must hold. 1) There must be two terms both of which are squares such as 4x 2 and 9x 4. 2) There must be a minus sign between the two terms. Difference of squares 1)x 2 – 25 2) 4x 2 – 81 3) a 2 4) 9m Not a difference of squares 1)x 2 – 24 2) 4x 2 – 15 3) -36 – x 2 Which are perfect squares? 1)4x 2 – 64 _____ 2) 9x 2 – 28 ______ 3) x 4 ____ 4) 25x 2 – 1 ________ 5) 16x _______

Name:________________________ Date:______________ 4 Chapter 6 Factoring Differences of Two Squares Factor Perfect squares 1 2 = = = = = = = = = = = = = = =225 1)x 2 – 4 = (x -2)(x +2) 2) x 2 – 9 = ___________ 3) y =_____________ 4) 4x = ___________ 5) m 6 – 16n 2 = ___________ 6) 36x 2 – 25y 6 = _______________ 7) 16x 2 – 25y 2 = ____________ 8) 25a 10 – 36b 8 = __________________ 9) 4y 2 – 49 = ______________ x2x2 2x -2x-4

Name:________________________ Date:______________ 5 Chapter 6 Lesson 3 and 4 Factoring x 2 +bx + c Standard Trinomial Squares Algebra Recognize and factor perfect squares of binomials. These can be factored using a diamond method. In the diamond you put the c term in the top and the b term in the bottom. Then ask yourself what two numbers add to b and multiply to equal c. 1)x x + 25 = For example 5 and 5 add to 10 and multiply to equal 25. The factors are then (x + 5)(x+ 5) 2) x 2 + 6x + 9 The numbers are 3 and 3 so the factorization of x 2 + 6x + 9 = (x + 3)(x + 3) c b 9 6

Name:________________________ Date:______________ 6 Chapter 6 Lesson 3,4 Factoring x 2 +bx + c More Examples. 3) x 2 – 14x + 49 = ________________ 4) x x + 64 = _________________ = (______)(______) ) x 2 + 7x + 12 = _____________ 6) x x + 36 =____________

Name:________________________ Date:______________ 7 Chapter 6 Lesson 3 and 4 Factoring second degree polynomials. 3) x 2 – 8x + 15 =__________________ 4) x 2 – 9x + 20 = _________________ 5) x 2 – 7x + 12 =________________ 6) a 2 + 7ab + 10b 2 = ______________________ 7) m 2 + 8mn + 15n 2 =____________ 8) a 2 + 5ab + 6b 2 =_______________

Name:________________________ Date:______________ 8 Chapter 6 Lesson 5 Factoring second degree polynomials. Standard Remember!!! Algebra I 11.0 Factoring second degree polynomials. Always check for common factors in the terms and factor them out first!! Factor 6x x ) Make a diamond but put a·c in the top spot and b in the bottom spot. so 6·20 = 120 2) Figure out the factors. In this case 20 and 3 work. 3) Create a box. Factor out like terms. 3x + ? 2x ? Try different possibilities for ? 10 and 2 work so the factors are (3x + 10) and (2x + 2) 3x +10 2x x 2 20x 3x20 6x 2 20x 6x20 ax 2 + bx + c

Name:________________________ Date:______________ 9 Chapter 6 Lesson 5 Factoring second degree polynomials. Example 2: Example 3: Factor: 6x 2 + 7x + 2 Any common factors? No So make diamond. 6 x 2 = 12 What numbers work? Make box. Factor out common factors. 2 x + ? 2x + 1 3x 3x +? +2 Try possibilities for ? 2 and 1 work so the factors are (3x + 2)(x + 1). Factor 8x x – 3 Any common factors? No. 12 and -2 work. so make box 2x + ? 4x + - ? 2x and -1 work and so the 4x factors are (4x-1)(2x-3) x 2 3x 4x2 6x 2 3x 4x x 2 12x -2x-3 8x 2 12x -2x-3

Name:________________________ Date:______________ 10 Chapter 6 Lesson 5 Factoring second degree polynomials. 8m 2 – 8m – 6 Can you factor out anything? Yes. 2 So 2(4m 2 -4m -3). Now factor. 4 x -2 = -12 so use -6 and 2. Now build the box. 2m+? Pull out 2m common factors. ? What are the ? 2m -3 -3, 1 2m Put them in. -1 Then the factors have to include the 2 from the beginning of the problem so the answer is 2(2m -3)(2m + 1). 1) 21a x + 2 =_______________________________ Factor Try These m 2 -6m 2m-3 4m 2 -6m 2m-3

Name:________________________ Date:______________ 11 Chapter 6 Lesson 5 Factoring second degree polynomials. Try These Remember to check for what can be factored out first! 2) 8x x + 3 3) 3x 2 -21x–+36 Remember to factor out a 3 first! ____________________________ 4) 4a 2 + 2a -6 __________________________ 5) 6m mn – 9n 2 _____________________________

Name:________________________ Date:______________ 12 Chapter 6 Lesson 7 Factoring Standard Rules Agebra I Standard 11.0 Apply basic factoring techniques of trinomials. 1. Always look for a common factor to remove by factoring it out. 2. If there is a not a number in front of the first term then use the diamond to factor the trinomial. 3. If there is a number in front of the first term then use the diamond with a ·c in the top box. Then use the area box to factor out common factors and figure out the missing numbers. Factor: 1. 10x 2 – 40x Factor out 10x so 10x(x 2 – 4) then factor so the factors are 10x(x – 2)(x + 2) 2. t 2 – 16 No common factor so use diamond. and factors are (x – 4)(x + 4)

Name:________________________ Date:______________ 13 Chapter 6 Lesson 7 Factoring 3. 2a 4 + 8a 3 + 6a 2 Factor out 2a 2 (________) Use diamond. Factors are 2a 2 (______)(________) 4. 2x 2 -11x + 12 No common factor so use diamond. ______________________ 1) 3m 4 – 3 _______________ 2) x 6 + 8x _______________ 3)8x 3 – 200x ________________ 4) y 5 – 2y 4 -35y 2 Try These

Name:________________________ Date:______________ 14 Chapter 6 Lesson 8. Solving Equations by Factoring Standard Zero Principle Add -1 to both sides. Divide both sides by 5 Algebra I Standard 14 Solve a quadratic equation by factoring. If two numbers have a product of zero then one of them must be zero. If a·b = 0 then either a = 0 or b = 0. We can use this principle to solve equations. Example 1: (5x + 1)(x – 7) = 0 Either 5x + 1 = 0 or x – 7 = 0 Solve each of these individually. 5x + 1 = 0orx – 7 = x = -1 x = 7 x = -1/5 or The solutions for this equation are x = -1/5 and 7. Example 2: x(2x – 9) = 0 so x = 0 or 2x – 9 = x = 9 x = 9/2 The solutions are 9/2 and 0

Name:________________________ Date:______________ 15 Chapter 6Lesson 8. Solving Equations by Factoring Try These 1) 2) 3) 4) Sometimes you have to factor first and then solve. Add 16 to both sides (x – 3)(x – 4) = 0 so x- 3 = 0 or x-4 = 0 solutions are ________________ (x – 7)(x - 3) = 0 so __________or__________ solutions are _______________________ y(3y – 7) = 0 so _________or__________ solutions are _____________________ (4t + 1)(3t – 2) = 0 so __________or_____________ solutions are _______________________ Example 1: x 2 – 8x = -16 x 2 – 8x + 16 = 0 (x - 4)(x – 4) = 0 x = 4 is only ans

Name:________________________ Date:______________ 16 Chapter 6 Lesson 8. Solving Equations by Factoring Solve Try These x – 4 = 0or x – 4 = 0 solution is 4 Example 2: x 2 + 5x + 6 = 0 (________)(_______) then _________or _____________ solutions are ____________________ 1) x 2 – x – 6 = 0 solutions are ______________ 2) m 2 – m = 56 solutions are _____________ 3) x 2 – 3x = 28 solutions are __________________

Name:________________________ Date:______________ 17 Chapter 6 Lesson 8. Solving Equations by Factoring Finding the Roots of a polynomial Steps To find the root or zero of a polynomial 1) set the polynomial equal to zero 2) Factor the polynomial 3) Set each factor equal to zero 4) Solve each equation. Example 1: Find the roots of the polynomial x 2 – 5x 1) x 2 – 5x = 0 2) x(x – 5) = 0 3) x = 0or x – 5 = 0 4) Solutions are x = 0 or x = 5 Example 2: Find the roots of 4x 2 – 25 1) 4x 2 – 25 = 0 2) (2x – 5)(2x + 5) = 0 3) 2x – 5 = 0or2x+ 5 = 0 2x = -5or2x = 5 x = -5/2orx = 5/2 -

Name:________________________ Date:______________ 20 Chapter 6 Lesson 8. Solving Equations by Factoring Try These1) x 2 + 6x + 9 __________________ _________or __________ 2) x 2 + 4x __________________ ______ or ________ ______ or _________ 3) 25x 2 – 16 ________________ __________________ ______ or ________ -

Name:________________________ Date:______________ 18 Chapter 6 Lesson 9 Using Equations that factor Area problems Try These Example 1: The width of a rectangular card is 2cm less than the length. The area is 15 cm 2. Draw a rectangle and label. l(l – 2) = 15 l 2 – 2l = 15 l 2 – 2l – 15 = 0 (l -5)(l + 3) = 0 l = 5 or l = -3 Only l = 5 works because distance can not be negative. The width is l – 2 = 3. 1) The area of a rectangular area is 10cm 2 The length is 3ft longer than the width. Find the length and width of the rectangle. 2) The length of a rectangle is 5 cm greater than the width. The area of the rectangle is 84 m 2. Find the length and width. 15cm Algebra I 10.0 Solve word problems using techniques with polynomials..Algebra I 14.0 Solve a quadratic equation by factoring. l w=l-2