Chapter 8: Factoring

Chapter 8 : Factoring Fill in the titles on the foldable Prime Factoring & Factor a monomial 8.1 Greatest Common Factor (GCF) 8.1 Factor Using Distributive Property 8.2 Factory by Grouping 8.2 Zero Product Property 8.2 Factoring Trinomials – x2 + bx + c 8.3 Factoring Trinomials – ax2 + bx + c 8.4 Factoring Differences of Squares 8.5 Factoring Perfect Squares 8.6 Square Root Property 8.6 Rational Expressions 11.2

8.1 Prime factoring and factor a monomial (top) Prime # = factors only include 1 and itself Composite # = more than two factors Ex: Prime factor 90 Prime numbers: 1, 3, 5, 7, 11, 13, 17, 23, 29, 31, 37 …. 90 2 45 3 15 3 5 2 x 3 x 3 x 5 = 2 x 32 x 5

8.1 Prime factoring and factor a monomial (bottom) Factor a monomial = list all factors separately (no exponents) Ex: -66pq2 Ex: 38rs2t -66 38 66 -1 2 19 2 33 2 x 19 x r x s x s x t 3 11 -1 x 2 x 3 x 11 x p x q x q

8.1 Greatest Common Factor (top) GCF = the largest factor that is in all the given monomials 1. factor all monomials 2. circle all common factors 3. Multiply all the circled factors

8.1 Greatest Common Factor (bottom) Ex: 84 & 70 Ex: 36x2y & 54xy2z 36 84 70 54 2 18 2 42 2 35 2 27 2 9 2 21 5 7 3 9 3 3 3 7 3 3 2 x 2 x 3 x 7 2 x 2 x 3 x 3 x x x x x y 2 x 5 x 7 2 x 3 x 3 x 3 x x x y x y x z 2 x 7= 14 2 x 3 x 3 x x x y= 18xy

8.2 Factor Using Distributive Property (top) Find the GCF of the monomials Write each term as a product of the GCF and the remaining factors Simplify the remaining factors

8.2 Factor Using Distributive Property (bottom) Ex: 12a2 + 16a Ex: 3p2q – 9pq2 + 36pq -9 36 12 16 -1 9 2 18 2 6 2 8 3 3 2 9 2 3 2 4 3 3 2 2 x 3 x p x p x q 2 x 2 x 3 x a x a -1 x 3 x 3 x p x q x q = 2 x 2 x a =4a 3 x p x q = 3pq 2 x 2 x 2 x 2 x a 2 x 2 x 3 x 3 x p x q 4a(3a) + 4a(4) = 3pq(p) + 3pq(-3q) + 3pq(12) = 4a(3a + 4) 3pq(p - 3q + 12)

8.2 Factor by Grouping (top) Group the terms (first two and last two) Find the GCF of each group Write each group as a product of the GCF and the remaining factors Combine the GCFs in a group and write the other group as the second factor

8.2 Factor by Grouping (bottom) Ex: 4ab + 8b + 3a + 6 Ex: 3p – 2p2 – 18p + 27 (4ab + 8b)(+ 3a + 6) (3p – 2p2 )( – 18p + 27) -18 4 6 27 8 -1 18 2 2 2 3 3 9 2 4 2 9 3 3 2 2 3 3 2 x 2 x a x b 3 x a 3 x p = 3 = p -1 x 2 x 3 x 3 x p =4b 2 x 2 x 2 x b 2 x 3 2 x p x p = 9 3 x 3 x 3 4b(a + 2) +3 (a + 2) p(3 – 2p) + 9(-2p + 3) (4b + 3)(a + 2) (p + 9)(-2p + 3)

8.2 Zero Product Property (top) Roots = the solutions to the equation When an equation is factored and equal to zero: Set each factor equal to zero and solve for the variable

8.2 Zero Product Property (bottom) Ex: 7f2 – 35f = 0 Ex: (d – 5)(3d + 4) = 0 -35 d – 5 = 0 3d + 4 = 0 7 x f x f = 7f + 5 + 5 - 4 - 4 -1 x 5 x 7 x f -1 35 3d = -4 d = 5 5 7 /3 /3 d = -4/3 7f(f) + 7f(-5) 7f(f – 5) = 0 Roots are d = 5 and -4/3 f – 5 = 0 7f = 0 + 5 + 5 /7 /7 f = 5 f = 0 Roots are f = 0 and 5

8.3 Factoring Trinomials – x2 + bx + c (top) Get everything on one side (equal to zero) Split into two groups ( )( ) = 0 Factor the first part x2 (x )(x ) = 0 Find all the factors of the third part (part c) Fill in the factors of c that will add or subtract to make the second part (bx) Foil to check your answer Use Zero Product Property to solve if needed

8.3 Factoring Trinomials – x2 + bx + c (bottom) Ex: x2 + 6x + 8 Ex: r2 – 2r - 24 Ex: s2 – 11s + 28 = 0 8 1, 8 2, 4 (x )(x ) 24 1, 24 2, 12 3, 8 4, 6 28 1, 28 2, 14 4, 7 (r )(r ) (s )(s ) (x + 2)(x + 4) (s- 4)(s - 7) = 0 (r + 4)(r - 6) FOIL x2 + 2x + 4x + 8 x2 + 6x + 8 FOIL s2 – 7s – 4s + 28 s2 – 11s + 28 FOIL r2 – 6r + 4r - 24 r2 - 2x - 24 s – 4 = 0 s – 7 = 0 +4 +4 +7 +7 s = 4 s = 7 s = 4 and 7

8.4 Factoring Trinomials – ax2 + bx + c (top) Get everything on one side (equal to zero) Put the first part in each set of parentheses Find product of the first and last parts Find the factors of the product Fill in the pair of factors that adds or subtracts to the second part Remove the GCF from one set of parentheses Write what is left of the that group as one factor and then the other group as the other factor if you can’t factor = prime (use the zero product property to solve if needed)

8.4 Factoring Trinomials – ax2 + bx + c (bottom) Hint: find the gcf to pull it out and make the numbers smaller if possible Ex: 5x2 + 13x + 6 Ex: 10y2 - 35y + 30 = 0 2 x 6 = 12 5 x 6 = 30 5(2y2 - 7y + 6) = 0 1, 12 2, 6 3, 4 (5x )(5x ) 1, 30 2, 15 3, 10 5, 6 5(2y )(2y )=0 (5x + 10)(5x + 3) 5(2y - 4)(2y - 3)=0 5x: x 5 = 5 2 x y x y 10: 2 5 = 2 2 x 2 (5x + 10) 5(x + 2) (2y - 4) 2(y - 2) (x + 2)(5x + 3) 5(y - 2)(2y - 3) = 0 Solve for y. y – 2 = 0 2y – 3 = 0 y = 2 and 1.5

8.5 Factoring Differences of Squares (top) Factor each term Write one set of parentheses with the factors adding and one with the factors subtracting Foil to check your answer Hint: find the gcf to pull it out and make the numbers smaller if possible Ex: n2 - 25 Ex: 9x3 – 4x n x n 5 x 5 x(9x2 – 4) (n + 5)(n - 5) x[ 3x x 3x 2 x 2] x(3x + 2)(3x - 2)

8.5 Factoring Differences of Squares (bottom) Ex: 5x3 + 15x2 – 5x - 15 Ex: 121a = 49a3 -121a -121a 5[x3 + 3x2 – x – 3] 0 = 49a3 – 121a 5[ (x3 + 3x2)( – x – 3)] 0 = a(49a2 – 121) 0 = a(7a x 7a 11 x 11) 0 = a(7a + 11)(7a - 11) x x x x x -1 x x = x2 = -1 3 x x x x -1 x 3 a = 0 7a + 11 = 0 7a - 11 = 0 -11 -11 +11 +11 5[ x2(x + 3) - 1(x + 3)] 7a = -11 7a = 11 /7 /7 /7 /7 5[(x2 – 1)(x + 3)] 5[(x x x 1 x 1)(x + 3)] a= -11/7 a = 11/7 5(x + 1)(x - 1)(x + 3) a = -11/7, 0, and 11/7

8.6 Factoring Perfect Squares (top) Perfect Square Trinomial: Is the first term a perfect square? Is the last term a perfect square? Does the second term = 2 x the product of the roots of the first and last terms? If any of these answers is no- it is not a perfect square trinomial

8.6 Factoring Perfect Squares (bottom) Ex: (x – 7)2 Ex: (a – 4)2 x2 – 14x + 49 a2 – 8a + 16 Ex: 9y2 – 12y + 4 1. 9y2 = 3y x 3y yes 2. 4 = 2 x 2 yes 3. 2(3y x 2) = 2(6y) = 12y yes (3y – 2)2

8.7 Square Root Property Ex: (y – 8)2 = 7 Ex: (b – 7)2 = 36 b – 7 = 6 +8 +8 +8 +8 +7 +7 +7 +7 b = 13 b = 1

11.2 Rational Expressions (top)

11.2 Rational Expressions (bottom)