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Chapter 8: Factoring. Fill in the titles on the foldable. 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.
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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 b – 7 = -6 +8 +8 +8 +8 +7 +7 +7 +7 b = 13 b = 1