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Chapter 8: Factoring

Chapter 8: Factoring. Fill in the titles on the foldable. 8.2 Greatest Common Factor (top). Find the GCF of the terms Write the GCF then the remaining part f each term in parentheses. 8.2 Greatest Common Factor (bottom). Ex: 12a 2 + 16a. Ex: 3p 2 q – 9pq 2 + 36pq.

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Chapter 8: Factoring

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  1. Chapter 8: Factoring

  2. Fill in the titles on the foldable

  3. 8.2 Greatest Common Factor (top) • Find the GCF of the terms • Write the GCF then the remaining part f each term in parentheses

  4. 8.2 Greatest Common Factor (bottom) Ex: 12a2 + 16a Ex: 3p2q – 9pq2 + 36pq 2 x 2 x 3 x a x a x 3 x p x p x q = 2 x 2 x a =4a -1 x 3 x 3 x p x q x q 2 x 2 x 2 x 2 x a 3 x p x q = 3pq 2 x 2 x 3 x 3 x p x q 4a(3a + 4) 3pq(p - 3q + 12)

  5. 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

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

  7. 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. 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

  9. 8.4 Factoring Trinomials – ax2 + bx + c (top) • Get everything on one side (equal to zero) • Find product of the first and last parts • Find the factors of the product • Rewrite the ax2 then fill in the pair of factors that adds or subtracts to make the second part followed by c • Factor by grouping • if you can’t factor = prime (use the zero product property to solve if needed)

  10. 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 ( ) ( ) 5x2 + 3x + 10x + 6 1, 30 2, 15 3, 10 5, 6 5[(2y2 -3y) + (-4y + 6)]=0 x(5x + 3) + 2(5x + 3) 5[y(2y - 3) + -2(2y - 3)]=0 (x + 2)(5x + 3) 5(y - 2)(2y - 3) = 0 Solve for y. y – 2 = 0 2y – 3 = 0 y = 2 and 1.5

  11. 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 Ex: n2 - 25 n x n 5 x 5 (n + 5)(n - 5)

  12. 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) 5[ x2(x + 3) + - 1(x + 3)] 0 = a(7a x 7a 11 x 11) 0 = a(7a + 11)(7a - 11) 5[(x2 – 1)(x + 3)] a = 0 7a + 11 = 0 7a - 11 = 0 5[(x x x 1 x 1)(x + 3)] -11 -11 +11 +11 7a = -11 7a = 11 5(x + 1)(x - 1)(x + 3) /7 /7 /7 /7 a= -11/7 a = 11/7 a = -11/7, 0, and 11/7

  13. 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? • The third term (c) must be positive • Use the sign of the second term • If any of these answers is no- it is not a perfect square trinomial

  14. 8.6 Factoring Perfect Squares (bottom) Ex: x2 – 14x + 49 Ex: a2 – 8a - 16 x x x 7 x 7 a x a 4 x 4 2 x x x 7= 14x 4 x 4 = 16 but it is a negative 16 so it can’t be a perfect square (x – 7)2 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

  15. Sum and Difference of Cubes(top) • a3 + b3 = (a + b)(a2 – ab + b2) • a3 - b3 = (a - b)(a2 + ab + b2) Same - Opposite - Always Positive To remember the signs: SOAP

  16. Sum and Difference of Cubes(bottom) Ex: x3 + 8 Ex: 27x3 – 64y3 Ex: 1000y3 – 216 Ex: 125a3 + 27b3 x x x 2 2 2 3x 3x 3x 4y 4y 4y x2 - 1x2x + 22 (3x - 4y) (9x2 + 12xy + 16y2) (x + 2) (x2 – 2x + 4) (10y - 6) (100x2 + 60y + 36) (5a + 3b) (25a2 – 15ab + 9b2)

  17. Quadratic Form(Bottom) Ex: x4 + 3x2 + 2 Ex: x4 – 16 Ex: x - + 21 Ex: x4 + 7x2 + 12 (x2 )(x2 ) (x2 )(x2 ) (x2 + 4)(x2 - 4) (x2 + 1)(x2 + 2) (x2 + 1)(x2 + 2)(x2 – 2) ( - 3)( - 7) (x2 )(x2 ) (x2 + 3)(x2 + 4)

  18. Combinations of Factoring Types(Top) • First look for the GCF and factor out if possible • Next look for patterns (perfect squares, difference of squares or sum and difference of cubes) • If no patterns appear factor the trinomial like normal

  19. Combinations of Factoring Types(bottom) Ex: 4x2 – 100 Ex: 3x2 – 3x – 60 Ex: 8x6 – 64x3 Ex: 8x3 – 32x 4(x2 – 25) 3(x2 – x - 20) 20 1, 20 2, 10 4, 5 4(x + 5)(x – 5) (x + 4)(x – 5) 8x3 (x3 – 8) 8x(x2 – 4) 8x3(x – 2)(x2 + 2x + 4) 8x(x + 2)(x – 2)

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