O slideshow foi denunciado.
Seu SlideShare está sendo baixado. ×

MAT1033.7.3.ppt

Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Carregando em…3
×

Confira estes a seguir

1 de 22 Anúncio
Anúncio

Mais Conteúdo rRelacionado

Mais de ErlenaMirador1 (20)

Mais recentes (20)

Anúncio

MAT1033.7.3.ppt

  1. 1. Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 7.3 - 1
  2. 2. Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 7.3 - 2 Factoring Chapter 7
  3. 3. Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 7.3 - 3 7.3 Special Factoring
  4. 4. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 4 7.3 Special Factoring Objectives 1. Factor a difference of squares. 2. Factor a perfect square trinomial. 3. Factor a difference of cubes. 4. Factor a sum of cubes.
  5. 5. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 5 7.3 Special Factoring The Difference of Squares Difference of Squares x2 – y2 = (x + y)(x – y)
  6. 6. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 6 7.3 Special Factoring EXAMPLE 1 Factoring Differences of Squares Factor each polynomial. There is a common factor of 2. 2n2 – 50 = 2(n2 – 25) Factor out the common factor. (a) 2n2 – 50 = 2(n + 5)(n – 5) Factor the difference of squares.
  7. 7. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 7 7.3 Special Factoring EXAMPLE 1 Factoring Differences of Squares Factor each polynomial. (b) 9g2 – 16 9g2 – 16 = (3g)2 – (4)2 A2 B2 = (3g + 4)(3g – 4) (A + B)(A – B) (c) 4h2 – (w + 5)2 4h2 – (w + 5)2 = (2h)2 – (w + 5)2 A2 B2 = (2h + w + 5)(2h – [w + 5]) (A + B) (A – B) – – = (2h + w + 5)(2h – w – 5)
  8. 8. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 8 7.3 Special Factoring Caution CAUTION Assuming no greatest common factor except 1, it is not possible to factor (with real numbers) a sum of squares, such as x2 + 16.
  9. 9. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 9 7.3 Special Factoring Perfect Square Trinomial Perfect Square Trinomial x2 + 2xy + y2 = (x + y)2 x2 – 2xy + y2 = (x – y)2
  10. 10. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 10 7.3 Special Factoring EXAMPLE 2 Factoring Perfect Square Trinomials Factor each polynomial. Here 9g2 = (3g)2 and 49 = 72. The sign of the middle term is –, so if 9g2 – 42g + 49 is a perfect square trinomial, the factored form will have to be (3g – 7)2. (a) 9g2 – 42g + 49 Take twice the product of the two terms to see if this is correct. 2(3g)(–7) = –42g This is the middle term of the given trinomial, so 9g2 – 42g + 49 = (3g – 7)2.
  11. 11. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 11 7.3 Special Factoring EXAMPLE 2 Factoring Perfect Square Trinomials Factor each polynomial. If this is a perfect square trinomial, it will equal (5x + 8y)2. By the pattern described earlier, if multiplied out, this squared binomial has a middle term of 2(5x)(8y), which does not equal 60xy. Verify that this trinomial cannot be factored by the methods of the previous section either. It is prime. (b) 25x2 + 60xy + 64y2
  12. 12. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 12 7.3 Special Factoring EXAMPLE 2 Factoring Perfect Square Trinomials Factor each polynomial. since 2(n – 4)9 = 18(n – 4), the middle term. (c) (n – 4)2 + 18(n – 4) + 81 = [ (n – 4) + 9 ]2 = (n + 5)2,
  13. 13. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 13 7.3 Special Factoring EXAMPLE 2 Factoring Perfect Square Trinomials Factor each polynomial. The result is the difference of squares. Factor again to get (d) c2 – 6c + 9 – h2 = (c – 3 + h)(c – 3 – h). (c2 – 6c + 9) – h2 = (c – 3)2 – h2 Since there are four terms, we will use factoring by grouping. The first three terms here form a perfect square trinomial. Group them together, and factor as follows.
  14. 14. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 14 7.3 Special Factoring Difference of Cubes Difference of Cubes x3 – y3 = (x – y)(x2 + xy + y2)
  15. 15. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 15 –5a a3 –125 7.3 Special Factoring EXAMPLE 3 Factoring Difference of Cubes Factor each polynomial. Recall, x3 – y3 = (x – y)(x2 + xy + y2). = a3 – 53 = (a – 5)(a2 + 5a + 52) = (a – 5)(a2 + 5a + 25) (a) a3 – 125 Check: = (a – 5)(a2 + 5a + 25) Opposite of the product of the cube roots gives the middle term.
  16. 16. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 16 7.3 Special Factoring EXAMPLE 3 Factoring Difference of Cubes Factor each polynomial. Recall, x3 – y3 = (x – y)(x2 + xy + y2). = (2g)3 – h3 = (2g – h) [ (2g)2 + (2g)(h) + h2) ] = (2g – h)(4g2 + 2gh + h2) (b) 8g3 – h3
  17. 17. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 17 7.3 Special Factoring EXAMPLE 3 Factoring Difference of Cubes Factor each polynomial. Recall, x3 – y3 = (x – y)(x2 + xy + y2). = (4m)3 – (3n)3 = (4m – 3n) [ (4m)2 + (4m)(3n) + (3n)2 ] = (4m – 3n)(16m2 + 12mn + 9n2) (c) 64m3 – 27n3
  18. 18. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 18 7.3 Special Factoring Sum of Cubes Sum of Cubes x3 + y3 = (x + y)(x2 – xy + y2)
  19. 19. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 19 7.3 Special Factoring Note on Signs NOTE Difference of Cubes x3 – y3 = (x – y)(x2 + xy + y2) Sum of Cubes x3 + y3 = (x + y)(x2 – xy + y2) The sign of the second term in the binomial factor of a sum or difference of cubes is always the same as the sign in the original polynomial. In the trinomial factor, the first and last terms are always positive; the sign of the middle term is the opposite of the sign of the second term in the binomial factor.
  20. 20. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 20 7.3 Special Factoring EXAMPLE 4 Factoring Sums of Cubes Factor each polynomial. Recall, x3 + y3 = (x + y)(x2 – xy + y2). = n3 + 23 = (n + 2)(n2 – 2n + 22) = (n + 2)(n2 – 2n + 4) (a) n3 + 8 = (4v)3 + (3g)3 = (4v + 3g) [ (4v)2 – (4v)(3g) + (3g)2 ] = (4v + 3g) (16v2 – 12gv + 9g2) (b) 64v3 + 27g3
  21. 21. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 21 7.3 Special Factoring EXAMPLE 4 Factoring Sums of Cubes Factor each polynomial. Recall, x3 + y3 = (x + y)(x2 – xy + y2). = 2(k3 + 125) = 2(k3 + 53) = 2(k + 5)(k2 – 5k + 25) (c) 2k3 + 250 =
  22. 22. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 22 7.3 Special Factoring Factoring Summary Special Types of Factoring (Memorize) Difference of Squares x2 – y2 = (x + y)(x – y) Perfect Square Trinomial x2 + 2xy + y2 = (x + y)2 x2 – 2xy + y2 = (x – y)2 Difference of Cubes x3 – y3 = (x – y)(x2 + xy + y2) Sum of Cubes x3 + y3 = (x + y)(x2 – xy + y2)

×