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April 8, 2009 You need: textbook calculator HW Answers: p.614 17. A 21. (5w + 1)(w – 2)

April 8, 2009 You need: textbook calculator HW Answers: p.614 17. A 21. (5w + 1)(w – 2) 23. (2y – 5)(3y + 2) 24. Not factorable 28. (2z + 9)(2z + 7) 30. (2b – 1)(4b + 3) 31. (z + 10)(2z – 1). 10.8 Factoring out a GCF before factoring.

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April 8, 2009 You need: textbook calculator HW Answers: p.614 17. A 21. (5w + 1)(w – 2)

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  1. April 8, 2009 You need: textbook calculator HW Answers: p.614 17. A 21. (5w + 1)(w – 2) 23. (2y – 5)(3y + 2) 24. Not factorable 28. (2z + 9)(2z + 7) 30. (2b – 1)(4b + 3) 31. (z + 10)(2z – 1)

  2. 10.8 Factoring out a GCF before factoring To make factoring easier, you might be able to factor out a common constant, variable, or both. For example: 4x3 + 28x2 + 48x Do you notice any GCF (number and/or variable) that all three terms share? 4x3+ 28x2 + 48x 4x 4x 4x They all have a GCF of 4x – divide 4x out of each term.

  3. Now your trinomial looks like: 4x ( x2 + 7x + 12 ) You can’t just throw away the GCF you divided out. It is now part of your problem. Factor your trinomial inside the ( ). 4x (x + 3) (x + 4) You can now find the roots. Set each factor equal to zero…even the 4x term! 4x = 0 (x + 3) = 0 (x + 4) = 0 x = 0 x = -3 x = -4 Your roots to 4x3 + 28x2 + 48x are 0, -3, and -4(it’s a cubic function so need 3 roots – graph it on calculator to see picture)

  4. Here’s another example: • Factor 45x4 – 20x2 completely. • Find any GCF (variables and/or numbers) • Both numbers are divisible by 5 and x2;divide each term by 5x2. • 5x2 ( 9x2 – 4 ) • 2. See if you can factor 9x2 – 4…by Slide and Divide! • 5x2 (x2 – 36) {slide the 9 over to the 4 and mult.} • 5x2 (x – 6)(x + 6) {-6*6 = -36 and -6 + 6 = 0} • 3. Set every factor equal to zero to find the ROOTS. • 5x2 = 0 (x – 6) = 0 (x + 6) = 0 • x = 0 x = 6 x = -6 • The roots are 0, 6, and -6 - - another cubic function!

  5. Factoring by Grouping (to get 2 GCFs) X3 + 2x2 + 3x + 6 Sometimes you can break a polynomial into two binomials in order to factor out 2 GCFs. However, you must make sure both binomials that you create have something in common! X3 and 2x2 both have an x2 in common -> x2(x +2) 3x and 6 both have a factor of 3 in common -> 3(x + 2) Notice the common (x + 2) – that will be ONE of your factors in your final answer. The other factor is made by putting the 2 GCFs into ( ) as a factor -> (x2 + 3). Therefore, your final factoring answer is (x + 2)(x2 + 3).

  6. Are you confused yet again? Try reading pages 625-627, examples 1-7. *There’s a green box on p.628! Practice: p.629, #6-11 Homework: p.629, #19 & 20 factor out GCF only #21-28, 29, 31 factor out GCF, trinomial, and solve for roots

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