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Factoring Activity
3.09
Greatest Common
Factor(GCF)
 Let’s try and factor 3x^3 + 27x^2 + 9x
 List the factors and find the GCF from
that each has in common.
3- 1 and 3
27- 1,3,9, and 27
9- 1,3, and 9
 They all have 3 in common.
Greatest Common
Factor(GCF)
 Put the GCF on the out side a pair of
parentheses: 3x()
 Divide the GCF from each of the
expressions.
3x^3÷3x x^2
27x^2÷ 3x  9x
9x÷ 3x  3
 3x(x^2+9x+3)
Special Products
 A binomial is considered a difference of
squares when both of the terms are perfect
squares
 a +/- b= (a - b)( a+ b)
  x^2 - 25
Both x^2 and 25 are perfect squares
 The square root of x^2 is x and the square
root of 25 is five
 (x-5)(x+5)
Factoring Trinomials
 -m^2-10m-16
 First you need to find the GCF and take
it out. Here it’s -1
-(m^2+10m+16)
 Now it’s time to find the factors
 -(m+ )(m+ )
 Factors of 10 and 16, let’s try 8 and 2
Factoring Trinomials
 -(m+8)(m+2)
FOILING gives us the original problem.
Factoring by Grouping
  x^3 − 3x^2 + 9x − 27
 There is no GCF between the four. So you
have to split them
 (x^3-3x^2) and (9x-27)
With the first the GCF you can take out is x^2.
x^2(-3 + x) . The second one you can take out
9. 9(x-3)
 X^2(x-3)+9(x-3)
 (x^2+9)
Sum and Difference of Cubes
 The formulas for this are
a3 + b3 = (a + b)(a2 – ab + b2) 
a3 – b3 = (a – b)(a2 + ab + b2)
 Let’s try factoring x^15 – 64
 The cubed roots are x^15 and 4
respectively.
 Use the formula
Sum and Difference of Cubes
a3 – b3 = (a – b)(a2 + ab + b2)
 x^15 – 64= (x^5-4)(x^10+4x^5+16)
 So your factor would be
(x^5-4)
Sum and Difference of Cubes
a3 – b3 = (a – b)(a2 + ab + b2)
 x^15 – 64= (x^5-4)(x^10+4x^5+16)
 So your factor would be
(x^5-4)

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Factoring Activity 3.09

  • 2. Greatest Common Factor(GCF)  Let’s try and factor 3x^3 + 27x^2 + 9x  List the factors and find the GCF from that each has in common. 3- 1 and 3 27- 1,3,9, and 27 9- 1,3, and 9  They all have 3 in common.
  • 3. Greatest Common Factor(GCF)  Put the GCF on the out side a pair of parentheses: 3x()  Divide the GCF from each of the expressions. 3x^3÷3x x^2 27x^2÷ 3x  9x 9x÷ 3x  3  3x(x^2+9x+3)
  • 4. Special Products  A binomial is considered a difference of squares when both of the terms are perfect squares  a +/- b= (a - b)( a+ b)   x^2 - 25 Both x^2 and 25 are perfect squares  The square root of x^2 is x and the square root of 25 is five  (x-5)(x+5)
  • 5. Factoring Trinomials  -m^2-10m-16  First you need to find the GCF and take it out. Here it’s -1 -(m^2+10m+16)  Now it’s time to find the factors  -(m+ )(m+ )  Factors of 10 and 16, let’s try 8 and 2
  • 6. Factoring Trinomials  -(m+8)(m+2) FOILING gives us the original problem.
  • 7. Factoring by Grouping   x^3 − 3x^2 + 9x − 27  There is no GCF between the four. So you have to split them  (x^3-3x^2) and (9x-27) With the first the GCF you can take out is x^2. x^2(-3 + x) . The second one you can take out 9. 9(x-3)  X^2(x-3)+9(x-3)  (x^2+9)
  • 8. Sum and Difference of Cubes  The formulas for this are a3 + b3 = (a + b)(a2 – ab + b2)  a3 – b3 = (a – b)(a2 + ab + b2)  Let’s try factoring x^15 – 64  The cubed roots are x^15 and 4 respectively.  Use the formula
  • 9. Sum and Difference of Cubes a3 – b3 = (a – b)(a2 + ab + b2)  x^15 – 64= (x^5-4)(x^10+4x^5+16)  So your factor would be (x^5-4)
  • 10. Sum and Difference of Cubes a3 – b3 = (a – b)(a2 + ab + b2)  x^15 – 64= (x^5-4)(x^10+4x^5+16)  So your factor would be (x^5-4)