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By Caleb Reagor, Brendan Inson, and Jack Guenther Flugenheiman (Need Help? Look at vocabulary at very end) Chapter 6 Tutorial
Chapter Opener http://algebraiperiod6.pbworks.com/Flugenheiman-Chapter-6-Opener
Section 6.1 http://algebraiperiod6.pbworks.com/Flugenheiman-Section-1
Section 6.2 In this section you will learn how to factor trinomials with the form of x²+bx+c The coefficient will only be 1 for the squared variable
Section 6.2 You may remember that factoring is writing a product into factors For example the factors of x²+3x+2 are (x+1) and (x+2) x times x equals x² and it can not be x² and 1 because that would get rid of the 3x in the middle There is 3x+2 because positive factors of two are 2 and 1 and the sum of them is 3 for the 3x
Section 6.2 Practice Problems Factor two binomials out of the problems x²+7x+10 x²+8x+12
Section 6.2 Answers (x+5)(x+2) (x+2)(x+6)
Section 6.2 There are negatives such as: x²-3x+2=(x-2)(x-1) x²+1x-2=(x-1)(x+2) x²-1x-2=(x+1)(x-2) There are some times perfect square trinomials x²+2x+1=(x+1)(x+1)=(x+1) ²
Section 6.2 Practice problems x²-3x+2 x²+3x-18 x²-3x-18 x²+6x+9
Section 6.2 Answers (x-2)(x-1) (x+6)(x-3) (x+3)(x-6) (x+3) ²
Section 6.2 There are also ones with the fourth power x⁴+3x²+2=(x²+2)(x²+1) There are also ones with two variables x²+2xy+y²=(x+y)(x+y) or (x+y)²
Section 6.2 Practice Problems x⁴+2x²+1 x²+4xy+2y²
Section 6.2 Answers (x²+1) ² (x+2y) ²
Section 6.3 http://voicethread.com/share/947591/
Section 6.4 In this section you will learn how to factor binomials of the form ax²+bx+c by grouping
Section 6.4 Go back to section 6.1 for the basics of factoring ax ²+bx+c with “a”=1 This section is an extended method of this but to have “a” larger then 1 An example of this is 6x²+11x+3 Before you start you always need to see if it is a perfect square trinomial or if it has a greatest common factor- this problem has none
Section 6.4 This is how to solve 6x ²+11x+3 To solve the problem you multiply 6x ² and 3 to equal 18x² You then find factors of 18x² that equal 11x 9x plus 2x equals 11x; when multiplied, they equal 18x²
Section 6.4 You take those factors and put it into the equation So instead of 6x ²+11x+3 you do 6x²+9x+2x+3 You then group them into groups with greatest common factors; (6x²+9x)+(2x+3) Then you factor out the greatest common factors of the grouped parts, try to make the inside problem the same, for example (6x ²+9x)+(2x+3)=3x(2x+3)+1(2x+3) You then add/subtract the greatest common factors; you then multiply that equation by the other same terms (3x+1)(2x+3)
Section 6.4 Practice Problems 10x²+21x+2 6x²+32x+10 7x²+17x+6
Section 6.4 Answers 10x²+20x+x+2=(10x²+20x)+(x+2)=10x(x+2)+1(x+2) finally equaling (10x+1)(x+2) 2(gcf)(3x²+15x+x+5)=2(3x²+x)+(15x+5)= 2[x(3x+1)+5(3x+1)] finally equaling 2(x+5)(3x+1) 7x²+14x+3x+6=(7x²+14x)+(3x+6)=7x(x+2)+3(x+2) finally equaling (7x+3)(x+2)
Section 6.4 You can do this for perfect square trinomials and negative problems such as  4x²-8x+4=(2x-2)² since that this is a prime you do not have to do all of the work ax²+bx-c in this you do the same thing you do for a positive number but the final equation looks like this (dx+e)(fx-g) and that is the same for ax²-bx-c but the negative number without a variable is larger then the positive
Section 6.4 Practice Problems 9x²+12x+4 8x²+63x-8 8x²-63x-8
Section 6.4 Answers (3x+2)² (8x-1)(x+8) (8x+1)(x-8)
Section 6.5 http://voicethread.com/share/965601/
Vocabulary http://algebraiperiod6.pbworks.com/Flugenheiman-Vocabulary

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Chapter 6 Tutorial

  • 1. By Caleb Reagor, Brendan Inson, and Jack Guenther Flugenheiman (Need Help? Look at vocabulary at very end) Chapter 6 Tutorial
  • 4. Section 6.2 In this section you will learn how to factor trinomials with the form of x²+bx+c The coefficient will only be 1 for the squared variable
  • 5. Section 6.2 You may remember that factoring is writing a product into factors For example the factors of x²+3x+2 are (x+1) and (x+2) x times x equals x² and it can not be x² and 1 because that would get rid of the 3x in the middle There is 3x+2 because positive factors of two are 2 and 1 and the sum of them is 3 for the 3x
  • 6. Section 6.2 Practice Problems Factor two binomials out of the problems x²+7x+10 x²+8x+12
  • 7. Section 6.2 Answers (x+5)(x+2) (x+2)(x+6)
  • 8. Section 6.2 There are negatives such as: x²-3x+2=(x-2)(x-1) x²+1x-2=(x-1)(x+2) x²-1x-2=(x+1)(x-2) There are some times perfect square trinomials x²+2x+1=(x+1)(x+1)=(x+1) ²
  • 9. Section 6.2 Practice problems x²-3x+2 x²+3x-18 x²-3x-18 x²+6x+9
  • 10. Section 6.2 Answers (x-2)(x-1) (x+6)(x-3) (x+3)(x-6) (x+3) ²
  • 11. Section 6.2 There are also ones with the fourth power x⁴+3x²+2=(x²+2)(x²+1) There are also ones with two variables x²+2xy+y²=(x+y)(x+y) or (x+y)²
  • 12. Section 6.2 Practice Problems x⁴+2x²+1 x²+4xy+2y²
  • 13. Section 6.2 Answers (x²+1) ² (x+2y) ²
  • 15. Section 6.4 In this section you will learn how to factor binomials of the form ax²+bx+c by grouping
  • 16. Section 6.4 Go back to section 6.1 for the basics of factoring ax ²+bx+c with “a”=1 This section is an extended method of this but to have “a” larger then 1 An example of this is 6x²+11x+3 Before you start you always need to see if it is a perfect square trinomial or if it has a greatest common factor- this problem has none
  • 17. Section 6.4 This is how to solve 6x ²+11x+3 To solve the problem you multiply 6x ² and 3 to equal 18x² You then find factors of 18x² that equal 11x 9x plus 2x equals 11x; when multiplied, they equal 18x²
  • 18. Section 6.4 You take those factors and put it into the equation So instead of 6x ²+11x+3 you do 6x²+9x+2x+3 You then group them into groups with greatest common factors; (6x²+9x)+(2x+3) Then you factor out the greatest common factors of the grouped parts, try to make the inside problem the same, for example (6x ²+9x)+(2x+3)=3x(2x+3)+1(2x+3) You then add/subtract the greatest common factors; you then multiply that equation by the other same terms (3x+1)(2x+3)
  • 19. Section 6.4 Practice Problems 10x²+21x+2 6x²+32x+10 7x²+17x+6
  • 20. Section 6.4 Answers 10x²+20x+x+2=(10x²+20x)+(x+2)=10x(x+2)+1(x+2) finally equaling (10x+1)(x+2) 2(gcf)(3x²+15x+x+5)=2(3x²+x)+(15x+5)= 2[x(3x+1)+5(3x+1)] finally equaling 2(x+5)(3x+1) 7x²+14x+3x+6=(7x²+14x)+(3x+6)=7x(x+2)+3(x+2) finally equaling (7x+3)(x+2)
  • 21. Section 6.4 You can do this for perfect square trinomials and negative problems such as 4x²-8x+4=(2x-2)² since that this is a prime you do not have to do all of the work ax²+bx-c in this you do the same thing you do for a positive number but the final equation looks like this (dx+e)(fx-g) and that is the same for ax²-bx-c but the negative number without a variable is larger then the positive
  • 22. Section 6.4 Practice Problems 9x²+12x+4 8x²+63x-8 8x²-63x-8
  • 23. Section 6.4 Answers (3x+2)² (8x-1)(x+8) (8x+1)(x-8)