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# Chapter 6 Tutorial

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Learn Alegebra 1

Learn Alegebra 1

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### Transcript

• 1. By Caleb Reagor, Brendan Inson, and Jack Guenther
Flugenheiman
(Need Help? Look at vocabulary at very end)
Chapter 6 Tutorial
• 2. Chapter Opener
http://algebraiperiod6.pbworks.com/Flugenheiman-Chapter-6-Opener
• 3. Section 6.1
http://algebraiperiod6.pbworks.com/Flugenheiman-Section-1
• 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
(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
(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
(x²+1) ²
(x+2y) ²
• 14. Section 6.3
• 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
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