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
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
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)
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