This chapter discusses factoring trinomials and binomials of various forms. Section 6.1 introduces factoring trinomials of the form x^2 + bx + c where the coefficient of x^2 is 1. Section 6.2 expands on this to include factoring trinomials with negative coefficients and perfect square trinomials. Section 6.3 provides a voice thread on factoring. Section 6.4 teaches how to factor trinomials where the coefficient of x^2 is greater than 1 by grouping like terms. It also addresses factoring perfect square trinomials and trinomials with negative coefficients. Practice problems with solutions are provided throughout to illustrate the techniques.
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