Review of Factoring Techniques - PowerPoint Presentation

1. Review of Factoring Techniques

2. Factoring?Factoring is a method to find the basic numbers and variables that made up a product.(Factor) x (Factor) = ProductSome numbers are Prime, meaning they are only divisible by themselves and 1 and not factorable.

3. When factoring trinomials, we always try GCF method first!!!!Number of termsFactoring Technique2Difference of 2 Squares3Sum and Product Method4 or 6Grouping

4. GCF Method

5. Warm-UP - Distribute each problem:

6. GCF Method is justdistributing backwards!!

7. Factor out the GCF for each polynomial:Factor out means you need the GCF times the remaining parts.a) 2x + 4y 5a – 5b 18x – 6y 2m + 6mn 5x2y – 10xy 2(x + 2y)6(3x – y)5(a – b)5xy(x - 2)2m(1 + 3n)Greatest Common Factorsaka GCF’sHow can you check?

8. Review: What is the GCF of 25a2 and 15a?5aLet’s go one step further…1) FACTOR 25a2 + 15a.Find the GCF and divide each term25a2 + 15a = 5a( ___ + ___ )Check your answer by distributing.5a3

9. 2) Factor 18x2 - 12x3.Find the GCF6x2Divide each term by the GCF18x2 - 12x3 = 6x2( ___ - ___ )Check your answer by distributing.32x

10. 3) Factor 28a2b + 56abc2.GCF = 28abDivide each term by the GCF28a2b + 56abc2 = 28ab ( ___ + ___ )Check your answer by distributing.28ab(a + 2c2)a2c2

11. Factor 20x2 - 24xyx(20 – 24y)2x(10x – 12y)4(5x2 – 6xy)4x(5x – 6y)

12. 5) Factor 28a2 + 21b - 35b2c2GCF = 7Divide each term by the GCF28a2 + 21b - 35b2c2 = 7 ( ___ + ___ - ____ )Check your answer by distributing.7(4a2 + 3b – 5b2c2)4a25b2c23b

13. Factor 16xy2 - 24y2z + 40y22y2(8x – 12z + 20)4y2(4x – 6z + 10)8y2(2x - 3z + 5)8xy2z(2 – 3 + 5)

14. Method #2Difference of Two Squaresa2 – b2 = (a + b)(a - b)

15. What is a Perfect SquareAny term you can take the square root evenly (No decimal)25361x2y4

16. Difference of Perfect Squares x2 – 4 =the answer will look like this: ( )( )take the square root of each part: ( x 2)(x 2)Make 1 a plus and 1 a minus: (x + 2)(x - 2 )

17. FACTORINGDifference of PerfectSquaresEX:x2 – 64How:Take the square root of each part. One gets a + and one gets a -.Check answer by FOIL.Solution:(x – 8)(x + 8)

18. Example 19x2 – 16(3x + 4)(3x – 4)

19. Example 2x2 – 16(x + 4)(x –4)

20. Ex 3 36x2 – 25(6x + 5)(6x – 5)

21. More than ONE MethodIt is very possible to use more than one factoring method in a problemRemember:ALWAYS use GCF first

22. Example 12b2x – 50xGCF = 2x2x(b2 – 25) 2nd term is the diff of 2 squares2x(b + 5)(b - 5)

23. Example 232x3 – 2xGCF = 2x2x(16x2 – 1) 2nd term is the diff of 2 squares2x(4x + 1)(4x - 1)

24. FactoringTRINOMIALSUsing Sum and Product Methodax2 + bx + c

25. Example 1: x2 + 11x + 24Their sum equals the middle term of the trinomial.When factoring these trinomials the factors will be two binomials: (x + )(x + )We know that the first terms of each binomial must be x because the first term of the trinomial is x2 and x  x = x2. The challenge is to find the last term of each binomial. They must be chosen so that they will cause the coefficient of the middle term and the last term of the trinomial to work out. (That’s 11 and 24 in this case.)+= 11Their product of those same numbers equals the last term of the trinomial.= 24

26. List the factors of 24:Example 1: x2 + 11x + 241242123846124SUM = 25212SUM = 1438SUM = 1146SUM = 10It is the factors 3 and 8 that produce a sum of 11 AND a product of 24 so they must be the last terms of each binomial.(x + 3)(x + 8)

27. If we multiply these factors using FOIL, we get the polynomial that we started with.(x + 3)(x + 8)= x2 + 8x + 3x + 24(x)(x) = x2(x)(8) = 8x(3)(x) = 3x(3)(8) = 24As we look at the 4 terms above, it becomes apparent why the sum of the last terms in each binomial must be equal to the middle term of the trinomial.(x + 3)(x + 8)= x2 + 8x + 3x + 24= x2 + 11x + 24

28. Example 2: a2 + 16a + 28Factors of 28:12821447SUM = 16a  a = a2 so they are the first terms of each binomial and the factors 2 and 14 make a sum of 16 so the are the last terms of each binomial.= (a + 2)(a + 14)

29. Example 3: y2 + 2y + 1Factors of 1:11y2 + 2y + 1= (y + 1)(y + 1)= (y + 1)(y + 1)Factors of 1:11Sometimes there is only 1 pair of factors to consider.Example 4: m2 + 3m + 1Factors of 1:11In this example the factors available do not make a sum of 3 which means that the trinomial can’t be factored.Example 5: p2 + 23p + 120Factors of 120:1120260340430524620815101211202603404305246208151012= (p + 8)(p + 15)= (p + 8)(p + 15)In this example there are many pairs of factors to consider. Most examples will have fewer than these. The trick is in being able to quickly find all of the factors of c.p2 + 23p + 120SUM = 2SUM = 3SUM = 23

30. In each of the preceding examples the signs of the terms in the trinomials were always positive. Now we will observe examples where the signs can be negative.Example 6: x2 + 5x + 6 Factors of 6:1623SUM = 5Factors of 6:1623= (x + 2)(x + 3) Example 7: x2 + 5x - 6 Factors of -6:-1+6-2+3SUM = 5When looking for the factors of a negative number, one must be positive and the other negative. If at the same time their sum is positive, then the factor that is bigger must be the positive one.-1+6-2+3= (x - 1)(x + 6)

31. REVIEW OF RULES FOR SIGNSSign of bigger number(+) + (+) = (+)(+) + (-) = (-) + (+) = (-) + (-) = (-)( )ADDITION(+)(+) = (+)(+)(-) = (-)(-)(+) = (-)(-)(-) = (+)MULTIPLICATIONExample 8: x2 - 5x - 6 Factors of -6:SUM = -5+1-6+2-3= (x + 1)(x - 6) When both the product and sum are negative, the factors have opposite signs but this time the bigger factor will be negative.

32. Example 9: x2 - 5x + 6 Factors of 6:SUM = -5-1-6-2-3= (x - 2)(x - 3) When looking for factors of a positive number when the sum is negative, both factors will be negative.Example 10: x2 - 5x - 36 Factors of -36:1-362-183-124-96-6SUM = -51-362-183-124-96-6= (x + 4)(x - 9)

33. Here we go! 1) Factor y2 + 6y + 8Use your factoring chart.Do we have a GCF?Is it a Diff. of Squares problem?Now we will learn Trinomials! You will set up a table with the following information.Nope!No way! 3 terms!Product of the first and last coefficientsMiddlecoefficientThe goal is to find two factors in the first column that add up to the middle term in the second column.We’ll work it out in the next few slides.

34. Here are some hints to help you choose your factors.1) When the last term is positive, the factors will have the same sign as the middle term.2) When the last term is negative, the factors will have different signs.

35. 2) Factor 5x2 - 17x + 14 Create your table.-1, -70-2, -35-7, -10Multiply Add+70 -17Product of the first and last coefficientsMiddlecoefficient-71-37-17Signs need to be the same as the middle sign since the product is positive.Replace the middle term.5x2 – 7x – 10x + 14Group the terms.

36. (5x2 – 7x) (– 10x + 14)Factor out the GCFx(5x – 7) -2(5x – 7)The parentheses are the same! (x – 2)(5x – 7)Hopefully, these will continue to get easier the more you do them.

37. Factor x2 + 3x + 2(x + 2)(x + 1)(x – 2)(x + 1)(x + 2)(x – 1)(x – 2)(x – 1)

38. Factor 2x2 + 9x + 10(2x + 10)(x + 1)(2x + 5)(x + 2)(2x + 2)(x + 5)(2x + 1)(x + 10)

39. Factor 6y2 – 13y – 5(6y2 – 15y)(+2y – 5)(2y – 1)(3y – 5)(2y + 1)(3y – 5)(2y – 5)(3y + 1)