Math A-100 IUSB Chapter 6.5 detailed notes – Week 10 Sec 6.5 – is a collection of the various factoring questions from Sec 6.1 – 6.4 Recall the Steps for Factoring: 1) Look for a GCF (greatest common factor). Pull it out if one is there. This ill make whatever is left easier to work with because the numbers will be smaller. (sec 6.1) 2) Look at the number of terms a) If there are 4 terms, use Grouping (Sec 6.1) b) If there are 3 terms, use the “ac” method (Sec 6.2 & 6.3) c) If there are 2 terms, it is a special case (Sec 6.4) NOTE: if a polynomial cannot factor, we label it as PRIME Special Circumstances “Special Circumstances” is the phrase I use for questions you will see in Math M-107, but not here in MY A-100 class, even though they are introduced in our textbook. Special Circumstance #1: “Substituting” Sometimes the expression you are given is very confusing to look at. In these case, 1) 2) 3) 4) replace the confusing part with something you are more familiar/comfortable with re-write the expression, using the “substitution” factor the “new” expression put the original part back in and simplify Example 1: (x + 5)2 – 4(x + 5) - 21 How do you deal with the ( )s??? You could distribute, to get: x2 + 10x + 25 – 4x – 20 – 21 x2 + 6x – 16 then simplify it to: which factors to (x + 8)(x – 2) OR 1) replace (x + 5) with some other variable, let’s use “y” 2) (x + 5)2 – 4(x + 5) – 21 becomes y2 – 4y – 21 3) Factoring gives us: (y -7)(y + 3) 4) Putting (x + 5) back in for “y” gives you: (x + 5 -7)(x + 5 + 3) simplifying gives you (x – 2)(x + 8) the same answer Example 2: x10 + 13x5 + 36 How do you deal with x10?? a. re-write the first term as (x )2 b. Look at the middle term for ideas 1) Re-write x10 as (x5)2 becomes (x5)2 + 13x5 + 36 2) Replace x5 with y, (x5)2 + 13x5 + 36 becomes (y)2 + 13y + 36 3) Factoring (y)2 + 13y + 36 gives you (y + 9)(y + 4) 4) Put x5 back in: (y + 9)(y + 4) becomes (x5 + 9)( x5 + 4) Final answer: (x5 + 9)( x5 + 4) Example 3: x1/2 - 7x1/4 - 30 How do you deal with x1/2?? c. re-write the first term as (x )2 d. Look at the middle term for ideas 5) Re-write x1/2 as (x1/4)2 becomes (x1/4)2 + 13x1/4 + 36 6) Replace x1/4 with y, (x1/4)2 - 7x1/4 - 30 becomes (y)2 - 7y - 30 7) Factoring (y)2 - 7y - 30 gives you (y - 10)(y + 3) 8) Put x1/4 back in: (y - 10)(y + 3) becomes (x1/4 - 10)( x1/4 + 3) Final answer: (x1/4 - 10)( x1/4 + 3) Special Circumstance #2 – The Sum of and Difference of Cubes We just saw how the Difference of Squares and the Sum of Squares are different. What about Cubes, i.e. when you have x 3? Both the Sum of and the Difference of Cubes factor, and they do in a particular way. The trick is to re-write the expression as cubes, i.e. ____3 to be able to see what numbers we will actually use in the factoring The basic “factoring” looks like this: Sum of cubes: (1st term)3 + (2nd term)3  (1st term + 2nd term)*[(1st)2 – (1st)(2nd) + (2nd)2 Difference of cubes: (1st term)3 - (2nd term)3  (1st term - 2nd term)*[(1st)2 + (1st)(2nd) + (2nd)2 NOTE: when it is factored, whatever is used in the original expression, + or -, is used in the first set of ( ), but the opposite one in used in the 2nd set of ( ). This give us a new acronym to help you decide whether to use a “+” or “-“ sign: SOAP SOAP -> Same Opposite Use same in 1st ( ) Always Positive In 2nd ( ), use opposite between the first two terms; third term is always positive Example 1: x3 + 64 Re-write the expressions as cubes: x3 + 64  x3 + 43 1st = x 2nd = 4 x3 + 43  (x + 4)*[ x2 - (x)(4) + 42] S O AP  (x + 4)(x2 - 4x + 16) NOTE: both sets of ( )s are prime, and will not break down further Example 2: x3 - 125 Re-write the expressions as cubes: x3 - 125  x3 - 53 1st = x 2nd = 5 x3 - 53  (x - 5)*[ x2 + (x)(5) + 52] S O AP  (x - 5)(x2 + 5x + 25) Example 3: 8x3 + 343 Re-write the expressions as cubes: 8x3 + 343 (2x)3 + 73 1st = 2x 2nd = 7 (2x)3 + 73 (2x + 7)*[(2x)2 - (2x)(7) + 72] S O AP  (2x + 7)(4x2 - 14x + 49) Example 4: 27x3 - 512 Re-write the expressions as cubes: 27x3 - 512 (3x)3 + 83 1st = 3x 2nd = 8 (3x)3 - 83 (3x - 8)*[(3x)2 + (3x)(8) + 82] S O AP  (3x - 8)(9x2 + 24x + 64) Special Circumstance #3 - Thought provoking” questions You may be confronted (in the HW) with some tricky questions which will require some “out of the box” thinking. Example 1: Factor x2 + 6x – y2 + 9 a. 4-terms  use Grouping  but Grouping does not work x2 + 6x – y2 + 9 break into x2 + 6x and – y2 + 9 factoring gives you x(x + 6) and -1(y2 – 9)  -1(y – 3)(y + 3)  ( ) not the same  does not work b. Try rearranging: rearrange x2 + 6x – y2 + 9 into x2 + 6x +9 – y2 x2 + 6x +9 factors into (x + 3)(x + 3)  (x + 3)2 replacing this into the question: x2 + 6x +9 – y2 becomes (x + 3)2 – y2  difference of squares factoring now gives you: [(x + 3) + y][(x + 3) – y] cleaning it up a bit, you have (x + 3 + y)(x + 3 – y) Example 2: Word questions You have a rectangle whose height is 5x and whose length is 5x + y. Find the area of the shaded region. ///// //// ////// y ///// /// 5x ///// 5x + y Area of shaded = Area of total – area of unshaded Area of total = Length x height = (5x + y)(5x) area of unshaded = (5x + y)(y) Substituting in: Area of shaded = (5x + y)(5x) – (5x + y)(y) You can see that you have a GCF  (5x + y), so factor it out Area of shaded = (5x + y)(5x – y) NOTE: this could potentially be taken one more step: Difference of squares: 25x2 - y2 Example 3: Unrecognized factors Factor x9 + 125x12 Taking the basics steps: 1. Find & factor out the GCF: GCF = x9 Factoring out: x9(1 + 125x3) - DONE! 2. Not so fast!!! 125x3 is a cube, AND SO IS 1 1 = 12 = 13 etc. So, 1 + 125x3 is the sum of cubes: 1 + 125x3  13 + (5x)3  (1 + 5x)( 12 – 5x + 5x2) 3. Fully factored: x9(1 + 5x)( 12 – 5x + 5x2)