a. You can t do the simple trick of finding two integers that multiply to give 6 and add to give 5 because the a (a = 4) is not equal to one.

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1 FACTORING TRINOMIALS USING THE AC METHOD. Factoring trinomial epressions in one unknown is an important skill necessary to eventually solve quadratic equations. Trinomial epressions are of the form a 2 + b + c and they can sometimes be factored into two binomials using real numbers. You will learn later that there are other numbers that are not real, and that in fact all trinomials can be factored into two factors. And don t forget: factoring is just un-multiplying! In fact, factoring is really just dividing in other words. 2. You likely have learned several methods already: factoring out common factors, perfect squares factoring, difference of squares factoring, grouping, trial and error, etc. 3. Factoring epressions of the form a 2 + b + c can often be complicated if the a (a not equal to one). 4. Eample: Try your known methods of factoring the trinomial epression: a. You can t do the simple trick of finding two integers that multiply to give 6 and add to give 5 because the a (a = 4) is not equal to one. b. You might stumble upon the grouping method: and notice that could be written as: 4 2 ( 3 +8) 6 so: so: (4 3) + 2(4 3) which is bouquets of (4 3) plus 2 bouquets of (4 3) so: ( + 2) bouquets of (4 3); so ( + 2)(4 3). Notice I like to call bracketed terms bouquets because that is what they really are! My wife loves her bouquets of 4 s less three. Last week I bought her of them and this week I bought her 2 of them. So she got ( + 2) bouquets of (4 3) or MA20SP_A_FactorACMethod.doc Revised:

2 2 THE AC METHOD OF FACTORING 5. The AC Method will work some times for many trinomial polynomial epressions to help you more easily find the four terms that can be readily factored by grouping. Let s try Step : Multiply the a*c: ac = 24 Step 2: Find the factors (that multiply together) of 24 and that add to give +5. Notice that 3 and +8 work. Step 3: Rewrite the original epression with the smaller number times first (without consideration of its sign; the one with the smaller absolute value as we call it) and the larger number times net in place of the original term; Notice that is still 5. Step 4: Factor the first two terms: (4 3) and factor the second two terms 2(4 3) Step 5: Do the final factoring. (4 3) + 2(4 3) both have (4 3) in common; so (4 3)( + 2) is the final answer. Step 6: Check your answer by multiplying out again (F.O.I.L) (4 3)( + 2) = PRACTICE EXAMPLES: 6. Factor: ac = 72 which has factors of 36 and 2, 9 and 8, etc. 9 and 8 add to give 7 though! So rewrite as: (smallest factor: 8 first) Factor the first two terms and the last two terms: 2(3 + 4) + 3(3 + 4) Do the final factoring since both parts have a in common; thus: (3 + 4)*(2 + 3) Check the answer by multiplying!

3 3 7. Eample 2: Factor ac = 30 which has factors of 5, 2 and 5, 6 and 6, 5, etc... 6 and 5 add to give the b of though! So rewrite as: Factor the first two terms and the last two terms: 5(2 + ) + -3(2 + ) Do the final factoring since both parts have a in common; thus: (5 3)*(2 + ) Check the answer by multiplying! It works! 8. One for you to try. Factor: (hint: 32 and 2 are the ac factors) The answer is: (8 + )(2 + 4) but notice you should have factored out a GCF constant factor of two first to make the factoring less onerous, so really the fully factored answer is: 2(8 + )( + 2). Multiply it out to check it!

4 4 FACTORING USING GEOMETRY (ALGEBRA TILES) 9. Some students like to see a more visual way of factoring. Your nieces and nephews in Grade 5 and 6 probably learn it this way using Algebra Tiles. In fact this is the original way that ancient mathematicians factored trinomials. This is just a demonstration of the method, you can research it further on your own if you want. 0. Factor the trinomial The ancients would draw squares of size * (ie: 2 ) in the dirt, and rectangles of length and width one, and tiny squares of length one by one. You have to pretend you do not know how long a length of is; it is rubber. long 2 long wide 2. So would be scratched in the dirt as: And then they noticed they could group them together like this: 2 2 Which is really two groups of

5 5 4. But both of these groups could be re-grouped like this: Which can be epressed mathematically as 2 bunches of (+3) by (+); in other words: 2((+3)(+)) or 2(+3)(+) So: = 2(+3)(+) Of course, negative numbers were a bit perpleing for the ancients! So they didn t bother with negative numbers. But you may find your nieces and nephews are taught that red tiles are negatives, and green are positive and that a red cancels a green because they are opposites. It just complicates things, but it can be done. FACTORING IN TWO UNKNOWNS 5. Factoring in two unknowns is also possible using the AC Method with some adaptation. 6. Eample. Factor: y + 2y 2. Pretend this is in a form a 2 + by + cy 2 ac = 4 and factors 4 and add to 5 so: y + 4y + 2y 2. So: (2 + y) + 2y(2 + y) So: ( + 2y)(2 + y)

6 6 SOME PRACTICE ON YOUR OWN 7. Try these on your own. The answers are given. Factor: a b Ans: (3 + 4)( + ) c Ans: (3 2)(5 + ) d Ans: (6 + 7)( 2 + 3) Ans: (5 + 2)(3 0) Simplify by factoring the numerator e f Ans: ( + 3) Ans: (3 + )