How to factor x x y y

Term Conversions. ALGEBRA terms 2. Binomial formulas. INTERNET LIBRARY FOR SCHOOL MATHEMATICS. File No. Friedrich W.

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1 ALGEBRA terms Term transformations Binomial formulas Mostly in class 8 File No. 110 Friedrich W. Buckel Status: November 4, 008 INTERNET LIBRARY FOR SCHOOL MATHEMATICS

2 Contents FILE What are and what are terms terms 1 Summarizing and arranging terms 4 3 Multiplying and factoring out 8 Be careful when dealing with minus signs 11 Be careful when dealing with powers 11 4 Multiplying brackets 15 5 More complicated tasks 18 Products with more than brackets 1 FILE Binomial Formulas 1 FILE Preliminary exercise 1 6. The 1st binomial formula 6.3 The. Binomial formula The 3rd binomial formula Devilish minus signs Mixed tasks Even more complicated terms 18 7 Factoring 1 FILE Level: Factoring 1 7. Inversion of the binomial formulas 3 Training task Factoring with parentheses and binomial formula 10 Training task Square extension 3 Training task Factoring with any brackets FILE Calculation of (a + b) n / Pascal's triangle 1 10 Calculation of (a + b + c) 1

3 110 Term Transformations Preliminary Exercise 6 Binomial Formulas A basic skill in algebra is the multiplication of parenthesized terms. Therefore, here is a preliminary exercise: () (c +) ac + a a + b d = d + bc + b d So four products are formed: Each addend in the first bracket is multiplied by each addend in the second bracket. Here is an even more challenging exercise: (a + 3b) (6a 4b) = a 6a + a (4b) + 3b 6a + 3b (4b) or faster like this: (a + 3b) (6a 4b) = a 6a a 4b + 3b 6a 3b 4b = 1a 8ab + 18ba 1b = 1a + 10ab 1b When multiplying two brackets, there are three special cases in which the result is so important that it has to be remembered. These three cases were therefore given special names. They are called the binomial formulas. Because they are so important, you have to be able to derive them and not just know what they look like: Friedrich Buckel

4 110 Term Transformations 6. The 1st Binomial Formula Derivation of the Formula It is about the calculation of the term (a + b), which is actually called (a + b) (a + b). We use the method that is always used to multiply parentheses: (a + b) (a + b) = a + ab + ba + b = a + ab + b That was all. Therefore we remember: () a + b = a + ab + b In words: a and b are each squared, i.e. a + b. Some stop here and forget that there is a third summand: + from, the double product! First example: Double product () x + 3 = x + 3x + 9 Square of the 1st summand Square of the. Addend That was the first example that is calculated in detail as follows: (a) (x + 3) = x + x = x + 6x + 9 (b) (3x + 5) = (3x) + 3x = 9x + 30x + 5 Explanation: Now it says 3x instead of a, so a = (3x) = 9x. Then comes the double product, i.e. 3x times 5 and the whole thing taken twice, i.e. again, results in 30x. After all, b = 5 is b = 5. This formula can also be represented with large placeholders: (+) = + + With this scheme it is possible to convert more complicated terms: (c) (5a + 7b) =? Try it yourself! Write the summands in our scheme! Friedrich Buckel

5 110 Term transformations 3 (5a + 7b) = 5a + 5a 7b + 7b = + + 5a 70 from 49b This scheme helps when students get confused because in the binomial formula it says (a + b) and it suddenly appears here (5a + 7b). The a and b in the formula are only placeholders for the 1st and. Summands of the term to be calculated! (d) (3+ z) = z + (z) = 9+ 1z + 4z (e) () () () 3y + 1x = 3y + 3y 1x + 1x = 9y + 7xy + 144x. (f) () () () x + y = x + xy + y = x + xy + y 4 4 (g) () () 4 4x + 5 = 4x + 4x = 16x + 40x + 5 (h) (3x + 8x) = (3x) + 3x 8x + (8x) = 9x + 48x + 64x 3 4 (i) () () () xy + xy = xy + xy xy + xy = xy + xy + xy because ( ) () () 4 xy = xy xy = xxyy = xy etc. j) (15x + 16x) = (15x) + 15x 16x + (16x) = 5x x + 56 x because (3) x = xx = (xxx ) (xxx) = x etc. Training exercise 1 Use the 1st binomial formula: (a) (m + n) (b) (3a + 4b) (c) (7x + 15) (d) (5c 6d) 15x 4y + (e) (+) 1 3 (f) (x +) (g) (3x +) 1 (h) (8b + a) 1 (i) (+) 3 16 x 4x solution a few pages further. Friedrich Buckel

6 110 Term Transformations Die. Binomial formula It's about the calculation of the term (a b), which is actually called (a b) (a b). We apply again the method that one always uses for the multiplication of brackets and gets: (a b) (a b) = a + a (b) + (b) a + (b) = a ab + b That was all. Therefore we remember: () a-b = a -ab + b In words: a and b are each squared, i.e. a + b. Some stop here and forget that there is a third summand: - ab. It is called the double product! This is subtracted here because it is also called a b. Examples (a) (4x 1) = (4x) 4x 1+ 1 = 16x 8x + 1 (b) (4a 9b) = (4a) 4a 9b + (9b) = 16a 7ab + 81b (c) () () () 3x 5y = 3x 3x 5y + 5y = 9x 30xy + 5y (d) () () 4 x 4 = xx 4+ 4 = x 8x + 16 (e) (3x 5x) = (3x) 3x 5x + ( 5x) = 9x 30x + 5x 4 3 (f) () 10 8z = z + 64z This was now without any intermediate calculation! (g) (3ab a b) = 9a b 1 ab + 4ab training task turn the. binomial formula an: (a) (5a c) (b) (7a b) (c) (0x 5) (d) (ab 4) (e) (4a) 1 (g) (4x) 1 3 (h) (uv) 4 (f) (1 a 1 b) x 6 3 (i) () Solution a few pages further. Friedrich Buckel

7 110 Term Transformations The 3rd Binomial Formula It is about the calculation of the term (a + b) (a b). We apply again the method that one always uses to multiply brackets and gets: (a + b) (ab) = a + a (b) + b a + b (b) = a ab + ab b = ab That was all . Therefore we remember: Now there is no double product, because the mixed products and ba are eliminated! Examples (a) (3a + b) (3a b) = (3a) (b) = 9a 4b (b) (3a b) (3a + b) = (3a) (b) = 9a 4b Did you discover that Exercise (a) is identical to (b): It doesn't matter whether the bracket with the minus sign is the first or the second bracket! (c) () () () 8x 3 8x + 3 = 8x 3 = 64x 9 (d) (4x 5) (5 + 4x) Attention, now you have to take a closer look, because the brackets do not only differ by the minus sign . The order of the summands is also different. So you have to change this first. You can only do this in the plus brackets, because there applies 5 + 4x = 4x + 5: () () () () () 4x x = 4x 5 4x + 5 = 4x 5 = 16x 5 (e) Caution Trap: (3x 4z) (3z + 4x) This exercise does NOT fit the 3rd binomial formula, because there are other summands in the second bracket! In order to be able to use the 3rd binomial formula, the problem should be called like this: (3x 4z) (3x + 4z) = 9x 16z. Now the intermediate step has been left out again. You usually do it in your head. + = = (f) () () () 4 x 3 x 3 x 3 x 9 (g) (7p + 4q) () (4q + 7p) () 4q 7p = 4q 7p = 16q 49p (h) ( ) () () () () () xy xz yx + zx = xy xz xy + xz = xy xz = xyxz Training exercise 3 (a) (cd) (c + d) (b) (3a b) (3a + b ) (c) (5x + 1) (5x 1) 1 1 (d) (4a) (4a +) (e) (ab) (a + b) (f) (4 x) (x + 4) (g ) (1 x 3) (1 x 3) + (h) (8u v) (8u + v) (i) (6+ x) (6+ x) () () a + b ab = a -b Friedrich Buckel

8 110 Term transformations Devilish minus signs (for very stupid, malicious tasks) There are terms that do not really fit a binomial formula, but are suitable for it after a little transformation. This is then due to the sign. A few preliminary exercises: If you swap the two numbers in a difference, then the sign of the result changes: 5 3 = but 3 5 = = 7 but 5 1 = - 7 Next we look at the following calculations: (5 3 ) = 5+ 3 (1 5) = Here the number 1 in front of the brackets (you don't have to write the 1) is multiplied into the brackets. If we read the last two equations from right to left, we get: And these equations teach us: 5+ 3 = (5 3) = (1 5) If you put a minus sign in brackets, this changes all signs in the brackets. a) (5+ 3) (5+ 3) Application For the 3rd binomial formula we would like (5 3) (5+ 3). We achieve this by removing -1 from the 1st bracket. () () () () (= = 5 3) = (5 9) = 16 b) () () () () (= = 1 5) = (144 5) = 119 c) (a + b) (a + b) = (a + b) (ab) = (ab) = a + b () () d) (3x + 4) (3x + 4) = (3x 4) (3x + 4) = (3x) 4 = 9x 16 = 9x + 16 Friedrich Buckel

9 110 Term transformations 7 This method also helps with this exercise: (a b) (a + b) =? The factor (-1) is excluded from the first bracket: This changes the calculation as follows: Examples (ab) = (a + b) a) (x 4) (x + 4) = (x + 4) (x + 4) = (x + 4) = (x + 8x + 16) = x 8x 16 b) (5a b) (b + 5a) = (5a + b) (b + 5a) = (5a + b) = (5a + 0ab + 4b ) = 5a 0ab 4b The following was used here: (b + 5a) = (5a + b), because with a sum you can swap the summands! c) (1rs + 8r) (8r 1rs) = (8r + 1rs) (8r + 1rs) = (8r + 1rs) = ((8r) + 8r 1rs + (1rs)) = (64r + 19r s + 144r s ) = 64r 19r s 144r s Here first the first sum was swapped and then from the. Brackets -1 excluded! This task should also be treated as follows: Examples () () () () () () -ab a + b = - a + b a + b = - a + b = - a + ab + b = -a - ab-b (a + b) (ab) =? You have to recognize that the signs in the first bracket are exactly the opposite of those in the second bracket. So we change this by excluding -1 from the 1st bracket: () () () () () () -a + b ab = - ab ab = - ab = - a -ab + b = -a + ab-b a) (3 5a) (3+ 5a) = (3 5a) (3 5a) = (3 5a) = (9 30a + 5a) = 9+ 30a 5a b) (4u 7v) (7v 4u) = (4u 7v) (7v + 4u) = (4u 7v) (4u 7v) Now you have to know that 7v + 4u = 4u 7v, because here too there is a sum in which the summands can be swapped. The minus sign does not show a difference but a sign! = (4u 7v) = (16u 56uv + 49v) = 16u + 56uv 49v Friedrich Buckel

10 110 Term transformations 8 c) (15a 5b) (5b 15a) = (15a 5b) (5b + 15a) = (15a 5b) (15a 5b) = (15a 5b) = (5a 750ab + 65b) = 5a + 750ab 65b d) (1uv 4v) (1uv + 4v) = (1uv 4v) (1uv 4v) = (1uv 4v) = (144u v 96uv + 16v) ​​= 144u v + 96uv 16v There is a fourth task that works in this way can be: Examples (ab) =? This means in detail: (a b) (a b) We extract the factor (-1) from each bracket. This then comes twice in front of the bracket, and because (1) (1) = + 1, this is omitted again, so the following applies: (-a -b) = (a + b) !!! a) (3x 6) = (3x + 6) = 9x + 36x + 36 b) (5x x) = (5x + x) = 5x + 0x + 4x 3 4 = a + ab + bc) () () ( ) 13ab 5c 13ab 5c = 13ab + 5c = 169a b + 130abc ​​+ 5c Training exercise 4 (very important!) Form these terms by excluding -1 and then use binomial formulas. a) (r 5s) (r 5s) b) (4a + b) (4a b) c) (5x 3) (5x 3) d) (x + 5) (x + 5) e) (from 3) f ) (ab + bc) g) (5x + x) (x 5x) h) (6a 5b) (5b 6a) i) (b 5a) (5b a) j) (3a b + ba) (ba 3a b) k) (z 3w) (z + 3w) l) (ab + ba) (ab ba) Friedrich Buckel

11 110 Term Transformations Mixed Exercises Here again the three binomial formulas: () a + b (ab) (+) (-) = a + ab + b = a -ab + babab = a -b and the minus sign rules a (ab ) = - (b-) (-a - b) = - (a + b) Training task 5 Use the appropriate formulas: (a) (3m 8n) (3m 8n) 1 + (b) () x 5 (c ) (3x + 4) (d) (7u + v) (v 7u) (e) (4a 7b) (4a 7b) (f) (8x + y) (g) (8x y) (y 8x) + ( h) (17a b) (17a b) (i) (u) (j) (65a + 15a) (k) () (6x 3x 3x 6x) (l) (5x +) (5x) (m) (1x + 3w) (1x 3w) (n) (8uv + 3vw) (8uv 3vw) (o) (pc 15rst) (p) (z + 3w) Friedrich Buckel

12 110 Term transformations 10 Here again, exercises 1 to 4 (as an exercise sheet to repeat) Training exercise 1 (a) (m + n) (b) (3a + 4b) (c) (7x + 15) (d) (5c 6d) 15x 4y + (e) (+) 1 3 (f) (x +) (g) (3x +) 1 (h) (8b + a) 1 (i) (+) 3 16 Training task x 4x (a ) (5a c) (b) (7a b) (c) (0x 5) (d) (from 4) (e) (4a) 1 (g) (4x) 1 3 (h) (uv) (f) () 4 ​​4 a 4 (i) () Training exercise 3 bx 6 3 (a) (cd) (c + d) (b) (3a b) (3a + b) (c) (5x + 1) (5x 1) 1 1 (d) (4a) (4a +) (e) (ab) (a + b) (f) (4 x) (x + 4) (g) (1 x + 3) (1 x 3) (h ) (8u v) (8u v) (i) (6 x) (6 x) Training exercise 4 (very important!) Form these terms by excluding -1 and then use binomial formulas. a) (r 5s) (r 5s) b) (4a + b) (4a b) c) (5x 3) (5x 3) d) (x + 5) (x + 5) e) (from 3) f ) (ab + bc) g) (5x + x) (x 5x) h) (6a 5b) (5b 6a) i) (b 5a) (5b a) j) (3a b + ba) (ba 3a b) k) (z 3w) (z + 3w) l) (ab + ba) (ab ba) Friedrich Buckel

13 110 Term transformations 11 Solutions to training tasks 1 to 5 Training task 1 Apply the 1st binomial formula: Solution (a) (m + n) (b) (3a + 4b) (c) (7x + 15) (d) (5c 6d) 15x 4y + (e) (+) 1 3 (f) (x +) (g) (3x +) 1 (h) (8b + a) 1 (i) (+) 3 16 x 4x ( a) () m + n = m + mn + n (b) (3a + 4b) = (3a) + 3a 4b + (4b) = 9a + 4ab + 16b (c) () 7x + 15 = 49x + 10x + 5 (d) () 5c + 6d = 5c + 60cd + 36d (e) () 15x + 4y = 5x + 10xy + 16y x + = x + x + = x + x + 4 4 (f) () () () ( g) () () 3 () 3x + = 3x + x + = 9x + 4x + Attention: Here again the principle applies: Be brief before multiplying. So in the double product you first cut out the 3 !! a (h) () () () 8b + a = 8b + b a + a = 64b + ab + The same applies here: One abbreviates the number 16 in the double product! x + 4x = x + x 4x + 4x = x + 4x + 16x 4 1 (i) () () () Friedrich Buckel

14 110 Term Transformations 1 Turn the. binomial formula to: solution training exercise (a) (5a c) (b) (7a b) (c) (0x 5) (d) (from 4) (e) (4a) 1 (g) (4x) 1 3 ( h) (uv) 4 (a) () 5a c = 5a 10ac + c 1 1 (f) () 4 4 (b) () 7a b = 49a 8ab + 4b (c) () 0x 5 = 400x 1000x + 65 (d) () from 4 = from 8ab + 16 a 4 (i) () (e) (4a) = () () 4a + (4a) = 4+ 16a + 16a 1 1 (f) () from = a ab + b = a ab + bx (g) () 1 4x = 16x x + = 16x + (h) (1 3) uv = u uv + v = u uv + v Make sure to shorten here and not first 6 16 (i) () x 6 = xx + 36 = x 8x bx 6 3 write in the double product! Here we have this situation again: First we shorten by 3! The master of algebra shows himself in the early abbreviation Friedrich Buckel

15 110 Term transformations 13 Solution of training exercise 3 (a) (cd) (c + d) (b) (3a b) (3a + b) (c) (5x + 1) (5x 1) 1 1 (d) (4a ) (4a +) (e) (ab) (a + b) (f) (4 x) (x + 4) (g) (1 x 3) (1 x 3) + (h) (8u v) ( 8u + v) (i) (6+ x) (6+ x) (a) (cd) (c + d) = cd (b) (3a b) (3a + b) = 9a 4b (c) (5x + 1) (5x 1) = 5x 1 (d) () () (e) () () 4a 4a + = 16a aba + b = ab 4 (f) () () () () 4 xx + 4 = 4 x 4+ x = 16 x (g) () () x + 3 x 3 = x 9 4 (h) () () () 8u v 8u + v = 8u v = 64u v (i) (6+ x ) (6+ x) = (x + 6) (x 6) = 4x 36 Friedrich Buckel

16 110 Term Transformations 14 Training Exercise 4 (very important!) Convert these terms by factoring out -1 and then using binomial formulas. Solution a) (r 5s) (r 5s) b) (4a + b) (4a b) c) (5x 3) (5x 3) d) (x + 5) (x + 5) e) (from 3) f) (ab + bc) g) (5x + x) (x 5x) h) (6a 5b) (5b 6a) i) (b 5a) (5b a) j) (3a b + ba) (ba 3a b ) k) (z 3w) (z + 3w) l) (ab + ba) (ab ba) a) (r 5s) (r 5s) = - (r 5s) (r + 5s) = (4r 5s) = 4r + 5s b) (4a + b) (4a b) = - (4a + b) (4a + b) = (4a + b) = (16a + 16ab + 4b) = 16a 16ab 4b c) () () () () () 5x 3 5x 3 = -5x + 3 5x 3 = 5x 9 = 5x + 9 d) () () () () (4) 4 x + 5 x + 5 = - x + 5 x 5 = x 5 = x + 5 or you swap the sums in both brackets: (x + 5) (x + 5) = (5+ x) (5 x) = 5 x 4 () () e) (from 3 ) = (ab + 3) = ab + 3 = 4a b + 1ab + 9 Here, -1 has been excluded from both brackets (which are hidden in the square), which cancels itself out again. f) (ab + bc) = (bc ab) = b c ab c + a b Here the two terms have been exchanged. g) () () () () () 3 4 5x + x x 5x = x 5x x 5x = x 5x = 4x 0x + 5x If you swap both terms in the first bracket, you get two identical brackets. So you don't have to exclude anything here: x and x have the same coefficients (prefixes), so there is no reason to change. Friedrich Buckel

17 110 Term transformations 15 h) (6a 5b) (5b 6a) = (6a 5b) (+ 5b + 6a) = (6a 5b) (6a + 5b) In this exercise you have to analyze two things: 1. Are the signs different and we cling from the. Bracket -1 off. then the order is not correct, so we swap the. Brackets the summands. Now the 3rd binomial formula is suitable for use: = (36a 5b) = 36a + 5b i) (b 5a) (5b a) This problem does not fit the 3rd binomial formula, because once it is 5a, then a. You can only multiply the 1st bracket with the second quite normally: (b 5a) (5b a) = 10b + 4ab 5ab + 10a = 10a 1ab 10b j) () () [() 3a b + baba 3a b = - - 3ab ba] (b a + 3a b) Here I have pulled -1 out of both brackets, which is canceled again. So you can go on like this: () () (= 3a b b a b a + 3a b = 3a b b a) (3a b + ba) Now I still have the. Brackets the summands reversed! Now the 3rd binomial formula fits! = 9a b 4b a 4 4 k) (z 3w) (z + 3w) = (z) (3w) = 4z 9w Use -z for a in the 3rd binomial formula. l) (ab + ba) (ab ba) = 0 because ab = ba, so every bracket has the value 0! Fallen in? Friedrich Buckel

18 110 Term transformations 16 Exercise 5 Use the appropriate formulas: Solution (a) (3m 8n) (3m 8n) x (b) () (c) (3x + 4) (d) (7u + v) (v 7u) (e) (4a 7b) (4a 7b) (f) (8x + y) (g) (8x y) (y 8x) + (h) (17a b) (17a b) (i) (u) (j) (65a + 15a) (k) () (6x 3x 3x 6x) (l) (5x +) (5x) (m) (1x + 3w) (1x 3w) (n) (8uv + 3vw) ( 8uv 3vw) (o) (pc 15rst) (p) (z + 3w) (a) (3m + 8n) (3m 8n) = 9m 64n 1 1 = (b) () () x 5 xx 5+ 5 = x 5x + 5 (c) () 3x + 4 = 9x + 4x + 16 (d) (7u + v) (v 7u) = (v + 7u) (v 7u) = 4v 49u (e) () (4a 7b ) - () (4a + 7b) () 4a 7b = 4a 7b = 16a 49b = 16a + 49b Here you had to get -1 from the. Exclude brackets so that the 3rd binomial formula is easy to use. (f) (8x + y) = (y 8x) = 4y 3xy + 64x I swapped the summands in brackets. (g) () () () () () + = = = + 8x y y 8x 8x y 8x y 8x y 64x 3xy 4y Here you had to recognize that both brackets are identical. In the second bracket only the summands were swapped, which I then reversed. No minus sign had to be pulled out for this! (h) (17a b) (17a b) = - (17a b) (17a + b) = (89a b) = 89a + b By excluding -1, the. Bracket from -17a as in the first bracket + 17a. Now you can see the 3rd binomial formula immediately.(i) (u) = (16 5u) = 56 5u u (see note f) = 5016 = 800 (j) (65a + 15a) = (65a 15a) = (50a) = 500 a Friedrich Buckel

19 110 Term Transformations 17 Did anyone want to use a binomial formula? Here you have to recognize that you can subtract 65a and 15a. By the way, I previously pulled -1 out of both brackets, which was canceled out in effect. But you don't have to do that, there is another way: (65a + 15a) = (50a) = 500 a (k) (6x 3x) (3x 6x) = - (6x 3x) (3x + 6x) = - (6x 3x) ​​= (36x 36x + 9x) = 36x + 36x 9x Here you had to recognize that the terms 6x and 3x in the. Brackets have the opposite signs. This was fixed by factoring out -1. Now only the order of the summands does not match, but the brackets are otherwise the same, so they can be written together as (6x 3x)! (l) (5x +) (5x) = (5x +) (5x +) = (5x +) = (5x) = 5x 0x + 4 Did you recognize that the second bracket is the same as the first, except for the order of the Summands! You can swap these and then you have (5x +). This can be calculated immediately with the 1st binomial formula: (5x +) = (5x) + (5x) + = 5x 0x + 4. Or, as I showed it first, you can pull -1 out of both brackets, which is in effect again, ie one simply changes both signs! (m) (1x + 3w) (1x 3w) = - (1x + 3w) (1x + 3w) = - (1x + 3w) = (144x + 7xw + 9w) = 144x 7xw 9w By excluding -1 from the . Both brackets became identical! (n) (8uv + 3vw) (8uv 3vw) = (8uv) (3vw) = 64u v 9v w Here you can use the 3rd binomial formula directly with a = -8uv and b = 3vw! (o) (pc 15rst) = (pc + 15rst) = 4p c + 60pcrst + 5r s t (p) (z + 3w) = (z 3w) = z 6zw + 9w Friedrich Buckel

20 110 Term Transformations Even More Complicated Terms !!! EXAMPLES a) () () [] [] a b a + b = a ab + b a + ab + b a ab b a ab b 4ab = + = Here you have to have at least the. Put the term in brackets after using the 1st binomial formula because there is a minus sign in front of it. So all three summands have to be subtracted. If you forget the brackets, you only subtract the first summand. This is wrong and would look like this: (ab) (a + b) = a ab + ba + ab + b = b (FALSE!) B) () () (+) [] 3x 4 3x 4 3x 4 = 9x 4x x 16 = 9x 4x x + 16 = 4x + 3 Here only the transformation result of the term was put in (square) brackets, because only this is subtracted. I'm just using the square brackets to make them stand out more. You can also use round brackets instead. c) () () () (4 a + b a b = a + a b + b a a b + b) = a + a b + b a + a b b = 4ab. Here, too, it was important to put brackets for the intermediate result of the term and then to reverse all signs for the subtraction (red). The first 4 4 brackets around a + a b + b are unnecessary. I only use it for visual reasons! a) (4a b) (ab) (ab) Training task 6 + b) (x 4) (x +) c) (u 8v) (u + 8v) (8v + u) (8v u) d) (a + b ) (a + b) e) (ab) (ab) abababf) () (+) () g) (x +) (x) h) (3a 4b) (4b 3a) (3a + 4b) (3a 4b) Friedrich humpback

21 110 Term transformations 19 a) (4a b) (a + b) (ab) = solution of training exercise 6 [] [] = 16a 16ab + 4b 4a + 4ab + b 4a 4ab + b = 16a 16ab + 4b 4a 4ab b 4a + 4ab b = 8a 16ab + bb) (x 4) (x +) = x 4 8x + 16 [x 4 + 4x + 4] 4 4 = x 8x + 16 x 4x 4 = 1x + 1 c) (u 8v) (u + 8v) (8v + u) (8v u) [] = 4u 64v 64v u = 4u 64v 64v + u = 5u 18v d) () () () () a + b a + b = a + ab + b a + b = a + a b + a b + ab + ab + b = a + 3a b + 3ab + be) () () () () abab = a ab + bab = aaba b + ab + ab b = a 3a b + 3ab bf ) (ab) (a + b) (ab) = a 4 b 4 [a 4 ab + b 4] = aba + abb = abbg) (x +) (x) = (x + 4x + 4) (x 4x + 4) x 4x 4x 4x 16x 16x 4x 16x 16 = = x 8x + 16 ATTENTION Problem (g) can be solved faster if you consider the following: (x +) (x) = (x +) (x +) (x) (x) = (x 4) (x 4) () 4 = x 4 = x 8x + 16 Or shorter: [] [] 4 (x +) (x) = (x +) (x) = x 4 = x 8x + 16 That is the optimal solution. You can do it when you realize that you can summarize the existing 4 terms more skillfully. h) (3a 4b) (4b 3a) (3a + 4b) (3a 4b) The blue and red brackets are identical except for the order of the summands. The factor - 1 is extracted from both: - (3a 4b) (4b + 3a) + (3a + 4b) (3a + 4b) = - (3a 4b) (3a + 4b) + (3a + 4b) (9a 16b ) + (9a + 4ab + 16b) = 9a 3b = + 4ab + 16b + 9a + 4ab + 16b Friedrich Buckel

22 + +! +! + +! +! +: +! +:! + +:! : +! + (+! + (! + + (! (+! Terms and term transformations Numbers, variables 1 and meaningful combinations of them with the help of arithmetic symbols or brackets) are called terms. Examples of terms: 3;; a 6; 9 (4 + 9 ); a² b³; x² +1 Term transformation: = 9; 8 = 16; 3a + a = 4a; 7b b = 9b Equivalent: 4 + a = a + 4; r 6 = 6 r Spellings: ab = ab = ba; 4 (a + b) = 4 (a + b); 1a = a; 0 a = 0 If the value of the variable is known, the value of the term can also be calculated! Example with solution: Put in the term 3 + x for x one after the other the numbers 3, 4 and 8. Then calculate the value of the term. Solutions: = 6; = 7; = Exercise: Calculate the term value. a) Set the numbers 8, 9 one after the other in the term x + 4 for x and 5! 1, 13, 9 b) Put in the term a + b for a the number 7 and for b the number 34! 48 c) Put in the term 3c y for c the number 6 and for y the number 17! 1 d) Put in the term 5x m + 9 for x the number 3 and for m the number 1! 18. Exercise: Compare the values ​​of the following terms: a) 4a + 4b and 4 (a + b) Substitute the number for a and the number 5 for b. 8/8 b) 3 (x) and 3x 6 Substitute the number 8 for x. 18/18 c) x (4x +) and x 4x Substitute the number 5 for x. 1/1 terms with variables are called equivalent if they result in the same value after inserting numbers for the variables. 3rd task: Term transformations addition / subtraction Summarizing 3 identical variables. Identical variables are added or subtracted by adding or subtracting their prefixes. a) a + 5a + 8a = 15a b) 5x + 7x + 9x = 1x c) 10y + 1y + 30y = 5y d) 40a 3a 9a = 8a e) 14x 7x 9x = af) 4y 19y = 15y g) 8a + 5a 4a = 9a h) 70x + 5x x 6x = 67x i) 10y + 1y + 30y 4y 6y y = 40y j) 1a + 3b + 13a + 14b = 5a + 17b k) 3y 1y + 34g 1g = 11y + 13gl) 6a + 13b 3a 15 b + 40a = 31a bm) 4x + 6y + 8z 6x 9y 1z + 50x n) 3a 6b + 7c 10a 4b 7c = 40x 3y 4z = 7a 10b + 0c 4. Loosen the brackets (from the inside out ) and then summarize: a) 1x + (5 3x) = 9x + 5 b) 18a (3b + 5a) = 13a 3b c) (15x + 6y) + (8x 17y) = 3x 11y d) (5m 14n ) (4m 7n) = 17m + 13n e) 1a (3b 15a) = 7a + 3b f) 5 (1 y + 1 7) 7 y = 8 + 5y g) (5a 3b) (8a + 5b) + (7a b) (3a 8b) = 1a 1b h) 40m + [7m (5n + 3)] = 47m 5n + 3 i) 35y + [8y (4x 8)] + [9 (5x 3)] = 43y + 1x + 0 1 variabilis (lat), changeable 3 Note: 4 5 = 9; = 50 9 = + 41 !! term, expression 4a 5a = 9a; 50a 4a 5a = 50a 9a = + 41a !!

23 5. First calculate the following products: a) (+3) (+5) +15 b) (5) (+6) 30 c) (+5) (6) 30 d) (7) (9) +63 e) (+ 7a) (+ 9b) 63ab f) (6x) (+ 7y) 4xy g) (+ 4s) (r) 8rs h) (6a) (5b) + 30ab i) (4x) (5y) ( 7a) 140axy j) (3a) (b) (1c) (4d) + 4abcd 6. Term transformation: Multiplication and division: Factor times sum (multiply): A factor is multiplied by a sum by adding each summand in brackets with multiplied by the factor. (The sign rules must be observed!) Factor Sum Sum Factor 5 (xy) = 5 x 5 y (x + y) 5 = 5 x + 5 y 6.1 Write without brackets (multiply!): Example: 1 (3m + 4n) = 1 3m + 1 4n = 36m + 48n 10a + 1b a) 5 (a + b) 5a + 5b b) 6 (3m + 4n) 19m + 4n c) (x + 3y) 5 10x + 15y d) (5a + 6b) e) 4 (ab) 4a 4b f) (4m n) 8m 4n g) (3x 7y) 3 9x 1y h) (3a 9b) 5 15a 45b 6. Multiply the brackets from and then summarize: Example: (a + b) + 3 (3a + 4n) = a + b + 3 3a +3 4n = a + b + 9a + 1n = 11a + b + 1n a) 8 (a + b) + 6 (m + 8b) 8a + 1m + 56b b) (x + 9y) 5 + (4x b) x + 45y 4b c) (5) (ab) (6a 3n) 17a + 5b + 6n d) () (x 8y) (3a 9y) 5 4x + 61y 15a 7th division: Example: (30m + 45n): (15) = m 3n a + 3b 3m + 4n x + 1y 0.5a + 4b a) (4a + 6b): b ) (30m + 40n): 10 c) (10x + 5y): 5 d) (1a + 8b): e) (0.5a 0.8b): f) (7m 3n): () g) (7x 9y ): 3 h) (8a 16b) :( 4) 0.5a 4b + 3.5m + 1.5n 13.5x 4.5y a + 4b 8. Combination of exercises 4., 6. and 7.: a) 10a + [4 + (4a + 8)] + 5 (a + 6) = 19a + 18 b) 1x + [10 (x + 7)] + 6 (x) = 16x 9 c) 11b [(b 3) ] + 6 (b 3) = 19b 19 d) 0c [8 (c + 3)] 6 (c 4) = 18c +14 e) 13x + [5 + (x 3) + (3x +)] 3 (x ) = 11x + 10 f) 33y + [7 + (3y 5) + (1y + 6)] 4 (y) + (10y + 0): = 45y Determine a term for the side length u of the figure shown. First label sides of the figure of the same length with the same variables: (u = circumference) a) u = a + 4b b) u = a + bc) u = a + b + cbbbaaaaacaabacbb 10. Solve the following equation: a) 5 x 36 = b) 7 xx = 5 x

24 Multiplying sums «The rectangle as a term converter Multiplication of sums Signs Binomial formulas Exercises Solutions« The rectangle as a term converter Exercise: Determine the area of ​​the given rectangle in two different ways: Length, width a) a d + eb) a + b d + ec) a + b + c d + e + f Answer: Plan figure: a) a (d + e) ​​= ad + aeb) (a + b) (d + e) ​​= ad + ae + bd + bec) ( a + b + c) (d + e + f) = ad + ae + ae + bd + be + be + cd + ce + ce «Multiplication of sums Note: You multiply two sums by multiplying them in terms and forming the total. «Sign Consider the area of ​​the yellow rectangle:

25 The following applies: (a - b) (d - c) = (+ a - b) (-c + d) = ad - ac - bd + bc Note: When multiplying in terms of terms, the usual sign rules apply. «Binomial formulas Once again special rectangular areas: The relationships can also be achieved by known algebraic transformations: (a + b) ² = (a + b) (a + b) = a² + ab + ab + b² = a² + ab + b² (a - b) ² = (ab) (ab) = a² - ab - ab + b² = a² - ab + b² All binomial formulas: (a + b) ² = a² + ab + b² (a - b) ² = a² - ab + b² (a + b) (a - b) = a² - b² «Exercises 1. BF (binomial formulas) simple: a) (x + y) ² b) (x-1) ² c) (z +1) ² d) (z +) (z-)

26. BF with numerical factors: a) (x + 3y) ² b) (, 5x - 0.5y) ² c) (1, u - 1.5v) (1, u + 1.5v) d) (8a - 0, 5b) (8a + 0.5b) 3rd BF with several factors: a) (5xy - 3y) ² b) (1a + 15xy) ² c) (1.1uv - 1.5uw) ² d) (8ab - 9bc ) (8ab + 9bc) 4th BF with factor in front of the brackets: a) (x + 1) ² b) a (a - b) ² c) ab (a + ab) ² d) 0.5a² (a - b ) (a + b) 5th BF with powers: a) (5x² - 6y) ² b) (a³ + 5xy²) ² c) (1.7uv² - 1.5u²w³) ² d) (8a² - 9b²) (8a² + 9b²) 6th terms with BF: a) (u² - 9v²) + (u² + 9v²) b) (x + y) (x - y) - (x + y) ² c) (4a + 5b) ² - ( 5a + 4b) ² d) (8a - 9b) (9a - 8b) + 7ab 7. Calculate the area of ​​the square as the sum of its partial areas Example: side length (u + v); A = (u + v) (u + v) = u + uv + va) Side length: a + 3b b) Side length: u - 3v c) Side length: a + b + cd) Side length: x + 3y + 4z 8. Calculate the area of ​​the rectangle as the sum of its partial areas a) Length: a - x; Width: a + x b) length: u - 3v; Width u + 3v c) length: 3a - (b + 4c); d) length: (x + y) - (u + v); Width: 3a + (b + 4c) Width: (x + y) + (u + v) 9th theorem of Pythagoras. In the case of a right triangle, the sum of the square areas above the cathetus is equal to the square area above the hypotenuse (the purple area corresponds to the yellow). Calculate the area c of the square above the hypotenuse if the sides have the following side lengths: a) a = x + y; b = x - y; b) a = u - v; b = u + v; c) a = (x + y + z); b = (x - y); d) a = (x + 3y) b = (x - 3y); «Solutions 1. BF simple: a) x² + xy + y² b) x² - x + 1 c) 4z² + 4z + 1 d) z² - 4th BF with numerical factors:

27 a) 4x² + 1xy + 9y² b) 6.5x² -, 5xy + 0.5y² c) 1.44u² -, 5v² d) 64a² - 0.065b² 3rd BF with several factors: a) 65x²y² - 150xy² + 9y² b) 144a² + 360axy + 5x²y² c) 1,1u²v² - 3,3u²vw +, 5u²w² d) 64a²b² - 81b²c² 4th BF with factor in front of the bracket: a) x² + 4x + b) a³ - a²b + ab² c) a³b + 4a³b² + a³b³ d) 0.5a³ - a²b² 5th BF with powers: a) 5x 4-60x²y + 36y² b) 4a 6 + 0a³xy² + 5x²y 4 c), 89u v 4-5,1u³v²w³ +, 5u 4 w 6 d) 64a 4-81b 4 6. Terms with BF: a) 8u v 4 b) -xy - y² c) -9a² + 9b² d) 7a² + 97ab - 7b² 7. Calculate the area of ​​the square as the sum of its partial areas a) A = ( a + 3b) = 4a + 1ab + 9b b) A = (u - 3v) = 4u - 1uv + 9v c) A = (a + b + c) = a + b + c + ab + ac + bc d) A = (x + 3y + 4z) = 4x + 9y + 16z + 1xy + 16xz + 4yz 8. Calculate the area of ​​the rectangle as the sum of its partial areas a) A = (a - x) (a + x) = a - 4x b) A = (u - 3v) (u + 3v) = 4u - 9v c) length: 3a - (b + 4c); d) length: (x + y) - (u + v); Width: 3a + (b + 4c) Width: (x + y) + (u + v) 9. Pythagoras: a) (x + y) ² + (xy) ² = x² + xy + y² + x² - xy + y² = x² + y² b) (uv) ² + (u + v) ² = u² + v² c) (x + y + z) ² + (xy) ² = x² + y² + z² + xy + xz + yz + x² -xy + y² = x² + y² + z² + xz + yz d) is still being processed!

28 Folding test binomial teaching formulas I First fold the sheet of paper along the line and then solve the following problems. Once all the tasks have been completed, the results are compared and the number of correct tasks noted. Factor! Solutions 1. 64x² - 48x + 9 = (8x - 3) ². x² - 18x + 81 = (x - 9) ² 3. 49x² + 140x = (7x + 10) ² 4. 36x² + 84x + 49 = (6x + 7) ² 5. 9x² -4x + 16 = (-3x + 4) ² 6. 16x² + 56x + 49 = (4x + 7) ² x² - 160x + 64 = (10x - 8) ² 8. x² + 16x + 64 = (-x - 8) ² 9. 36x² + 4x + 4 = (-6x -) ² x² - 40x + 4 = (-10x +) ² x² + 36x + 9 = (-6x - 3) ² 1. 36x² + 10x = (-6x - 10) ² 13. 4x² + 8x + 49 = (x + 7) ² 14. 9x² + 60x = (-3x - 10) ² x² - 18x + 1 = (-9x + 1) ² Result: Maria Niehaves 00

29 Folding test binomial teaching formulas I First fold the sheet of paper along the line and then solve the following problems. Once all the tasks have been completed, the results are compared and the number of correct tasks noted. Multiply out! Solutions a) (5x -5) ² = 5x²-50x + 5 b) (7x +1) ² = 49x² + 14x + 1 c) (3x -) ² = 9x²-1x + 4 d) (3x +8) ² = 9x² + 48x + 64 e) (-9x -6) ² = 81x² + 108x + 36 f) (-10x +9) ² = 100x²-180x + 81 g) (-6x-10) ² = 36x² + 10x + 100 h) (-7x +9) ² = 49x²-16x + 81 i) (x -4) ² = 4x²-16x + 16 j) (3x -9) ² = 9x²-54x + 81 Result: Maria Niehaves 00

30 Fold-out card TERMUMFORMUNGEN A. Mattler, LWG 1 Summarizing the same summands Solutions 1. Calculate as a multiplication problem! = 6 = 5 6 = = 7 = 3 7 = = 14 = 4 14 = 56. Combine the same summands (see example)! Example: m + m + m = 3m m + m + m + 4m = 8m m + m + 3m + r + r + m = 6m + ra) b + b + b + b + b = 5b b) a + a + a + a + a + c + c + c = 5a + 3c c) a + z + 3a + z + z + z + b + a + b + z 5a + b + 6z d) 4a + 1y + 13z + 8a + 9y + a + a + a + 3y + y = 16a + 5y + 13z e) 3a from + 3b z + 3z a 4y = a + b 4y + zf) 13y 6y + 8y = 15y g) 1x y + x 6 + 3x = x + 4y h) xxy + 7y = 1 + 3x + 8y i) 3 k + m + 3k 8 + k = k + mj) 1,5a + 4 3,5a + b = 4 a + bk) 0, b + c, 9b + 1.1c + 0.03 = 4.03 +, 1b +, 1c l) 0.9a 1, a + 3.1b + c 4.1c = - 0.3a + 3, 1b - 3.1c m) 1 5 a + 4 b + 4 a + 7 b = 7 10 decreases) 0.5a + 1.375bab = a + b

31 Folded card TERMUMFORMUNGEN A. Mattler, LWG Summarizing the same factors Solutions If the same factors are multiplied, this product can be written as a power. Example: = 8³ xxxxx = x 5 a) aaaaaa = a 6 b) mmmmmmmmm = m 9 c) nnnn = n 4 d) = 1 4 e) ss = s² f) 0.3 0.3 0.3 0.3 = 0.3 4 g) -1 (-1) (-1) (-1) (-1) = (-1) 5 Summarize the same factors in the products as in the examples. (The multiplication symbol may be omitted) Example: xxx 7 = 7x³ 3 3 aaaaa = 9a 5 ccckkkkk = c³ k 5 = c³k 5 h) 4 4 xxxxx = 16 x 5 i) 5 5 ss = 5 s² j) rrrhhpppy = r³ h² p³ y = r³h²p³y k) ddggg = 8 d²g³ l) 0.1 0.1 xxxxxyyy = 0.01 x 5 y³ m) 1, 1, aaabbcccd = 1.44 a³b²c³d

32 Folded card TERM FORMING A.Mattler, LWG 3 Combining the same summands and factors Solutions Combine and calculate! a) a + a + a + a + a = 5a b) aaaaa = a 5 c) = 7 d) = 16 f) yyyy = y 4 g) xxyyy = x²y³ h) x + x + y + y + y + y = x + 4y i) a + a + a + a + a + a = 6a + 6 j) 4 4 ssskk = 16 s³k² k) abaaba = a 4 b² l) x + 4x x + y + y + y = x + 3y m) n + n + x + x + 13x 1, x = n + 13.8x n) xxx = 4x³ o) 3 5 aaacc = 15 a³c² p) xx 3 xyx = 6 x 4 yq) 1, aa 1, aam = 1.44 a 4 mr) (-1) aaccc (-1) = a²c³ s) (a) (-a) (-a) (-a) (-a) = (-a) 5 = - a 5 t) (-b) (-b) (-b) (-b) = (-b) 4 = b 4

33 Folded card TERMUMFORMUNGEN A. Mattler, LWG 4 Organize and summarize! Calculate if possible! Solutions a) 1.5x + y + 0.6x -, 3y =, 1x - 1.3y b) x 3 yyx = 1 x²y² c) eeee = - 4e d) x + x + x + 1.05x + xyyy = 5.05x - 3y e) 0.3 vvx 0.3 v = 0.09 v³x f) u + u 18u -13m + 1.3m = 8-16u - 11.7m g) uuu = 7 u³ h) 1 a 1 a 1 a = 1 a 3 i) = 1 7 j) m + m + m = 3m - 7 k) 0.1 nnn 6 0.1 = 0.06 n³ l) 0.1 + n + n + n , 1 = 6, + 3n m) x + x + x = xn) xxx = 16x³ o) r + r + r = 8 + 3r p) 4 4 rrr = 16 r³ q) 3a 4a = 1 a² r) 9a 7a = 63 a² s) 3b 4b 5b = 60 b³

34 Folded card TERMUMFORMUNGEN A. Mattler, LWG 5 Calculate. Note the calculation rules! Solutions = = = = = = = = 135 Could one summarize the above problems without calculating them? Yes, proceed as follows: = or with variables: + ³ = ³ = 4 + ³ x + xxx + x + x + x = x + x³ + x + x + x = x + x + x + x + x³ = 4 x + x³ = 4x + x³ a) y + yyyy + y + y + y = 4y + y 4 b) z + z + zz + z + z = 4z + z² c) xxx + x + x + x + x = 4x + x³ d) y + y + y + y + yyy + y = 5y + y³ e) z + zzzzzz = z + z 6

35 Fold-out card TERMUMFORMUNGEN A. Mattler, LWG 6 Combining the same summands Solutions The same summands can be combined, but they are only the same if the power is also the same. Example: x² + x³ + 4x² + 7x³ = 6x² + 8x³ Combine the same summands!

36 Fold-out card TERMUMFORMUNGEN A. Mattler, LWG 7 Identical summands can be summarized Solutions Summands are the same if they agree in the variable and their power.

37 Folded card TERMUMFORMUNGEN A. Mattler, LWG 8 Addition and subtraction brackets Solutions If there is a plus in front of the bracket, it can simply be left out; if there is a minus in front of the bracket, each character in the bracket changes. In the case of multiplication brackets, each term in the brackets is multiplied by the number in front of the bracket and the signs are taken into account.

38 Folded card TERMUMFORMUNGEN A. Mattler, LWG 9 Addition and subtraction brackets Solutions a) x + x + x + xxxx + ax + ax = 4x + x 4 + ax + ax = 4x + x 4 + ax b) a (x) xax (3 a) ax = a-ax-xa-3x + ax-ax = a - 4x - ax c) ab + ab + a + b + a + b + a (1 b) + b (4 a) = ab + from + a + b + a + b + a-from + 4b-from = 3a + 6b d) on + on + a + a + m + on + on = on + on + a + a + m + on + am = a + m + 4am e) u (a + b) 3 (a + b) + 0.5au 0.5bu = au + bu-3a-3b + 0.5au-0.5bu = - 3a - 3b + 1.5au + 0.5bu f) 1.5 (a + b + c) 0.5 (a + c) 0.5 (b + c) = 1.5a + 1.5b + 1.5c-0, 5a-0.5c-0.5b-0.5c = a + b + 0.5c g) x (ax + ab) + a (x² + bx c) = ax² + abx + ax² + abx-ac = ax² + abx - ac h) a (bc) (bc) + a (cb) = - ab + ac b + c + ac - ab = - ab + ac b + ci) 5 (a + b) 3 (abc) + a (b + c) = 5a + 5b-3a + 3b + 3c + ab + ac = a + 8b + 3c + ab + ac j) v (4 + b) + 5 (v + 1) v (v 8) = 4v + bv + 5v + 60-v² + 8v = 17v + bv - v² + 60 k) 1 4 (a + b + c) (a - b) = 1 4 a + 1 4 b + 1 4 c + 1 8 ab = 3 8 abcl) 0.375 (x + y + z) (xy) = 3 8 xyzxy = 1 xyz

39 Folded card TERMUMFORMUNGEN A. Mattler, LWG 10 Summarize! Solutions Malmarks can be omitted.

40 Folded card TERMUMFORMUNGEN A. Mattler, LWG 11 Brackets by brackets Solutions The following applies to the multiplication of two brackets: Multiply every term in the first bracket by every term in the second bracket and note the signs! Example: thus: xx + x (-7) + 4 x + 4 (-7) = x² - 7x + 4x 8 = x² - 3x - 8 a) (x + 3) (x) = x² - x + 3x 6 = x² + x- 6 b) (y 5) (y + 3) = y² + 3y - 5y -15 = y² - y -15 c) (z + 1) (3 + z) = 3z z² + z = z² + 4z + 3 d) (a) (a 6) = a² - 6a - a + 1 = a² - 8a + 1 e) (a 3) (b + 5) = ab + 5a - 3b - 15 f) (ab ) (a + 9) = a² + 9a from - 9b g) (x + y) (x - 13) = x² - 13x + xy 13y h) (a +) (a + b + 3) = a² + from + 3a + a + b + 6 = a² + 5a + b + ab + 6 i) (x + 4) (x + y + 9) = x² + xy + 9x + 4x + 4y + 36 = x² + 13x + 4y + xy + 36 j) (4 x) (x + y 5) = 4x + 4y 0 - x² - xy + 5x = - x² + 9x + 4y xy 0 k) (a + b) ² = (a + b) ( a + b) = a² + ab + b²

41 Folded card TERMUMFORMUNGEN A. Mattler, LWG 1 Fractional terms solutions Expand the given denominator! Example: 4 9 = = a = ax 7 a = 7x ax a) 5 1 = 60 b) 3 b = from 5 1 = b = 3a from c) a 3b = 3ab d) xc = abc a 3b = a² from xc = abx abc e) 4 9y = 45xy f) abc 7ac = 49abc 4 9y = 0x 45xy abc 7ac = 7ab²c 49abc g) 6 mn = 6mnx h) 1 4udo = 1udo² 6 mn = 36x 6mnx 1 4udo = 36o 1udo² i) 15 5eis = 5e²i²s² 15 5eis = 75eis 5e²i²s² j) 4 8a²bc = 4a²b²c² 4 8a²bc = 7bc 4a²b²c² k) 1xy 5ax = 40ax² 1xy 5ax = 96x²y 40ax² l) 3abc²²²² 9abc² 4 3abc² = 4ab²bc

42 Folded card TERMUMFORMUNGEN A. Mattler, LWG 13 Fractional terms solutions Find the main denominator, expand and calculate the fractions! Example: = = 3 0 abc + 3 ab = abc + 3c abc = + 3c abc a) = b) = = 11 0 c) 5 ax + 3 a = d) am + 4m = 5 + 3x ax 4a + 4m e) 7 5ax + 1 af) 3 7amx + am = 7 + 5x 5ax 3 + 14x 7amx g) ade = h) 4xy + 1 1x²y = 4ade + 40ade 36x + 1 1x²y i) 4 a²bc² + 7 a²b²c² = j) a xyz² + b x²y²z² 4b + 7 a²b²c² axy + b x²y²z² k) x ab²c² + yb = l) 1 u²vwz² + uv = x + abc²y ab²c² 1 + uwz² u²vwz² m) 5 7xy x²y = n) 3 63ab²c + x 9b = 5x + 5 + 7abcx 63ab²c

43 Folded card TERMUMFORMUNGEN A. Mattler, LWG 14 Fractional terms solutions When expanding, both sides must now be expanded. Example: 4ab + 3 5bc = main denominator is: 0abc Expand: a) 1 ab + 1 cd = b) a xy + b yz = cd + ab abcd az + bx xyz c) r 3uvw + s 6uv = d) a 5xy + b 7xz = r + sw 6uvw 7az + 5by 35xyz e) 3x ab²c + y a²b²c = f) 1 1ab² a²b = 3ax + y a²b²c 5a + 4b 60a²b² g) e 6ax + f 10bx = h) a ab²c + 3b a²bc = 5be + 3af 30abx a² + 3b² a²b²c i) 1 3abc bcd = j) ax 13ax bx = 96d + 45a 4abcd 5abx + 3a 65abx k) 1 a + b + 3 c = l) 1 x + y + z 5 = bc + ac + 3ab abc 5y + 10x + xyz 5xy m) a xy²z + b xz + y² = a + by² + xz xy²z

44 1.1.

45

46 1.1. binomial. Use formula!) D) (3u v) (u v) = (9u 1uv + 4v) (u v) = 9u 3 1u v + 16uv 4v 3 3