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Establishing terms with several variables

The goal of establishing terms with several variables is, for example, to express the area in numbers or letters (to establish a term). Therefore, the following example is given:

Now, there could be a task given which insists on figuring out the area of a frame in a term. The solution to this would be:

c ∙ a + c ∙ a + (b – 2a) ∙ a + (b – 2a) ∙ a      (the borders right and left and the borders on top and bottom, minus the intersection)

= 2ca + 2(b – 2a) ∙ a

= 2ca + 2(ba – 2a²)

= 2ca + 2ba – 4a²

Recapitulation of sum terms

A product is the short spelling for a sum. 4 ∙ a is consequently nothing more than a + a + a + a. That is why it is aloud to recapitulate sums when the same factor has to be multiplied by a certain number several times. Example: 4a + a (a = 1a) = 5a, because a + a + a + a + a = 5 ∙ a. 3a + 2ab can not be recapitulated because a + a + a + ab + ab are neither 5a nor 5ab.

Example:

4a + ⅓b + 5,5c + 3b – 5c + a                     arrange

= 4a + 1a + ⅓b + 3b + 5,5c + (-5)c            summarize

= 5a + 3⅓b + 0,5c

Bracket rules

It is learned in the 5^th grade how to dissolve brackets in context with the distributive rules. The bracket is dissolved, when all addend in the brackets are multiplied with the factor in front of the bracket. Hereto an example:

Dissolve the brackets and sum up as far as possible.

5x – (4y + 2x) – 2(8x – 2y)                (a bracket with a minus in front of it, is as if a -1 would stand in front of it.)

= 5x – 4y – 2x – 16x + 4y                arrange

= 5x – 2x – 16x – 4y + 4y                sum up

= - 13x 

Summing up of products

The summing up of products is imaginably simple: All numbers have to be multiplied and the variables need to be arranged, then ggf. Then the powers are made.

Example:

5x ∙ 2y ∙ x ∙ 4y ∙ x

= 5 ∙ 2 ∙ 4 ∙ x ∙ x ∙ x ∙ y ∙ y

= 40x³y²

Addition of products

The explanation is in summing up of sum terms already. Consequently, this is only a repetitive example:

1,5y ∙ 8yx + 9y² ∙ ⅔x² - 5xy²

= 1,5 ∙ 8 ∙ x ∙ y ∙ y + 9 ∙ ⅔ ∙ x ∙ x ∙ y ∙ y – 5 ∙ x ∙ y ∙ y

= 12xy² + 6x²y² - 5xy²

= 7xy² + 6x²y²

Multiplication of sums and differences

The multiplication of sums and differences is almost the same as the bracket rules.

Example:

(2x – 3y) ∙ (4a – 5b + c)

= 2x ∙ 4a + 2x ∙ (- 5b) + 2x ∙ c + (- 3y) ∙ 4a + (- 3y) ∙ (- 5b) + (- 3y) ∙ c

= 2 ∙ 4 ∙ a ∙ x + 2 ∙ (- 5) ∙ b ∙ x + 2 ∙ c ∙ x + (- 3) ∙ 4 ∙ a ∙ y + (- 3) ∙ (- 5) ∙ b ∙ y + (‑ 3) ∙ c ∙ y

= 8ax – 10bx + 2cx – 12ay + 15by – 3cy

Binomial formula

1. The first binomial formula: (a + b)² = a² + 2ab + b²

(a + b)²

= (a + b) ∙ (a + b)

= a ∙ a + a ∙ b + b ∙ a + b ∙ b

= a² + 2ab + b²

2. The second binomial formula: (a – b)² = a² - 2ab + b²

(a – b)²

= (a – b) ∙ (a – b)

= a ∙ a – a ∙ b – b ∙ a + (- b) ∙ (-b)

= a² - 2ab + b²

3. The third binomial formula: (a + b) ∙ (a – b) = a² - b²

(a + b) ∙ (a – b)

= a ∙ a – a ∙ b + b ∙ a + b ∙ (- b)

= a² - b²

Examples:

(2w + 5d)² = (2w)² + 2 ∙ 2w ∙ 5d + (5d)² = 4w² + 20wd + 25d²

(x – 4y)² = x² - 2 ∙ x ∙ 4y + (4y)² = x² - 8xy + 16y²

(3d – 3c)(3d + 3c) = (3d)² - (3c)² = 9d² - 9c²

Transformation of sums into products

Transforming sums into product by

     Making brackets (always factor out the biggest common factor!)

     Binomial formula

Example:

2xy + 10xyz – 20xz = 2x ∙ (y + 5yz -10z)

a² + 2ab + b² = (a + b)²

4a² - 20ab + 25b² = (2a)² - 2 ∙ 2a ∙ 5b + (5b)² = (2a – 5b)²

Copyright © 2005 Christian Franzki - Emkacom Group
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02/09/07