#### Chapter 1 Elementary Arithmetic

Section 1.3 Transformation of terms

# 1.3.2 Transformation of Terms

Dealing with terms gets interesting when we investigate the equality of two terms or simplify complicated terms.
##### Info 1.3.2

Two terms are equal if they can be transformed into each other by allowed transformations. Complicated terms can be simplified using calculation rules. Note in doing so:
1. Exponentiation precedes multiplication precedes addition.
2. If brackets are involved, the distributive property applies:

$a·\left(b±c\right)=a·b±a·c , \left(a±b\right)·c=a·c±b·c .$

3. With $d\ne 0$ we have: $\left(a±b\right):d=\frac{a}{d}±\frac{b}{d}.$
4. For expressions with nested brackets, first evaluate what's inside the innermost set of brackets with respect to the calculation rules and then work your way towards the outermost brackets.

##### Exercise 1.3.3
Remove the brackets from each of the following terms and simplify as far as possible:
1. $\left(1-a\right)·\left(1-b\right)$$=$
.
2. $5a-\left(2b-\left(2a-7b\right)+4a\right)-3b$$=$
.
The input must not contain any brackets.

##### Info 1.3.4

The three binomial formulas are:

$\left(a+b{\right)}^{2}={a}^{2}+2ab+{b}^{2} , \left(a-b{\right)}^{2}={a}^{2}-2ab+{b}^{2} , \left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2} .$

Here, for $a$ and $b$ both numbers and whole terms can be inserted.
##### Example 1.3.5
A few typical applications of the binomial formulas are:
• $\left(1+2x{\right)}^{2}={1}^{2}+2·1·2x+\left(2x{\right)}^{2}=1+4x+4{x}^{2}$.
• $\left(1+2x\right)\left(1-2x\right)={1}^{2}-\left(2x{\right)}^{2}=1-4{x}^{2}$.
• ${x}^{4}-1=\left({x}^{2}+1\right)\left({x}^{2}-1\right)=\left({x}^{2}+1\right)\left(x+1\right)\left(x-1\right)$, in this transformation it can be seen, that in the set of real numbers ${x}^{4}-1$ has only the roots $x=1$ and $x=-1$.
• $\left(1+x+y{\right)}^{2}={\left(\left(1+x\right)+y\right)}^{2}=\left(1+x{\right)}^{2}+2\left(1+x\right)y+{y}^{2}=1+2x+{x}^{2}+2y+2xy+{y}^{2}$.

##### Exercise 1.3.6
Simplify the following term using the second binomial formula:
$\left(-3x+4\right)\left(4-3x\right)$$=$ .

##### Example 1.3.7
The binomial formulas can be used to transform quadratic expressions cleverly. This is very useful if we want to calculate squares without any aid. For this, the number to be squared is split into a simple number (usually a power of ten) and the remainder:

$\begin{array}{ccc}\multicolumn{1}{c}{{103}^{2}}& =\hfill & \left(100+3{\right)}^{2}={100}^{2}+2·100·3+{3}^{2}=10609 ,\hfill \\ \multicolumn{1}{c}{{49}^{2}}& =\hfill & \left(50-1{\right)}^{2}={50}^{2}-2·50·1+{1}^{2}=2401 ,\hfill \\ \multicolumn{1}{c}{{61}^{2}-{59}^{2}}& =\hfill & \left(61-59\right)\left(61+59\right)=2·120=240 .\hfill \end{array}$

##### Exercise 1.3.8
Using the method described in Example 1.3.7, calculate ${1005}^{2}$$=$ .

In the following exercise section you can practise the transformation methods in several exercises.