## Content

### Algebraic manipulation

#### Simplification

An algebraic expression consists of pronumerals, numbers and operations. Only like terms can be added or subtracted.

#### Example

Simplify

\[ 3x^2-4xy+3x-7-5y+2yx+12x^2-8x+7. \]#### Solution

\(3x^2-4xy+3x-7-5y+2yx+12x^2-8x+7 = 15x^2-2xy-5x-5y\).Note that \(xy = yx\) and that it is usual, but not necessary, to write the terms in order of powers, often from highest to lowest.

#### The index laws

Indices are initially a convenient way to represent repeated multiplication. Once this is done, we can state a number of obvious laws that arise naturally to assist in simplification. The meaning of the indices can then be extended to include integers and rational numbers.

A fluency and comfort with the use of indices is essential before dealing with differential calculus, whose basic rules involve the manipulation of indices.

The index laws were covered in the TIMES modules:

- Negatives and the index laws in algebra (Years 7–8)
- Fractions and the index laws in algebra (Years 8–9)
- Indices and logarithms (Years 9–10) .

The index laws state that, for \(a,b\) non-zero and \(m,n\) positive integers,

\begin{alignat*}{2} a^m a^n &= a^{m+n} &\qquad\qquad\qquad a^m\div a^n &= a^{m-n}\\ (ab)^n &= a^nb^n & \left(\frac{a}{b}\right)^n &= \frac{a^n}{b^n}\\ (a^m)^n &= a^{mn}. \end{alignat*}These laws hold for any non-zero real numbers \(a\) and \(b\). The laws also hold when \(m,n\) are negative integers and rational numbers. Meaning can be given to \(a^m\) when \(m\) is an irrational number, but this is much harder and the question is best avoided at this stage.

The meaning of a negative, zero or rational index is summarised as follows.

For \(a\) positive and \(m,n\) positive integers,

\begin{alignat*}{2} a^0 &= 1 &\qquad\qquad a^{-n} &= \frac{1}{a^n}\\ a^{\frac{1}{2}} &= \sqrt{a} & a^{\frac{1}{n}} &= \sqrt[n]{a}\\ a^{\frac{m}{n}} &= \sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m. \end{alignat*}#### Example

Simplify

1 a | \(3x^3y^6 \times 5x^4yz^3\) | b | \(64a^4b^7c\div 4a^3b^2c\) |

2 a | \(27^{\frac{4}{3}}\) | b | \(17^0\times 25^{-\frac{1}{2}}\). |

#### Solution

1 a | \(3x^3y^6 \times 5x^4yz^3=15x^7y^7z^3\) | b | \(64a^4b^7c\div 4a^3b^2c=16ab^5\) |

2 a | \(27^{\frac{4}{3}}=(\sqrt[3]{27})^4=81\) | b | \(17^0\times 25^{-\frac{1}{2}}=\dfrac{1}{5}\). |