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# Algebraic manipulations

## Multiplying out brackets

When we expand (multiply) two brackets, we need to multiply all the pairs of terms from one bracket and the other.

We want to multiply out the expression $$(\Tred{x+3})(\Tblue{x-5}).$$

We muliply all the pairs and add them $$(\Tred{x+3})(\Tblue{x-5}) = \Tred{x}\times\Tblue{x} + \Tred{3}\times\Tblue{x} + \Tred{x}\times(\Tblue{-5}) +\Tred{3}\times(\Tblue{-5}).$$

We can then simplify the expression on the right to get $$(\Tred{x+3})(\Tblue{x-5}) = x^2 + 3x - 5x - 15 = \Tviolet{x^2 -3x -15}$$

With practice, you can do all the operations in your head.

We give a few additional examples.

\begin{align*} (\Tred{x+1})(\Tblue{x-1})=& x^2 + x - x - 1 = \Tviolet{x^2 - 1}\\ (\Tred{x+1})(\Tblue{y-1})=&\Tviolet{xy + y - x - 1}\\ \Torange{x}(\Tred{x+1})(\Tblue{x-1})=&\Torange{x}(x^2 - 1) = \Tviolet{x^3 - x} \end{align*}

## Expanding expressions

Algebraic expressions can sometimes be complicated!

$$(3\Tred{x} - 2)(2\Tred{x}^2 + \Tred{x} - 1) + 2$$

The expansion of an algebraic expression removes all the brackets and collects like terms. It is then quicker to evaluate the expression.

• Remove the brackets one by one by to find the products: \begin{align*} (\Tred{a}+\Tblue{b})(c+d) &= \Tred{a}(c+d) + \Tblue{b}(c+d) \\ &= \Tred{a}c + \Tred{a}d + \Tblue{b}c + \Tblue{b}d \end{align*}
• Get the signs right: \begin{align*} \Tred{\mathbf{+}}\times \Tred{\mathbf{+}} = \Tred{\mathbf{+}} &\qquad \Tred{\mathbf{+}}\times \Tblue{\mathbf{-}} = \Torange{\mathbf{-}}\\ \Tblue{\mathbf{-}}\times \Tblue{\mathbf{-}} = \Tred{\mathbf{+}} &\qquad \Tblue{\mathbf{-}}\times \Tred{\mathbf{+}} = \Tblue{\mathbf{-}} \end{align*}
• Rewrite products of variables using powers: $$a\times a = a^2,\quad a^2\times a = a\times a^2 = a^3$$
• Collect all the like terms.
$$\Tblue{2}(y-1) - 1= \Tblue{2}y + \Tblue{2}(-1) - 1 = 2y - 2 - 1 = 2y -3$$ \begin{align*} 2(\Tred{t}+1)(\Tred{t}-2) - 1 &= (2\Tred{t}+2)(\Tred{t}-2) - 1 \\ &= 2\Tred{t}\times \Tred{t} + 2\Tred{t} - 4\Tred{t} - 4 -1\\ &= 2\Tviolet{t^2} + (2-4)\Tred{t} + (-4-1) \\ &= 2\Tviolet{t^2} -2\Tred{t} -5 \\ \end{align*}

## Factorising expressions

Factorisation is the opposite of expansion. Factorisation involves finding the common factors of an expression and rewriting it as the product of simpler expressions.

The expression on the right is the factorisation of the expression on the left.

$$x^2 + x -2 = (x-1)(x+2)$$

To fully factorise an expression, you need to identify all the common factors. There is no overall rule.

$$3y+6 = \Tgreen{3}\times y + \Tgreen{3}\times 2 = \Tgreen{3}(y+2)$$ $$2a^2 -4a = \Tgreen{2a}\times a - \Tgreen{2a}\times 2 = \Tgreen{2a}(a-2)$$

Factorisation is very useful when you need to find the solutions to an equation!

## Methods of factorisation: inspection

When we factorise a quadratic expression, we need to put it "back into brackets".

If we want to factorise a quadratic expression $x^2+ax+b$, then we first look for a pair of numbers that multiply to give $b$, and add to give $a$.

We will try to factorise $x^2 - 7x +12$. We need numbers that multiply to give $12$, but that also add to give $-7$. This means that both the numbers we are looking for will be negative.

The pairs of negative numbers that multiply to give $12$ are $-1$ and $-12$, $-3$ and $-4$, and $-2$ and $-6$. The only pair that adds to give $-7$ is $-3$ and $-4$.

We now use these numbers to fill in the brackets. $$(x-3)(x-4) = x^2-7x+12$$

## Methods of factorisation: difference of two squares

Factorising an expression is simple if it is of the form $A^2 -B^2$. This is called the difference of two squares

$A$ and $B$ could be any algebraic expressions or numbers, so sometimes it is tricky to spot expressions of this kind.

Expressions of the form $A^2-B^2$ always factorise into the expression $(A+B)(A-B)$

1. Factorise $a^4-25$.

We spot that both $a^4$ and $25$ are squares. We take their square roots and fill in the difference of two squares formula.

$$a^4-25 = (a^2+5)(a^2-5)$$

2. Factorise $2xy^6-72x$.

At first these do not look like squares, so it seems that we cannot factorise using the difference of two squares.

But if we try factoring out the common factor of $2x$, the task becomes easier. \begin{align*} 2xy^6-72x = 2x(y^6-36) &=2x((y^3)^2-6^2)\\ &= 2x(y^3+6)(y^3-6) \end{align*}

## Factorisation of quadratic expressions

A quadratic expression is an expression involving a variable and its square.

$\Tred{t^2}-1$, $\Tred{x^2}+\Tred{x}-1$, $\Tred{y}-3$ are quadratic expressions.

$\Torange{t^3}-\Tred{t}$ and $\Torange{\sqrt{x}}+\Tred{x}-1$ are not quadratic expressions.

Factorisation and simplification of quadratic expressions use the following formulae.

\begin{align*} &(a+b)^2 = a^2 + 2ab + b^2,\\ &(a-b)^2 = a^2 - 2ab + b^2,\\ &(a+b)(a-b) = a^2 - b^2. \end{align*}

Here are applications of the rules:

$$x^2 - 1 = (x-1)(x+1),\quad (2x-3)^2 = 4x^2 -12x +9$$

The first rule can be used to compute squares of numbers $\ge 10$

\begin{align*} 12^2 = (\Tblue{10}+\Tgreen{2})^2 & = \Tblue{10}^2 + 2 \times \Tblue{10}\times \Tgreen{2} + \Tgreen{2}^2\\ & = 100 + 40 + 4 = 144\\ 18^2 = (\Tblue{20}-\Tgreen{2})^2 & = \Tblue{20}^2 - 2\times \Tblue{20}\times \Tgreen{2} + \Tgreen{2}^2\\ & = 400 - 80 + 4 = 324 \end{align*}

## Proof of the common quadratic expressions

These quadratic factorisations can be proved directly using expansion.

\begin{align*} \Tblue{(a+b)^2} &= (a+b)(a+b)\\ &= a^2 + ab + ba + b^2\\ &\Tblue{= a^2 + 2ab + b^2}\\\\ \Tblue{(a-b)^2} &= (a-b)(a-b) \\&= a^2 - ab - ba + b^2\\ &\Tblue{= a^2 - 2ab + b^2}\\\\ \Tblue{(a+b)(a-b)} & = a^2 - ab + ba - b^2 \\ &\Tblue{= a^2 - b^2} \end{align*}

## Algebraic fractions

An algebraic fraction is a fraction involving algebraic expressions. It is a fraction with letters and numbers, and is itself an algebraic expression.

Examples of algebraic fractions are $$\frac{\Tred{x}^2-\Tred{x}+1}{\Tred{x}+3},\quad\frac{1}{\sqrt{\Tred{y}} - 1}$$

An algebraic fraction works just like any other algebraic expression. You replace the variable by a value, and work out the answer.

The algebraic fraction $\displaystyle\frac{\Tred{x}-1}{\Tred{x}^2+1}$ evaluated when $\Tred{x} = 5$ is equal to $$\frac{\Tred{5}-1}{\Tred{5}^2+1} = \frac{4}{26} = \frac{2}{13}.$$

## Multiplying algebraic fractions

Multiplication and division of algebraic fractions follow the same rules as for numerical fractions. The numerator and the denominator are generally simplified.

• Multiplication: Numerators and denominators multiply. $$\frac{\Tblue{2x+1}}{\Tgreen{x-1}}\times\frac{\Tblue{x-2}}{\Tgreen{x+1}} = \frac{\Tblue{(2x+1)(x-2)}}{\Tgreen{(x-1)(x+1)}} = \frac{2x^2-3x-2}{x^2-1}$$
• Reciprocal: Numerator and denominator are swapped. $$\Big(\frac{\Tblue{x-2}}{\Tgreen{x^2+3}}\Big)^{-1} = \frac{\Tgreen{x^2+3}}{\Tblue{x-2}}$$
• Division: The numerator and denominator of the divisor are swapped and multiplied to the denominator and the numerator of the dividend. $$\frac{\Tblue{2t+1}}{\Tgreen{t-1}}\div\frac{\Tblue{t-2}}{\Tgreen{t+1}} = \frac{\Tblue{(2t+1)}\Tgreen{(t+1)}}{\Tgreen{(t-1)}\Tblue{(t-2)}} = \frac{2t^2+3t + 1}{t^2 -3t +2}$$

## Adding and subtracting algebraic fractions

Addition and subtraction of algebraic fractions follows the same rules as numerical fractions. The fractions are converted so that they have the same denominator and the numerators are then added or subtracted.

• Addition \begin{align*} \frac{1}{\Tred{a}}+\frac{2}{\Torange{a+1}} &= \frac{\Torange{(a+1)}}{\Tred{a}\Torange{(a+1)}} + \frac{2\Tred{a}}{\Tred{a}\Torange{(a+1)}}\\ &= \frac{\Torange{(a+1)} + 2\Tred{a}}{\Tred{a}\Torange{(a+1)}}\\ &= \frac{3a+1}{a^2+a} \end{align*}
• Subtraction \begin{align*} \frac{\Tblue{2x+1}}{\Tred{x-1}}-\frac{\Tgreen{x-2}}{\Torange{x+1}} &= \frac{\Tblue{(2x+1)}\Torange{(x+1)}}{\Tred{(x-1)}\Torange{(x+1)}} - \frac{\Tred{(x-1)}\Tgreen{(x-2)}}{\Tred{(x-1)}\Torange{(x+1)}}\\ &= \frac{\Tblue{(2x+1)}\Torange{(x+1)}-\Tred{(x-1)}\Tgreen{(x-2)}}{\Tred{(x-1)}\Torange{(x+1)}}\\ &= \frac{(2x^2+3x+1)-(x^2-3x+2)}{x^2-1}\\ &= \frac{x^2+6x+3}{x^2-1} \end{align*}