Linear Equations

Triple Brackets

Remember what we did for double brackets?

Well, all you need to do is to solve it like you would double brackets and ignore the other bracket until you have 2 brackets.

For example, $(1 + a)(2 + b)(3 * a)$

$\begin{aligned} (1 + a)(2 + b)(3 * a) &= (1 + a)(2 \color{red}{*3} + b \color{red}{*3} + b \color{red}{* a} + 2 \color{red}{* a}) = (1 + a)(6 + 3b + ba + 2a) \\ 6 \color{red}{* 1} + 3b \color{red}{*1} + ba \color{red}{*1} + 2a \color{red}{*1} + 6 \color{red}{*a} + 3b \color{red}{*a} + ba \color{red}{*a} + 2a \color{red}{*a} &=\\ 6 + 3b + ba + 2a + 6a + 3ba + ba^2 + 2a^2 &= 6 + 3b + 4ba + 8a + ba^2 + 2a^2 \end{aligned}$

You have to write a lot but it is fairly straightforward.

Factorising

Remember Highest Common Factors?

Well to factorise all you need to do is find the highest common factor of the expression and put it outside and all the other terms inside brackets.

$3x^2 + 6x = \color{red}{(3x)} (3x^2 \color{red}{(\div 3x)} + 6x \color{red}{(\div 3x)} ) = 3x(x^2+2)$

$10x^2 + 20x^3 = \color{red}{(5x^2)}(10x^2 \color{red}{(\div 5x^2)} + 20x^3 \color{red}{(\div 5x^2)}) = (5x^2 \color{red}{* 2})(2\color{red}{(\div 2)} + 4x\color{red}{(\div 2)}) = 10x^2(1 + 2x)$

You could draw a factor tree diagram then answer the question.

Difference of two squares rule

$a^2 - b^2 = (a+b)(a-b)$