When a factor is repeated in a denominator, the LCD must contain that factor to its highest power, and no higher. **Choosing the LCD** **Denominators** | **Highest power of each factor** | **LCD** (x+2)2 and x+2 | (x+2)2 | (x+2)2 3(x-1) and (x-1)2 | 3, (x-1)2 | 3(x-1)2 x2-6x+9 and x-3 | (x-3)2 | (x-3)2 Why: (x+2)2=(x+2)(x+2) is already a multiple of x+2, just as 9 is already a multiple of 3. Multiplying the denominators to get (x+2)3 gives a common denominator that is larger than needed. Method - Factorise each denominator and write repeated factors as powers. - Take each factor to its highest power to form the LCD. - Multiply each numerator by whatever its denominator is missing. - Combine, expand, collect and state the restrictions. Example Express x+5x2-6x+9- 1x-3 as a single fraction. x+5x2-6x+9- 1x-3= x+5(x-3)2- x-3(x-3)2 = x+5-(x-3)(x-3)2 = x+5-x+3(x-3)2 = 8(x-3)2, x 3 Check: the second numerator must be multiplied by x-3; leaving it as 1 is a common error. A reversed single bracket changes sign: 42-x= 4-(x-2)=- 4x-2. A squared reversed bracket does not: (2-x)2=(x-2)2, because squaring removes the minus sign.

Adding and subtracting with repeated factors

Secondary 2

When a factor is repeated in a denominator, the LCD must contain that factor to its highest power, and no higher.

Choosing the LCD

Denominators	Highest power of each factor	LCD
$(x+2)^2$ and $x+2$	$(x+2)^2$	$(x+2)^2$
$3(x-1)$ and $(x-1)^2$	$3$, $(x-1)^2$	$3(x-1)^2$
$x^2-6x+9$ and $x-3$	$(x-3)^2$	$(x-3)^2$

Why: $(x+2)^2=(x+2)(x+2)$ is already a multiple of $x+2$, just as $9$ is already a multiple of $3$. Multiplying the denominators to get $(x+2)^3$ gives a common denominator that is larger than needed.

Method

Factorise each denominator and write repeated factors as powers.
Take each factor to its highest power to form the LCD.
Multiply each numerator by whatever its denominator is missing.
Combine, expand, collect and state the restrictions.

Example

Express $\dfrac{x+5}{x^2-6x+9}-\dfrac{1}{x-3}$ as a single fraction.

$\begin{aligned}\dfrac{x+5}{x^2-6x+9}-\dfrac{1}{x-3}&=\dfrac{x+5}{(x-3)^2}-\dfrac{x-3}{(x-3)^2} \\ &=\dfrac{x+5-(x-3)}{(x-3)^2} \\ &=\dfrac{x+5-x+3}{(x-3)^2} \\ &=\dfrac{8}{(x-3)^2},\qquad x\ne3\end{aligned}$

Check: the second numerator must be multiplied by $x-3$; leaving it as $1$ is a common error. A reversed single bracket changes sign: $\dfrac{4}{2-x}=\dfrac{4}{-(x-2)}=-\dfrac{4}{x-2}$. A squared reversed bracket does not: $(2-x)^2=(x-2)^2$, because squaring removes the minus sign.

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Denominators	Highest power of each factor	LCD
\((x+2)^2\) and \(x+2\)	\((x+2)^2\)	\((x+2)^2\)
\(3(x-1)\) and \((x-1)^2\)	\(3\), \((x-1)^2\)	\(3(x-1)^2\)
\(x^2-6x+9\) and \(x-3\)	\((x-3)^2\)	\((x-3)^2\)