UPSKILL MATH PLUS

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Learn more*Excluded Value*

A value that makes a rational expression (in its lowest form) undefined is called an Excluded value.

Suppose the rational expression \(\frac{p(x)}{q(x)}\) is in its lowest form, then the value for which the expression becomes undefined is said to be its excluded value.

**:**

*Working rule to find the excluded value of a rational number***Step 1**: Simplify or factorise the numerator \(p(x)\) and the denominator \(p(x)\).

**Step 2**: Cancel out the common factors in the numerator and the denominator.

**Step 3**: Equate the lowest form of the denominator \(q(x)\) to zero.

**Step 4**: Thus, the obtained value for which the denominator becomes zero is the excluded value of that rational number.

Example:

Find the excluded value of the expression \(\frac{x^2 + 5x + 6}{(x + 2)(x - 5)}\).

**Solution**:

*: Factorise the numerator \(x^2 + 5x + 6\) by splitting the middle term.*

**Step 1**\(x^2 + 5x + 6\) \(=\) \(x^2 + 2x + 3x + 6\)

\(=\) \(x (x + 2) + 3 (x + 2)\)

\(=\) \((x + 2)(x + 3)\)

*: Rewrite the expression and cancel out the common factors.*

**Step 2**\(\frac{x^2 + 5x + 6}{(x + 2))(x - 5)}\) \(=\) \(\frac{(x + 2)(x + 3)}{(x + 2))(x - 5)}\)

\(=\) $\frac{\overline{)\left(x+2\right)}\left(x+3\right)}{\overline{)\left(x+2\right)}\left(x-5\right)}$

\(=\) \(\frac{x + 3}{x - 5}\)

*: Equate the lowest form of the denominator to zero.*

**Step 3**\(x - 5\) \(=\) \(0\)

Add \(5\) on both sides of the equation.

\(x - 5 + 5\) \(=\) \(0 + 5\)

\(\Rightarrow\) \(x\) \(=\) \(5\)

**: Write the excluded value.**

*Step 4*The rational expression \(\frac{x^2 + 5x + 6}{(x + 2)(x - 5)}\) is undefined when \(x\) \(=\) \(5\).

That is \(\frac{x^2 + 5x + 6}{0}\) \(=\) not defined, when \(x\) \(=\) \(5\).

Therefore, \(x\) \(=\) \(5\) is called an excluded value for the rational expression \(\frac{x^2 + 5x + 6}{(x + 2)(x - 5)}\).