Rational expressions
A rational expression is p(x)/q(x) with q(x) ≠ 0. Factorise first, cancel common factors, and use the LCM of denominators to add or subtract.
Rational expressions and excluded values
A rational expression is one polynomial divided by another, like (x + 1)/(x − 3). The bottom polynomial is the denominator. Division by zero has no meaning. A value of x that makes the expression undefined is an excluded value. In (x + 1)/(x − 3), x = 3 is excluded. Your textbook uses the expression in its lowest form: cancel common factors first, then put the denominator equal to zero and solve. A cancelled factor still cannot be zero, so note it as a condition. For example, (x + 2)/[(x + 2)(x − 1)] = 1/(x − 1), with x ≠ −2. Its excluded value is x = 1.
The four operations
| Operation | Rule | Example |
|---|---|---|
| Lowest form | Factorise top and bottom, then cancel common factors | (x² − 9)/(x² + 5x + 6) = (x − 3)(x + 3)/[(x + 2)(x + 3)] = (x − 3)/(x + 2) |
| Multiply | Factorise, multiply tops and bottoms, then cancel | (x² − 4)/(x² + 3x) × (x + 3)/(x − 2) = (x + 2)/x |
| Divide | Multiply by the reciprocal (the second expression turned upside down) | (x² − 16)/(x + 2) ÷ (x + 4)/(x² + 2x) = x(x − 4) |
| Add or subtract | Rewrite both over the LCM of the denominators, then add or subtract the tops | 1/x + 1/(x + 1) = (2x + 1)/[x(x + 1)] |
Worked example 1: excluded values
Find the excluded values of (x + 5)/(x² − 7x + 12).
- Factorise the denominator: x² − 7x + 12 = (x − 3)(x − 4), since 3 × 4 = 12 and 3 + 4 = 7.
- Check the lowest form: the numerator x + 5 shares no factor with (x − 3)(x − 4), so nothing cancels.
- Excluded values make the denominator zero: (x − 3)(x − 4) = 0.
- So x = 3 or x = 4.
Excluded values: x = 3 and x = 4
Worked example 2: add two rational expressions
Simplify x/(x² − 1) + 1/(x + 1).
- Factorise the denominators: x² − 1 = (x − 1)(x + 1); the second is (x + 1).
- LCM of the denominators = (x − 1)(x + 1).
- Rewrite the second: 1/(x + 1) = (x − 1)/[(x − 1)(x + 1)].
- Add the numerators: [x + (x − 1)]/[(x − 1)(x + 1)] = (2x − 1)/[(x − 1)(x + 1)].
- No common factor is left to cancel.
(2x − 1)/(x² − 1)
Worked example 3: a cycling word problem
Priya cycles 12 km to her aunt's house at x km/h. She returns at (x + 2) km/h. Write her total travel time as one rational expression, and find it when x = 4.
- Formula: time = distance ÷ speed.
- Going: 12/x hours. Returning: 12/(x + 2) hours.
- Total = 12/x + 12/(x + 2). The LCM of the denominators is x(x + 2).
- Total = [12(x + 2) + 12x]/[x(x + 2)] = (24x + 24)/[x(x + 2)] = 24(x + 1)/[x(x + 2)].
- When x = 4: 24 × 5/(4 × 6) = 120/24 = 5.
- Check: 12/4 = 3 hours and 12/6 = 2 hours, and 3 + 2 = 5 hours.
Total time = 24(x + 1)/[x(x + 2)] hours, which is 5 hours when x = 4
Sharing gives rational expressions. If ₹ 600 of snacks is shared by x friends, each pays 600/x; with one more friend, each pays 600/(x + 1).
2 marks: “Find the excluded values of (2x + 1)/(x² − 9).” Put x² − 9 = 0: (x − 3)(x + 3) = 0, so x = 3 or x = −3.
2 marks: “Reduce to lowest form: (x² − 25)/(x² + 2x − 15).” Factorise: (x − 5)(x + 5)/[(x + 5)(x − 3)]. Cancel (x + 5): (x − 5)/(x − 3).
5 marks: “Simplify (x² − 4)/(x² + 3x + 2) × (x² − 1)/(x² − 3x + 2).” Factorise all four: (x − 2)(x + 2)/[(x + 1)(x + 2)] × (x − 1)(x + 1)/[(x − 1)(x − 2)]. Every factor cancels. Answer: 1.
Factorise first, always. Then: multiply and cancel; divide by flipping; add over the LCM.