Simplifying a rational function by factorising: the domain stays that of the original
Mathematics Prep 3 Term 1, Unit 2, Lesson 3
GoalSimplify a rational fraction by cancelling common factors, and keep the domain of the original function.
New words
cancel · divide the top and the bottom by the same non-zero factorcommon factor · a factor that appears in both numerator and denominatorsimplest form · a fraction in which numerator and denominator have no common factorhole · the missing point of the graph at an x outside the domain
Explanation
To simplify a fraction, factorise the numerator and the denominator, then cancel the factors they share. Cancel factors, never terms: (x + 2) ÷ (x + 3) is not 2 ÷ 3. Before cancelling, write down the domain, because the simplified fraction can be defined at a place where the original function is not. For f(x) = (x² − 9) ÷ (x − 3), f(3) is undefined, yet after cancelling we get x + 3, which has a value at 3. The function f is still undefined at 3: its graph is the line y = x + 3 with a hole at the point (3, 6). So the domain is the one found from the original denominator, R − {3}. We write a small ring on the graph for the hole.
Worked examples
Simplify f(x) = (x² − x − 12) ÷ (x² − 9) and give its domain. Factorise: numerator (x − 4)(x + 3), denominator (x − 3)(x + 3). Domain from the original denominator: R − {3, −3}. Cancel (x + 3): f(x) = (x − 4) ÷ (x − 3). Check at x = 1: original (1 − 1 − 12) ÷ (1 − 9) = −12 ÷ (−8) = 3/2, simplified (1 − 4) ÷ (1 − 3) = 3/2 ✓.
In the example above, is f(−3) defined? The simplified rule gives (−3 − 4) ÷ (−3 − 3) = 7/6, but −3 is not in the domain of the original function, so f(−3) does not exist. The graph has a hole at x = −3. Only the original domain counts.
x
(x + 2) ÷ (x + 3)
2 ÷ 3
1
3/4
2/3
0
2/3
2/3
2
4/5
2/3
Is (x + 2) ÷ (x + 3) equal to 2 ÷ 3? Test x = 1: (1 + 2) ÷ (1 + 3) = 3/4, but 2/3 is different. We may cancel only factors that multiply the whole top and the whole bottom, never single terms of a sum.
Class activity
Catch the hole. Use a calculator to find f(x) = (x² − 9) ÷ (x − 3) for x = 2.9, 2.99, 2.999 and 3.001. What number do the answers approach? Why can the graph not have a point at x = 3?
Practice
1
Simplify (x² − 4) ÷ (x² + 4x + 4) and give the domain.
Simplify (x² − 5x) ÷ (x² − 25) and give the domain.
x(x − 5) ÷ ((x − 5)(x + 5)) = x ÷ (x + 5). Domain R − {5, −5}.
3
Simplify f(x) = (x² − 1) ÷ (x² − x). Then find f(2), and say whether f(1) exists.
(x − 1)(x + 1) ÷ (x(x − 1)) = (x + 1) ÷ x, domain R − {0, 1}. f(2) = 3/2. f(1) does not exist because 1 is not in the domain, even though (1 + 1) ÷ 1 = 2.
4
Is (x + 2) ÷ (x + 3) equal to 2 ÷ 3? Explain with x = 0 and x = 1.
At x = 0 both are 2/3, but at x = 1 the first is 3/4. One agreeing value is not enough; they are not equal.
5
Simplify (x − 3) ÷ (3 − x) and state the domain.
3 − x = −(x − 3), so the fraction is −1. Domain R − {3}.
