Simplifying algebraic fractions
Factorise the top and the bottom of an algebraic fraction, cancel the common bracket, and state the values of x the fraction cannot take.
What a learner can do afterwards
- Simplify algebraic fractions by factorising both parts, including (x² - 9)/(x² + 7x + 12)
- Cancel whole factors only, never single terms inside a sum
- State the excluded values that would make the denominator zero
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You have factorised expressions and solved zero-factor cases. Use those skills to simplify algebraic fractions. The numerator is the top expression; the denominator is the bottom expression. Factorise both parts, record every input that makes the original denominator zero, then cancel shared whole factors. Keep every original restriction.
Cancellation divides the entire numerator and denominator by the same nonzero factor. In (x² + 7x)/(x + 7), x is a factor of the entire numerator because x² + 7x = x(x + 7). But x is not a factor of the entire denominator x + 7. The shared factor is x + 7. Cancelling it gives x, with x ≠ -7. Factorisation reveals factors that were already present.
Deleting additive terms is different from dividing whole factors. Deleting x from (x + 8)/(x + 2) leaves 8/2 = 4. Check the allowed input x = 2: the original gives 10/4 = 2.5, not 4. One disagreement at an allowed input disproves equivalence.
A difference of squares uses x² - a² = (x - a)(x + a). For (4x² - 196)/(4x + 28), the top is 4(x - 7)(x + 7) and the bottom is 4(x + 7). The original bottom excludes -7. Cancel the numerical factor 4 and the shared bracket x + 7. The result is x - 7, with x ≠ -7.
For x² + bx + c, find a signed pair with sum b and product c. For ax² + bx + c, split bx using a pair with sum b and product ac, then group. For 2x² + 9x + 4, use 8 and 1: 2x(x + 4) + (x + 4) = (2x + 1)(x + 4). Dividing by x + 4 leaves 2x + 1, with x ≠ -4.
For (x² + 7x + 6)/(x² - 1), the pair 1 and 6 gives top factors (x + 1)(x + 6). The bottom is (x - 1)(x + 1). Exclude -1 and 1. Cancel x + 1 to get (x + 6)/(x - 1), keeping both exclusions. You must still record the exclusion from the cancelled factor. If no whole factor is shared, nothing cancels.
Find exclusions from the original denominator alone. Set each factor equal to zero and solve its case. A bottom x(x - 8) excludes 0 and 8. A repeated factor (x + 9)² excludes -9 once, since exclusions are distinct inputs. A numerator being zero does not exclude an input when the denominator is nonzero. For a requested product, multiply the distinct excluded values with their signs.
Opposite factors leave a minus sign: 7 - x = -(x - 7). For (x² - 9x + 14)/(7 - x), the top is (x - 2)(x - 7). Exclude x = 7. Rewrite the bottom as -(x - 7), then cancel x - 7. The result is -(x - 2) = 2 - x, with x ≠ 7.
Factor common numbers and powers of x from entire expressions. For x ≠ 0, dividing 15x and 20x by 5x gives 15x/(20x) = 3/4. A cubic can hide a quadratic factor: x³ - 64x = x(x² - 64). Thus (x³ - 64x)/(x² - 64) becomes x, with x ≠ -8 and x ≠ 8. Both original denominator factors still restrict the input.
To reconstruct a polynomial numerator, multiply the simplified expression by the original denominator. If the simplified expression is x + 2 and the original denominator is x - 3, the numerator is (x + 2)(x - 3). Expanding gives x² - 3x + 2x - 6 = x² - x - 6. The original fraction has x ≠ 3.
Cancel one copy of a factor at a time. In (x + 6)/(x + 6)², the denominator contains two copies of x + 6. Only one cancels with the numerator, leaving 1/(x + 6), with x ≠ -6. When the entire numerator cancels, its remaining factor is 1.
To evaluate, first check the original restrictions, then substitute into the simplified form and divide. The fraction (x² - 36)/(x² + 8x + 12) simplifies to (x - 6)/(x + 2), with x ≠ -6 and -2. Input 3 is allowed. Its value is (3 - 6)/(3 + 2) = -3/5 = -0.6.
Factorise the top and the bottom, cancel only whole factors, and state the excluded values from the original denominator.
This lesson, its practice questions and its flashcards were written for this dataset.
The additional questions, readings and flashcards were written for this dataset. The two linked rational-expression videos are by Khan Academy.
More to read
Division of whole factors
Cancelling a factor means dividing the entire numerator and denominator by that same nonzero factor. For example, dividing both parts of 18/30 by 6 gives 3/5.
In (x² + 10x)/(x + 10), the numerator is x(x + 10). The whole bracket x + 10 cancels, leaving x for inputs other than -10.
The factor x also belongs to the entire numerator, but it does not belong to the entire denominator x + 10. That is why x itself cannot be cancelled across this fraction. Factorisation helps identify the factors of each whole expression.
Keeping the original exclusions
Keep a record of the original denominator before simplifying. Its zeros are the excluded inputs, even if some factors later disappear.
For (x² - 100)/(x² + 13x + 30), factor the top as (x - 10)(x + 10) and the bottom as (x + 3)(x + 10). The original denominator excludes -3 and -10. Cancelling x + 10 leaves (x - 10)/(x + 3), with both exclusions retained.
The remaining denominator still displays the exclusion -3. The separate restriction must retain -10. The original and simplified forms agree at all allowed inputs.
Checking a sign after cancellation
Reversing a subtraction changes its sign. The expressions x - 11 and 11 - x are opposites because 11 - x = -(x - 11).
Consider ((x - 11)(x + 4))/(11 - x). Rewrite the denominator as -(x - 11). For x ≠ 11, cancel x - 11 and retain the minus sign. The result is -(x + 4), or -x - 4, with x ≠ 11.
Check an allowed input. At x = 0, the original gives (-11·4)/11 = -4. The result -x - 4 gives -4 too. Dropping the minus sign would give 4 instead.
Factor copies and substitution
A squared bracket contains two factor copies. In (x + 12)/(x + 12)², one denominator copy cancels with the numerator. The other remains, giving 1/(x + 12), with x ≠ -12.
Before substituting an input into any simplified expression, check it against the original denominator. A cancelled factor may hide a restriction.
A zero numerator is different from a zero denominator. At x = 7, (x - 7)/(x + 1) has value 0/8 = 0. At x = -1, its denominator is zero, so the expression is undefined. Division by zero is undefined even when the numerator is also zero.
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