P3 1.1 - Simplifying rational expressions

P3 1.1 - Simplifying rational expressions

Simplify algebraic fractions by factorising, combining them and dividing polynomials. Keep track of excluded inputs so that a shorter expression still describes the original fraction accurately.

Factorising and cancelling with exclusions

Can two fractions have the same simplified rule but different allowed inputs? Cancelling can hide a forbidden value, so simplification needs both algebra and a record of the original denominator.

For numbers, 6/9=(3×2)/(3×3)=2/36/9=(3\times2)/(3\times3)=2/3. We cancel a common factor, because we divide the whole numerator and denominator by the same nonzero number. Algebra works in the same way.

Rational expression

A rational expression is a quotient of polynomials, such as (x2−1)/(x+2)(x^2-1)/(x+2). It is defined only where its denominator is nonzero.

Factorise before cancelling. If x≠1x\ne1,

x2−1x−1=(x−1)(x+1)x−1=x+1.\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1.

The rule x+1x+1 has a value at x=1x=1, but the original fraction does not. The equality therefore retains x≠1x\ne1.

Worked example 1: retaining both exclusions

Question. Simplify x2−9x2+x−6\dfrac{x^2-9}{x^2+x-6}, stating its excluded inputs.

Plan. Factorise both quadratics, identify the denominator's zeros, then cancel the common factor.

x2+x−6=(x+3)(x−2)product −6, sum 1x2−9x2+x−6=(x−3)(x+3)(x+3)(x−2)difference of squares=x−3x−2cancel x+3\begin{aligned} x^2+x-6&=(x+3)(x-2) &&\text{product }-6\text{, sum }1\\ \frac{x^2-9}{x^2+x-6}&=\frac{(x-3)(x+3)}{(x+3)(x-2)} &&\text{difference of squares}\\ &=\frac{x-3}{x-2} &&\text{cancel }x+3 \end{aligned}

Answer. x−3x−2\dfrac{x-3}{x-2}, with x≠−3,2x\ne-3,2.

Check. At x=0x=0, the original gives (−9)/(−6)=3/2(-9)/(-6)=3/2; the simplified form gives (−3)/(−2)=3/2(-3)/(-2)=3/2. At x=−3x=-3, the original is still undefined.

Change only the numerator from x2−9x^2-9 to x2−4x^2-4. A different factor cancels, but the original denominator still forbids the same two inputs.

Worked example 2: changing which factor cancels

Question. Simplify x2−4x2+x−6\dfrac{x^2-4}{x^2+x-6}, stating its excluded inputs.

Plan. Use the same denominator factorisation and cancel only a whole common factor.

x2−4x2+x−6=(x−2)(x+2)(x+3)(x−2)=x+2x+3cancel x−2\begin{aligned} \frac{x^2-4}{x^2+x-6}&=\frac{(x-2)(x+2)}{(x+3)(x-2)}\\ &=\frac{x+2}{x+3} &&\text{cancel }x-2 \end{aligned}

Answer. x+2x+3\dfrac{x+2}{x+3}, with x≠−3,2x\ne-3,2.

Check. At x=0x=0, both forms give 2/32/3. The simplified rule gives 4/54/5 at x=2x=2, but that input remains excluded from the original fraction.

Products can be treated by factorising all numerators and denominators together. To divide by a rational expression, multiply by its reciprocal; the divisor itself must also be nonzero. For example,

xx+1÷x−2x+1=xx+1x+1x−2=xx−2,x≠−1,2.\frac{x}{x+1}\div\frac{x-2}{x+1} =\frac{x}{x+1}\frac{x+1}{x-2} =\frac{x}{x-2},\qquad x\ne-1,2.

Here −1-1 makes an original denominator zero, while 22 makes the divisor zero.

Watch out

Do not cancel terms across a sum: (x+3)/x≠3(x+3)/x\ne3. You can split it as 1+3/x1+3/x, for x≠0x\ne0. Cancellation removes factors that multiply the entire numerator and denominator.

Factorise both quadratics before shortening the next fraction.

Combining fractions

A common denominator lets you add or subtract whole numerators without changing the fractions' values.

For 1/6+1/41/6+1/4, the denominator 1212 works because it contains the necessary factors of both denominators. Likewise, factorising algebraic denominators reveals the factors that the common denominator must contain.

Combining rational expressions

For nonzero BB and DD,

AB±CD=AD±BCBD.\frac{A}{B}\pm\frac{C}{D}=\frac{AD\pm BC}{BD}.

If denominators share factors, using their lowest common multiple usually keeps the algebra shorter. Retain all original exclusions.

Not in the formula booklet: learn it.

In 1/[(x+3)(x−2)]+1/(x+3)1/[(x+3)(x-2)]+1/(x+3), the first denominator already contains both required factors. Only the second numerator needs multiplying by x−2x-2. Multiplying both denominators blindly would work but creates an unnecessary extra factor.

Worked example 3: a shared denominator factor

Question. Simplify 5xx2+x−6+xx+3\dfrac{5x}{x^2+x-6}+\dfrac{x}{x+3}, stating the exclusions.

Plan. Factorise the quadratic denominator and combine over (x+3)(x−2)(x+3)(x-2).

5xx2+x−6+xx+3=5x+x(x−2)(x+3)(x−2)same denominator=x2+3x(x+3)(x−2)expand and collect=x(x+3)(x+3)(x−2)=xx−2factorise, then cancel\begin{aligned} \frac{5x}{x^2+x-6}+\frac{x}{x+3} &=\frac{5x+x(x-2)}{(x+3)(x-2)} &&\text{same denominator}\\ &=\frac{x^2+3x}{(x+3)(x-2)} &&\text{expand and collect}\\ &=\frac{x(x+3)}{(x+3)(x-2)}=\frac{x}{x-2} &&\text{factorise, then cancel} \end{aligned}

Answer. xx−2\dfrac{x}{x-2}, with x≠−3,2x\ne-3,2.

Check. At x=1x=1, the original is −5/4+1/4=−1-5/4+1/4=-1, agreeing with 1/(1−2)=−11/(1-2)=-1.

Change just the plus sign to a minus. The minus must act on the whole second numerator, and the cancellation from Example 3 may disappear.

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