HA4 - Algebraic fractions

HA4 - Algebraic fractions

Can cancelling a factor make a forbidden input allowed? HA4 uses factorisation and fraction arithmetic to simplify algebraic fractions, multiply or divide them, and combine sums or differences. Keep the original restrictions alongside the simplified expression so that it represents the same allowed values.

Simplifying

A fraction's denominator cannot be zero. Factorising first makes both the forbidden inputs and the common factors visible.

Algebraic fraction and restrictions

An algebraic fraction has an algebraic expression in its numerator or denominator, such as (x+2)/(x−3)(x+2)/(x-3). A restriction excludes an input where an original denominator is zero. Cancelling a common non-zero factor simplifies the expression but does not restore inputs excluded by the original fraction.

Cancellation divides numerator and denominator by the same non-zero factor. This is why you may cancel factors of a product, but not isolated terms of a sum. Record restrictions before cancelling, especially when a denominator factor will disappear.

Worked example 1: a cancelled denominator factor

Question. Simplify x2−9x2+x−6\frac{x^2-9}{x^2+x-6}, stating all restrictions.

Plan. Factor numerator and denominator, exclude denominator zeros, then cancel the common factor.

x2−9x2+x−6=(x−3)(x+3)(x+3)(x−2)=x−3x−2,x≠−3,2\begin{aligned} \frac{x^2-9}{x^2+x-6} &=\frac{(x-3)(x+3)}{(x+3)(x-2)}\\ &=\frac{x-3}{x-2},\qquad x\ne-3,2 \end{aligned}

Answer. (x−3)/(x−2)(x-3)/(x-2), with x≠−3,2x\ne-3,2.

Check. At x=0x=0, both give 3/23/2. At x=−3x=-3, the original denominator is zero, so that input stays excluded even though the shorter expression could be evaluated there. At x=3x=3, the original fraction is zero and is allowed.

Now a nonmonic numerator and a difference of squares hide a linear common factor.

Worked example 2: factorising both sides

Question. Simplify 2x2+7x+34x2−1\frac{2x^2+7x+3}{4x^2-1}, with restrictions.

Plan. Factor the quadratic numerator and the difference-of-squares denominator.

2x2+7x+34x2−1=(2x+1)(x+3)(2x−1)(2x+1)=x+32x−1,x≠12,−12\begin{aligned} \frac{2x^2+7x+3}{4x^2-1} &=\frac{(2x+1)(x+3)}{(2x-1)(2x+1)}\\ &=\frac{x+3}{2x-1},\qquad x\ne\frac12,-\frac12 \end{aligned}

Answer. (x+3)/(2x−1)(x+3)/(2x-1), excluding both 1/21/2 and −1/2-1/2.

Check. At x=1x=1, the original gives 12/3=412/3=4, and the simplified form gives 4/1=44/1=4. The cancelled factor 2x+12x+1 was zero at −1/2-1/2, so its restriction remains.

An input making only the numerator zero is usually allowed; an input making any original denominator zero is not.

Watch out

You cannot cancel xx from (x+3)/(x+5)(x+3)/(x+5) to get 3/53/5: the xx's are terms in sums, not whole factors. At x=1x=1, the original is 4/64/6, not 3/53/5.

Factor before cancelling in a fresh fraction.

Products and quotients

For multiplication, multiply numerators together and denominators together, cancelling common factors when helpful. For division, first make sure the divisor is defined and is not zero, then multiply by its reciprocal.

Multiplying and dividing fractions

AB×CD=ACBD,AB÷CD=ADBC.\frac AB\times\frac CD=\frac{AC}{BD},\qquad \frac AB\div\frac CD=\frac{AD}{BC}.

For the product, require B≠0,D≠0B\ne0,D\ne0. For the quotient, also require C≠0C\ne0, because the original divisor C/DC/D must not be zero. Not in the formula booklet: learn the method.

List those conditions from the original expression before turning the second fraction over. A restriction can come from the divisor's numerator as well as from either original denominator. The numerator of the first fraction may be zero if every denominator is non-zero and the divisor is non-zero.

Worked example 3: multiplying factorised fractions

Question. Simplify x2−4x2−1×x+1x+2\frac{x^2-4}{x^2-1}\times\frac{x+1}{x+2}, giving restrictions.

Plan. Exclude zeros of the original denominators, factor and cancel across the product.

Original denominators give x≠−1,1,−2x\ne-1,1,-2.

(x−2)(x+2)(x−1)(x+1)×x+1x+2=x−2x−1\begin{aligned} \frac{(x-2)(x+2)}{(x-1)(x+1)}\times\frac{x+1}{x+2} &=\frac{x-2}{x-1} \end{aligned}

Answer. (x−2)/(x−1)(x-2)/(x-1), with x≠−2,−1,1x\ne-2,-1,1.

Check. At x=0x=0, the original product is 4×12=24\times\frac12=2, and the shorter form is (−2)/(−1)=2(-2)/(-1)=2. The disappeared denominator factors still leave their excluded inputs.

Change the operation to division. The second fraction's zero now matters independently.

Continue this lesson.

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