6.01g - Algebraic fractions

6.01g - Algebraic fractions

Algebraic fractions follow the same arithmetic rules as numerical fractions, but their denominators can rule out particular values of the variable. You will factorise before cancelling, multiply and divide fractions, and combine sums and differences over a common denominator. Keeping the original restrictions makes each simplification precise.

Cancelling factors and excluded values

If (x2−9)/(x+3)(x^2-9)/(x+3) simplifies to x−3x-3, can you now use x=−3x=-3?

Higher only. With numbers, (5×7)/5=7(5\times7)/5=7 because the common factor 55 is nonzero. With algebra, a common factor might be zero for a particular value. You can cancel it only while retaining the original denominator restriction.

Excluded value

An excluded value is a value for which the original expression is undefined. In a fraction, every value making a denominator zero is excluded. Cancelling a factor does not restore those values.

Factorise the whole numerator and denominator before cancelling. A factor multiplies an entire expression; a term is joined to another term by addition or subtraction. You may cancel common nonzero factors, not individual terms inside a sum.

Worked example 1: keeping a removed restriction

Question. Simplify (x2−9)/(x+3)(x^2-9)/(x+3), stating the excluded value.

Plan. Find the original restriction, then factorise the difference of squares.

x+3≠0⇒x≠−3original denominatorx2−9x+3=(x−3)(x+3)x+3factorise=x−3(x≠−3)cancel a nonzero factor\begin{aligned} x+3&\ne0\quad\Rightarrow\quad x\ne-3 && \text{original denominator} \\ \frac{x^2-9}{x+3} &=\frac{(x-3)(x+3)}{x+3} && \text{factorise} \\ &=x-3\quad(x\ne-3) && \text{cancel a nonzero factor} \end{aligned}

Answer. x−3x-3, with x≠−3x\ne-3.

Check. Multiplying x−3x-3 by x+3x+3 recovers x2−9x^2-9. At x=−3x=-3, the original would be 0/00/0, which is undefined; the simplified formula alone would conceal this.

Change only the sign in the denominator. A different factor cancels, and a different value is excluded.

Worked example 2: cancelling the other factor

Question. Simplify (x2−9)/(x−3)(x^2-9)/(x-3), stating the excluded value.

Plan. Exclude zero in the new denominator, then cancel its factor.

x−3≠0⇒x≠3new restrictionx2−9x−3=(x−3)(x+3)x−3factorise=x+3(x≠3)cancel\begin{aligned} x-3&\ne0\quad\Rightarrow\quad x\ne3 && \text{new restriction} \\ \frac{x^2-9}{x-3} &=\frac{(x-3)(x+3)}{x-3} && \text{factorise} \\ &=x+3\quad(x\ne3) && \text{cancel} \end{aligned}

Answer. x+3x+3, with x≠3x\ne3.

Check. (x+3)(x−3)=x2−9(x+3)(x-3)=x^2-9. At the allowed value x=−3x=-3, both the original and simplified expressions give zero; a zero numerator is allowed when the denominator is nonzero.

Quadratic denominators need factorising too. For example,

x2−9x2+5x+6=(x−3)(x+3)(x+2)(x+3)=x−3x+2,x≠−2,−3.\frac{x^2-9}{x^2+5x+6} =\frac{(x-3)(x+3)}{(x+2)(x+3)} =\frac{x-3}{x+2},\qquad x\ne-2,-3.

Both original denominator zeros stay excluded, even though only one remains visible. This answers the opening question: x=−3x=-3 is still forbidden.

Watch out

(x+3)/x≠3(x+3)/x\ne3. The xx in the numerator is a term, not a factor of the whole numerator. The valid rewrite is 1+3/x1+3/x, with x≠0x\ne0.

State the restriction before it disappears from view.

Products and quotients of algebraic fractions

Multiply algebraic fractions by multiplying their numerators and their denominators, factorising to spot any cancellation.

Division means multiplying by the reciprocal of the second fraction, just as (2/3)÷(4/5)=(2/3)×(5/4)(2/3)\div(4/5)=(2/3)\times(5/4). First check that both original fractions are defined. For division, also exclude any value making the divisor zero: you cannot divide by zero, even when neither original denominator is zero.

Worked example 3: a product with repeated factors

Question. Simplify x2−4x+3×x−2x+3\frac{x^2-4}{x+3}\times\frac{x-2}{x+3}, stating the excluded values.

Plan. Factorise the quadratic and multiply; cancel only if a factor is common to top and bottom.

x≠−3both denominatorsx2−4x+3×x−2x+3=(x−2)(x+2)(x−2)(x+3)(x+3)factor and multiply=(x−2)2(x+2)(x+3)2collect repeated factors\begin{aligned} x&\ne-3 && \text{both denominators} \\ \frac{x^2-4}{x+3}\times\frac{x-2}{x+3} &=\frac{(x-2)(x+2)(x-2)}{(x+3)(x+3)} && \text{factor and multiply} \\ &=\frac{(x-2)^2(x+2)}{(x+3)^2} && \text{collect repeated factors} \end{aligned}

Answer. (x−2)2(x+2)(x+3)2\frac{(x-2)^2(x+2)}{(x+3)^2}, with x≠−3x\ne-3.

Check. At x=0x=0, the original gives (−4/3)(−2/3)=8/9(-4/3)(-2/3)=8/9; the result gives 4×2/9=8/94\times2/9=8/9. The two numerator factors x−2x-2 do not cancel each other.

Change only multiplication to division. The reciprocal now puts matching factors on opposite sides of the fraction bar.

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