Partial fractions · Teacher reference · answers included

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Teacher reference: contains answers, conditions and proofs.

13 Partial fractions · g10-13-01 · 5–10 min · TEACHER · ANSWERS

Two denominators, one identity

How can substituting a forbidden value still find a coefficient?

Predict: Decompose (mx+n)/[(x−a)(x−b)]; predict what remains in the numerator identity at x=a.

Compare	Computed observation
Distinct linear factors First root a = 0; Second root b = -1; Numerator coefficient m = 0; Numerator constant n = 1	Polynomial part=0; A=1, B=-1; x≠0,-1
A numerator cancellation First root a = 1; Second root b = -1; Numerator coefficient m = 1; Numerator constant n = -1	Polynomial part=0; A=0, B=1; x≠1,-1
Repeated factors need a new form First root a = 1; Second root b = 1; Numerator coefficient m = 1; Numerator constant n = 2	This distinct-root model requires a≠b. Use the repeated-factor activity instead.

Conclusion: After clearing denominators, mx+n=A(x−b)+B(x−a) is a polynomial identity. Substitution there is valid even at a or b. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: a≠b; the original fractions require x≠a,b.

Justification: Putting x=a gives A=(ma+n)/(a−b); x=b gives B=(mb+n)/(b−a). These coefficients also satisfy A+B=m and −Ab−Ba=n.

Ask: Point explicitly to the expression in which the substitution is performed.

Watch for: Do not substitute a forbidden value into the original rational expression.

Extension: Recover A and B by coefficient comparison instead of special-value substitution.

Exit: Decompose 1/[x(x+1)].

1/x−1/(x+1), retaining x≠0,−1.

Before: Rational domains and coefficient identities · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

13 Partial fractions · g10-13-02 · 5–10 min · TEACHER · ANSWERS

Why include every repeated power?

Can one term represent every numerator over (x−a)²?

Predict: Compare A/(x−a)+B/(x−a)² with the incomplete proposal B/(x−a)².

Compare	Computed observation
Both powers are needed Repeated root a = 1; Numerator coefficient m = 1; Numerator constant n = 0	Repeated-factor coefficients: A=1, B=1; x≠1
First coefficient is zero Repeated root a = 1; Numerator coefficient m = 0; Numerator constant n = 2	Repeated-factor coefficients: A=0, B=2; x≠1
Second coefficient is zero Repeated root a = 1; Numerator coefficient m = 1; Numerator constant n = -1	Repeated-factor coefficients: A=1, B=0; x≠1

Conclusion: For numerator mx+n, coefficient comparison gives A=m and B=ma+n. Dropping A removes an entire variable term unless m=0. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: x≠a. The repeated-root form is used even if a numerator cancellation is possible.

Justification: Clearing denominators gives mx+n=A(x−a)+B. The x coefficient forces A=m; the constant then forces B=n+ma.

Ask: Show the incorrect graph as well as the corrected graph and ask for the missing coefficient.

Watch for: One numerical match does not prove an identity of rational expressions.

Extension: Generalise the template to denominator (x−a)³.

Exit: Decompose x/(x−1)².

1/(x−1)+1/(x−1)², with x≠1.

Before: Coefficient comparison · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

13 Partial fractions · g10-13-03 · 5–10 min · TEACHER · ANSWERS

Divide before decomposing

Where does the polynomial part come from?

Predict: Compare the degrees of x²+px+q and (x−a)(x−b); predict the missing polynomial term.

Compare	Computed observation
A nonzero remainder Numerator coefficient p = 2; Numerator constant q = 3; First root a = 1; Second root b = -1	Polynomial part=1; A=3, B=-1; x≠1,-1
Zero remainder, two holes Numerator coefficient p = 0; Numerator constant q = -1; First root a = 1; Second root b = -1	Polynomial part=1; A=0, B=0; x≠1,-1
A cancelled factor Numerator coefficient p = 0; Numerator constant q = 0; First root a = 0; Second root b = 1	Polynomial part=1; A=0, B=1; x≠0,1

Conclusion: The quotient is 1. Only the lower-degree remainder is decomposed into simple fractions. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: a≠b and x≠a,b. The numerator and denominator are monic quadratics.

Justification: Subtracting the denominator leaves (p+a+b)x+(q−ab). Thus the original ratio is 1 plus that remainder ratio; the distinct-root coefficient identities then verify the decomposition.

Ask: Ask what remains when x becomes large and the proper fractions become small.

Watch for: A sum of these proper fractions cannot reproduce the nonzero limiting polynomial part.

Extension: Explain the horizontal limiting value without assuming the graph proves it.

Exit: Rewrite x²/[x(x−1)] as a polynomial plus simple fractions.

1+1/(x−1), retaining x≠0,1.

Before: Polynomial division and simple fractions · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

13 Partial fractions · g10-13-04 · 5–10 min · TEACHER · ANSWERS

Does cancellation repair a hole?

Are equal values on a shared domain the same as equal domains?

Predict: Compare (x−a)(x+c)/[(x−a)(x−b)] with (x+c)/(x−b), especially at x=a.

Compare	Computed observation
A normal allowed input Cancelled root a = 1; Remaining denominator root b = -1; Numerator shift c = 2; Probe input x = 0	Original excludes 1,-1; the written simplified formula excludes -1.
The cancelled root Cancelled root a = 1; Remaining denominator root b = -1; Numerator shift c = 2; Probe input x = 1	Original excludes 1,-1; the written simplified formula excludes -1.
Both singularities are removable Cancelled root a = 1; Remaining denominator root b = -1; Numerator shift c = 1; Probe input x = -1	Original excludes 1,-1; the written simplified formula excludes -1.

Conclusion: Cancellation preserves values only at original allowed inputs. The simplified formula may extend the domain, but the original hole remains excluded. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: a≠b. Original domain excludes a,b; the written simplified formula still excludes b.

Justification: The factor x−a is nonzero on the original domain, so division by it is legal there. At x=a the original denominator vanishes regardless of the finite simplified value.

Ask: Ask pupils to write both domains before comparing the plotted curves.

Watch for: A removable discontinuity is not permission to redefine the original function silently.

Extension: Choose c=−b and classify what remains at both excluded inputs.

Exit: Is (x²−1)/(x−1) the same function as x+1 on R?

No; the original excludes 1, although the values agree elsewhere.

Before: Domains, cancellation and function equality · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.