Learning Objectives
Choose an efficient algebraic method from the structure of a problem.
Carry restrictions through simplification and equation solving.
Combine simultaneous equations with factor and remainder conditions.
Find and repair a missing case in an algebraic argument.
Prerequisites: Section 1B Lessons 01–06. This review uses only methods already introduced; all variables are real unless stated otherwise. All practice questions in this lesson are original, not official TMUA questions.
What This Lesson Is About
This review brings together algebraic identities, rational expressions, equations and polynomial division. The main task is to recognise what a question asks before selecting a calculation.
We first compare the available methods, then examine division steps that can lose solutions, and finally use reconstruction to check answers. The mixed practice requires you to make these choices independently.
1. Choosing an Algebraic Method
Similar-looking expressions can lead to different tasks. Decide whether you must prove an identity, find solutions, simplify a quotient or calculate a remainder.
1.1 Matching the requested result to a method
An identity asks for equality throughout a domain, so expansion and coefficient comparison are useful. An equation asks for inputs, so work with its solution set. A quotient asks you to inspect its denominator, while a remainder question may need only one substitution.
1.2 Knowing whether a full division is needed
If a polynomial must be factorised completely, a zero test is only the start: divide out the factor and inspect the quotient. If only the remainder is requested, a full division may be unnecessary. Write what the question asks before committing to a calculation.
Worked Example: An Identity Reveals a Parameter
Question: Find p such that x² + px + 6 = (x + 2)(x + 3) for every real x.
Method: Expand the product and compare the coefficient of x on both sides of the identity.
Expand the right side: x² + 5x + 6. Matching coefficients gives p = 5.
Checking at x = 0 would give 6 = 6 for every p and would not determine the parameter. The coefficient comparison uses the requirement “for every real x”.
2. Checking Before Dividing
Division is valid only when the divisor is nonzero. In equations, this can require a separate case rather than a single line of cancellation.
2.1 Separating zero and nonzero cases
Dividing by a variable expression requires it to be nonzero. If its zeros are possible in the original equation, either check them separately or use factorisation to avoid losing them. If it occurs in an original denominator, those zeros were already excluded.
2.2 Using the original expression to decide which inputs are allowed
This distinction explains why cancellation in a rational expression and division during equation solving need different checks. In both cases, the original statement decides which inputs are allowed.
Worked Example: A Lost Root
Question: A learner solves x(x − 3) = 0 by dividing by x and concludes x = 3. What is missing?
Method: Use the zero-product rule before any division, so neither possible zero factor is lost.
Division assumes x ≠ 0, but the original equation allows x = 0 and it satisfies the equation. The zero-product rule instead gives x = 0 or x − 3 = 0.
The complete solution set is {0, 3}. If the problem had explicitly stated x ≠ 0, the learner’s shorter argument would have been valid.
Worked Example: A Parameter That Cannot Be Cancelled Unconditionally
Question: Classify (a − 2)x = a − 2.
Method: Separate the case a = 2 before dividing by a − 2.
If a ≠ 2, divide by a − 2 to get x = 1. If a = 2, the equation is 0 = 0 and every real x works.
The value x = 1 is always a solution, but “always a solution” does not mean “always the only solution”.
3. Checking by Reconstructing the Original Problem
A result is easier to trust when it reproduces the information you started with. Choose a check that matches the kind of algebra you performed.
3.1 Expanding, multiplying back and substituting
For division, multiply the divisor by the quotient and add the remainder. For factorisation, multiply the factors. For a simultaneous system, substitute the entire ordered pair into both equations. These checks return to the original problem rather than repeating the same manipulation.
3.2 Checking both values and degree for a remainder
For a remainder of the form mx + c, check both supplied values and the degree requirement. The degree condition is part of the answer: infinitely many higher-degree polynomials could take the same two values.
Worked Example: A Remainder from Two Conditions
Question: P leaves remainder 2 on division by x − 1 and remainder 8 on division by x − 3. Find the remainder on division by (x − 1)(x − 3).
Method: Write a linear remainder and use the two stated polynomial values to determine its coefficients.
Write R(x) = mx + c. Then m + c = 2 and 3m + c = 8, so m = 3 and c = −1.
Thus R(x) = 3x − 1. It has degree below 2 and satisfies R(1) = 2 and R(3) = 8. Those checks establish the uniquely determined linear remainder.
Common Mistakes
Do not match coefficients when equality is required only at a single input.
Do not cancel a potential zero factor from an equation without handling that case.
An excluded input cannot return just because the simplified expression has a value there.
Two evaluated values determine a linear remainder only when the divisor has two distinct corresponding zeros.
4. Practice
These questions mix the methods from Section 1B. Name the relevant method, record any restrictions before simplifying, and check the final result in the original problem.
1. Find a and b if (x − 2)(x + a) = x² + x + b for every real x.
2. Simplify (x² − 4)/(x² − x − 2), stating every excluded input.
3. Solve (x² − 9)/(x − 3) = 6. A: x = 3. B: x = −3. C: no solution. D: every real x except 3.
4. Classify (k − 1)x = k² − 1 for every real parameter k.
5. Solve 2x + y = 7 and x − 2y = −4.
6. Solve y = x + 1 and y = x² − 1.
7. Divide x³ − 2x² + x + 5 by x² + 1, giving quotient and remainder.
8. Find p if 2x − 1 is a factor of 2x³ + px² − 5x + 2.
9. P(1) = 5 and P(−1) = −1. Find the remainder when P is divided by x² − 1.
10. For y = (ax + b)/(x − 1), where a and b are fixed real constants, determine when the value y = a is possible and state the corresponding x values.
11. A learner argues that x = 1 is the only root of x³ − 1 because x − 1 is a factor. Complete the reasoning to determine whether the conclusion is correct over the real numbers.
12. Which statement is always true for a real polynomial P? A: P(2) = 0 implies (x − 2)² is a factor. B: P(2) = 3 means the remainder on division by x − 2 is 3. C: P(2) = 0 implies 2x + 1 is a factor. D: A quadratic divisor always has a constant remainder. Give counterexamples to the false statements.
Worked Solutions
1. Comparing coefficients in an identity
Expand to x² + (a − 2)x − 2a. Match a − 2 = 1, so a = 3 and b = −6. Check: (x − 2)(x + 3) = x² + x − 6.
2. Cancelling factors with original restrictions
Factor: (x − 2)(x + 2)/[(x − 2)(x + 1)]. The simplified formula is (x + 2)/(x + 1), but the original excludes x = 2 and x = −1. At x = 0 both forms equal 2.
3. Checking a candidate against an excluded input
C. The original excludes 3. On its domain, cancellation gives x + 3 = 6, whose only candidate is the excluded value 3. There is no solution.
4. Separating a parameter case before division
Factor k² − 1 = (k − 1)(k + 1). If k ≠ 1, the unique solution is x = k + 1. If k = 1, the equation is 0 = 0 and every real x works. Substitution of x = k + 1 verifies the nonzero case.
5. Solving a linear simultaneous system
From the first, y = 7 − 2x. Substitute into the second: x − 2(7 − 2x) = −4, so 5x = 10 and x = 2; then y = 3. Check: 4 + 3 = 7 and 2 − 6 = −4.
6. Finding line–curve intersections
Equating the y values gives x² − x − 2 = (x − 2)(x + 1) = 0. The pairs are (2, 3) and (−1, 0). Their y values agree with both x + 1 and x² − 1.
7. Reconstructing a polynomial division
Q = x − 2 and R = 7, since (x² + 1)(x − 2) + 7 = x³ − 2x² + x + 5. The constant remainder has degree below 2.
8. Using a non-monic factor condition
Evaluate at x = 1/2: 1/4 + p/4 − 5/2 + 2 = (p − 1)/4. Setting this to zero gives p = 1. Check: 2x³ + x² − 5x + 2 = (2x − 1)(x² + x − 2).
9. Finding a remainder from two values
Write R(x) = mx + c. Then m + c = 5 and −m + c = −1. Adding gives c = 2, and then m = 3. The answer is 3x + 2, with the required values at 1 and −1 and degree below 2.
10. Treating the exceptional value in a formula
The original domain is x ≠ 1. Setting y = a and multiplying by x − 1 gives ax − a = ax + b, so b = −a is necessary. If b = −a, the original equals a for every x ≠ 1. If b ≠ −a, no x gives y = a. This handles the zero coefficient case directly, without dividing by y − a.
11. Proving that a cubic has only one real root
Division gives x³ − 1 = (x − 1)(x² + x + 1). The quotient is always positive: x² + x + 1 = (x + 1/2)² + 3/4 > 0. This identity follows by expansion and the nonnegativity of a square. Hence it has no real zeros, so x = 1 is the only real root. The conclusion is correct, but the original argument had not excluded roots of the quotient.
12. Testing universal claims with theorems and counterexamples
B is the remainder theorem. A fails for P(x) = x − 2, which has only one such factor. C fails for the same P, since P(−1/2) = −5/2 ≠ 0. D fails when P(x) = x³ is divided by x² + 1: the quotient is x and remainder is −x, which is linear.
5. Summary
- Choose the method that matches the requested identity, input, factor or remainder.
- Check domains and parameter zeros before division.
- Reconstruct the original expression or substitute the final values to verify the answer.
- Revisit identities for 1, rational restrictions for 2–3, parameters and systems for 4–6 and 10, and polynomial methods for 7–9 and 11–12.