Learning Objectives
Distinguish terms, factors, equations and identities.
Expand and factorise expressions using the distributive law.
Compare coefficients in a polynomial identity.
State the domain when simplifying a quotient.
Prerequisites: TMUA Section 1A arithmetic, powers and nonzero-denominator restrictions. All practice questions in this lesson are original, not official TMUA questions.
What This Lesson Is About
Algebra lets us describe calculations with variable quantities instead of one fixed set of numbers. This lesson explains how to read that notation, change the form of an expression without changing its value, and distinguish an identity from an equation that holds only for some inputs.
We begin with terms, factors and like terms. We then expand brackets, reverse that process by factorising, and use identities to determine unknown coefficients. The final section introduces the domain restrictions that become central when algebraic fractions are studied in the next lesson.
1. Reading Algebraic Expressions and Equalities
Before rearranging algebra, identify what each part represents. The same symbols can show a sum, a product or a condition on a variable; knowing which structure you are looking at tells you which operations are justified.
1.1 Terms, coefficients and factors: the parts of an expression
An expression names a value, such as 3x + 6. Its terms are the pieces added together: 3x and 6. A coefficient is the numerical multiplier of a variable term, so the coefficient of x here is 3. A factor multiplies an entire expression: writing 3x + 6 as 3(x + 2) exhibits the factors 3 and x + 2.
1.2 Like terms: combining the same variable quantity
Only like terms can be collected. The terms 4x and −x combine to 3x, while 4x and x² do not combine because they involve different powers. Write subtraction as adding a negative term if a sign is easy to lose.
The distinction is the same as counting unlike units: three copies of x and two copies of x make five copies of x, but a copy of x² is a different quantity. Substitute a number if you need to test whether a proposed combination is valid.
1.3 Equations and identities: some inputs or every allowed input
An equation asserts equality and may hold only for particular inputs. An identity holds for every input in its stated domain. For example, 2(x + 3) = 2x + 6 is an identity for real x; 2x + 3 = 9 is an equation whose only solution is x = 3.
Worked Example: Collect with the Signs Intact
Question: Simplify 4x − 2(3x − 5) + x.
Method: Distribute the negative multiplier to both terms before collecting the coefficients of x.
Distribute −2 to both terms: 4x − 6x + 10 + x. Collect the x terms to obtain −x + 10.
At x = 2, the original gives 8 − 2 + 2 = 8, and the simplified expression gives −2 + 10 = 8. This checks the arithmetic at one input; the distributive steps establish the identity for all real x.
2. Expanding Brackets: Distributing Every Product
Expansion turns a product of sums into a sum of products. This makes individual coefficients visible, which is useful for collecting like terms and comparing expressions. The area model offers a way to see why every cross-product must be included.
2.1 The distributive law: multiplying every term
The distributive law a(b + c) = ab + ac says that multiplication acts on each term in a sum. Applying it twice gives (a + b)(c + d) = ac + ad + bc + bd. Each term in the first bracket must multiply each term in the second.
2.2 The area model: connecting four products to one rectangle
For x > 0, a rectangle with side lengths x + 2 and x + 3 splits into four smaller rectangles. Their areas are x², 2x, 3x and 6, so adding the areas gives x² + 5x + 6. The area picture uses positive lengths, but the distributive identity is valid for every real x.
2.3 Squared brackets and conjugate brackets: adding or cancelling cross-products
Two useful identities follow by expansion: (a + b)² = a² + 2ab + b² and (a − b)(a + b) = a² − b². In a squared bracket, the middle term is present because the cross-products add; in conjugate brackets, they cancel.
For a squared bracket, the two cross-products are both ab, so together they contribute 2ab. For (a − b)(a + b), they are ab and −ab, so their sum is zero. These expansions explain the rules rather than leaving them as separate formulas to memorise.
Figure 1. For x > 0, the four areas add to (x + 2)(x + 3) = x² + 5x + 6. The diagram is schematic, not to scale.
Worked Example: A Sign-Sensitive Product
Question: Expand (2x − 3)(x + 4).
Method: Multiply each term of the first bracket by each term of the second, then combine only like terms.
The four products are 2x², 8x, −3x and −12. Collect the middle terms: 2x² + 5x − 12.
Check at x = 1: (−1)(5) = −5, and 2 + 5 − 12 = −5.
3. Factorisation: Rewriting a Sum as a Product
Factorisation is useful when a common multiplier is hidden in an expanded expression. A product can be easier to simplify or solve than a sum, so we now reverse expansion and look for factors that reconstruct the original.
3.1 Common factors and quadratic factors
Factorising rewrites a sum as a product. First look for a factor shared by every term. For a quadratic x² + bx + c, try (x + r)(x + s): expansion shows that r + s must equal b and rs must equal c.
A quadratic expression has highest power x². Expanding (x + r)(x + s) produces x² + (r + s)x + rs, which explains why the sum and product of r and s must match its coefficients.
3.2 Splitting the middle term: revealing a shared bracket
For a quadratic with a coefficient other than 1 on x², you can split the middle term and group. In 2x² + 5x − 3, the split 6x − x works because 6 + (−1) = 5 and 6 × (−1) = 2 × (−3). Grouping then reveals a shared bracket.
3.3 Verification: multiply the factors back
Always expand the proposed factors to check them. Not every quadratic has integer factors; failure to find them does not mean the expression is zero or that an equation has no real solution. General quadratic methods follow in Section 1C.
Worked Example: Split and Group
Question: Factorise 2x² + 5x − 3.
Method: Split the middle term so the two groups share x + 3, then take out that common bracket.
Write 2x² + 6x − x − 3 = 2x(x + 3) − (x + 3). Taking out the shared bracket gives (2x − 1)(x + 3).
Expanding gives 2x² + 6x − x − 3, recovering the original. The coefficient −1 in the second group is essential.
4. Identities and Domains: Stating Where Equality Holds
The first sections changed expressions using identities. We now make the phrase “for every input” precise: coefficient comparison relies on it, while a denominator restriction can reduce which inputs are allowed in the first place.
4.1 Coefficient comparison: determining a polynomial identity
A polynomial is a finite sum of constant multiples of nonnegative integer powers of x. When two such expressions agree for every real x, the coefficient of each power must match, not just the value at one chosen input.
If two polynomials are identical for every real x, their coefficients of corresponding powers agree. This gives equations for unknown coefficients. Equality at a single input is weaker and does not justify matching coefficients.
4.2 Domain restrictions: cancellation does not define a missing value
The domain is the set of inputs for which an expression is defined. In a quotient, a denominator of zero removes that input from the domain, even if a later simplification has no visible denominator.
Simplifying a quotient needs a domain statement. For example, (x² − 4)/(x − 2) = x + 2 for x ≠ 2, because the numerator factors as (x − 2)(x + 2). Cancelling the nonzero factor x − 2 does not define the original expression at x = 2.
Worked Example: Determine an Identity
Question: Find constants A and B such that (x + 2)(x + A) = x² + 7x + B for every real x.
Method: Expand the product and match coefficients because the equality is required for every real x.
Expansion gives x² + (A + 2)x + 2A. Match the x coefficients: A + 2 = 7, so A = 5. Match constants: B = 2A = 10.
Substituting the constants gives (x + 2)(x + 5) = x² + 7x + 10, confirmed by expansion.
Common Mistakes
Do not combine unlike powers: x + x² is not 2x³.
Distribute a negative multiplier to every term, including constants.
Do not omit 2ab from (a + b)².
A numerical check cannot replace a proof of an identity, and cancellation cannot restore an excluded input.
5. Practice
The first four questions practise expansion and factorisation. The final three ask you to justify an identity, preserve a domain restriction and evaluate a claimed rule. Check proposed factors by multiplying them out.
1. Simplify 3(2x − 1) − 2(x + 4).
2. Expand (x − 5)².
3. Factorise 3x² − 12x and 9x² − 16.
4. Factorise 3x² + 7x + 2.
5. Find a and b if (2x + a)(x − 1) = 2x² + bx − 3 for every real x.
6. Simplify (x² − 9)/(x + 3) and state its domain.
7. A student checks (x + 1)² = x² + 1 at x = 0 and calls it an identity. Explain the error and find all real x for which it holds.
Worked Solutions
1. Distribute before collecting like terms
6x − 3 − 2x − 8 = 4x − 11. At x = 2, both original and simplified expressions equal −3.
2. Retain both cross-products in a square
(x − 5)(x − 5) = x² − 10x + 25. At x = 5 both forms are zero; the cross-products are −5x and −5x.
3. Extract a common factor or a difference of squares
3x² − 12x = 3x(x − 4). Also 9x² − 16 = (3x − 4)(3x + 4). Multiplying each product recovers its original terms.
4. Split and group the middle term
Split 7x as 6x + x: 3x(x + 2) + (x + 2) = (3x + 1)(x + 2). Expansion gives 3x² + 6x + x + 2.
5. Match corresponding coefficients
Expansion gives 2x² + (a − 2)x − a. Constants require −a = −3, so a = 3; then b = a − 2 = 1. Check: (2x + 3)(x − 1) = 2x² + x − 3.
6. Cancel with the original domain intact
The numerator is (x − 3)(x + 3), so the quotient is x − 3 for x ≠ −3. At x = −3 the original denominator is zero.
7. Use expansion to distinguish an equation from an identity
A single successful input is not enough. Expansion reduces the equation to 2x = 0, so it holds only at x = 0. The input x = 1 disproves the identity because 4 ≠ 2.
6. Summary
- Read sums as terms and products as factors.
- Expand every product and reverse the process by extracting common factors or grouping.
- Use coefficient comparison only for a stated polynomial identity.
- Preserve the original domain when simplifying fractions.