An algebraic expression can have several useful forms. Expanding reveals terms to collect, while factorising reveals multiplication. A good rewrite preserves the value for every allowed input and serves a stated purpose. The shortest-looking form is not automatically the most useful one.
Here the goal is to choose and check equivalent expressions. We are not looking for values that make an expression zero. Every worked case starts with a reason to change its form, then checks that the transformation preserved both its value and any restrictions.
Decide what the new form needs to show
Use expanded and collected form when you want a single coefficient for each variable part or efficient repeated substitution. Use factorised form when you want to display a common factor or simplify a quotient of products. Before making changes, identify terms, factors and any denominator.
Terms are separated by addition or subtraction at the outer level. In 5x + 7, the terms are 5x and 7. Factors are quantities being multiplied: in 5(x + 7), the factors are 5 and the whole bracket. Confusing these structures is the source of many invalid cancellations.
Like terms have exactly matching variable parts, including powers. Thus 4ab and −9ab combine to −5ab. However, 4a and 4a² are not like terms. Matching coefficients alone does not make terms alike, and a²b differs from ab² unless a special input happens to make them equal.
Worked case 1: expand to make substitution shorter
Rewrite E = 3(2x − 5) − 2(x + 4) as a collected expression, then evaluate it at x = −2. Multiply every term in each bracket by the factor outside, carrying its sign: E = 6x − 15 − 2x − 8. Therefore E = 4x − 23.
The second bracket contributes −2x − 8, not −2x + 8. One way to make this explicit is to write +(−2)(x + 4), then distribute the factor −2. Collect the x terms together and the constant terms together only after both brackets have been expanded correctly.
At x = −2, the collected form gives 4(−2) − 23 = −31. Checking directly in the original gives 3(−4 − 5) − 2(−2 + 4) = 3(−9) − 2(2) = −27 − 4 = −31. At x = 1/2, both forms give −21. These checks are useful error detectors, while the distributive steps establish the identity for all real x.
A visual model for expansion and factorisation
For positive x, the rectangle’s common height is 3 and its adjoining widths are x and 4. The two areas are 3x and 12. Adding them gives 3x + 12, while multiplying the whole width by the height gives 3(x + 4). Reading from the product to the sum expands; reading backwards factorises.
The positive-length picture illustrates the rule but does not restrict the algebraic identity to positive x. The distributive law also works for negative and zero inputs, where the drawn pieces should no longer be interpreted as ordinary positive physical lengths. Use the algebra, rather than forcing a negative length into the picture.
Worked case 2: extract the complete common factor
Factorise F = 18a²b − 12ab² to expose everything shared by its two terms. The largest positive integer factor shared by 18 and 12 is 6. Both terms also contain at least one factor a and one factor b. Taking out 6ab leaves F = 6ab(3a − 2b).
Check each term by multiplication: 6ab × 3a = 18a²b and 6ab × (−2b) = −12ab². The remaining terms inside the bracket share no additional nonconstant monomial factor. Taking out only 6 would be valid but would not fully expose the common monomial factor.
At a = −1 and b = 2, the original expression is 18(1)(2) − 12(−1)(4) = 36 + 48 = 84. The factored form gives 6(−1)(2)[3(−1) − 2(2)] = (−12)(−7) = 84. At a = 0, both forms give zero. The identity still works there because factorising has not divided the original expression by a variable.
Worked case 3: simplify a quotient without restoring an excluded input
Simplify R = (x² − 9)/(x − 3). Start with the original denominator: x − 3 cannot equal zero, so x ≠ 3. Next use the difference of squares, x² − 9 = (x − 3)(x + 3). The quotient becomes [(x − 3)(x + 3)]/(x − 3).
On the allowed domain, x − 3 is nonzero, so dividing numerator and denominator by that complete factor leaves R = x + 3, for x ≠ 3. At x = −2, the original is (4 − 9)/(−5) = 1 and the simplified form is −2 + 3 = 1. At x = 5, both give 8.
At x = 3, the original gives 0/0, which is undefined. The unrestricted expression x + 3 has a value of 6 there, but that value does not belong to the original quotient. Keep the restriction beside the simplified answer; erasing it changes the set of allowed inputs.
Use a counterexample to reject a false rewrite
The expressions 2(x + 3) and 2x + 6 are equivalent by distribution. The expressions 2(x + 3) and 2x + 3 are not: at x = 0 they give 6 and 3. One disagreement is enough to disprove an identity.
Agreement at one value proves much less. The expressions x² and x agree at x = 0 and x = 1 but differ at x = 2. Use valid symbolic transformations to establish an identity, then use substitution as a secondary check. Also, (x + 6)/x cannot become 1 + 6 by crossing out the x terms. For x ≠ 0 it is 1 + 6/x, because the denominator divides both numerator terms.
Three solved practice tasks
Practice 1: collect before repeated evaluation
Rewrite 5(2t + 1) − 3(t − 4), then evaluate at t = −3. Expansion gives 10t + 5 − 3t + 12 = 7t + 17, useful because only one multiplication remains for each input. At −3 the value is −21 + 17 = −4. Directly, 5(−5) − 3(−7) = −25 + 21 = −4.
Practice 2: reveal a shared product
Factorise 14p²q + 21pq² completely. The common factor is 7pq, giving 7pq(2p + 3q). At p = −2 and q = 1, this equals (−14)(−1) = 14. Expansion returns both original terms, and direct substitution gives 56 − 42 = 14.
Practice 3: retain the restriction
Simplify (y² + 5y)/y and say whether y = 0 is allowed. The numerator factors as y(y + 5), so the answer is y + 5, for y ≠ 0. Factorised form is useful because it reveals the cancellable product. At y = −2, both forms give 3. Zero remains excluded because the original denominator would be zero.
Frequently asked questions
Is expanding always the best way to simplify?
No. Expand to combine terms or evaluate efficiently; factorise to expose multiplication. Choose the form that serves the stated purpose.
Why does a minus before a bracket affect every term?
It multiplies the whole bracket by negative one, or by the full negative coefficient present. Distribute that factor to each term.
Can x and x² be collected?
No. Their variable parts differ. You can factor x from x + x², but cannot add the coefficients as if both terms were x.
Does checking two inputs prove an identity?
No. Different expressions may agree at selected values. A valid algebraic derivation establishes the identity; substitution helps catch errors.
Can I cancel part of a sum?
Not as an isolated term. First identify a factor multiplying the entire numerator and denominator. A denominator dividing a sum applies to every term.
Why keep a restriction after its denominator disappears?
The restriction comes from the original expression. Cancellation establishes equality only where the cancelled factor is nonzero.
Does factorising an expression mean solving it?
No. Factorisation rewrites the same quantity as a product. Solving requires an equation or another condition specifying what values are wanted.
Subject references
The original questions, diagrams and solutions above use standard mathematical definitions and methods. These source-authored references support the underlying facts.
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