Expanding, factoring and algebraic identities

An identity agrees for every permitted input

An identity states that two expressions have the same value for every permitted value of their variables. In this lesson the expressions are polynomials, so their inputs may be any real numbers unless a context narrows the domain. The statement 2(x + 3) = 2x + 6 is an identity in x: the distributive property explains why it holds for every real x.

An equation such as 2(x + 3) = 10 instead asks which permitted x make a particular equality true. Expanding its left side replaces an expression by an equal expression; finding its solutions still requires equation-solving steps. A factored expression is a way to write the same value, not an instruction to set the expression equal to zero.

Multiply every term and keep its sign

The distributive property gives a(b + c) = ab + ac and a(b − c) = ab − ac for real a, b and c. A negative multiplier acts on every term too: −3(2x − 5) = −6x + 15. Subtracting a whole bracket means multiplying that bracket by −1, not changing only its first term.

For two brackets, each term in the first multiplies each term in the second. Thus (x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6. There are four products before like terms are collected. Leaving out either middle product changes the expression.

Factor a common multiplier without excluding zero

Factorisation uses distributivity in reverse: 6x + 9 = 3(2x + 3). The common factor must multiply every term in the bracket. Expanding the proposed product checks whether the expression has been preserved; a useful common factor need not be the largest one unless the task asks for that.

A common factor may contain a variable: x² + 3x = x(x + 3) for every real x, including zero. This is an identity, not division of an equation by x. At x = 0, both expressions are zero, so factoring has not removed that permitted input. Dividing an equation by x, or cancelling x in a fraction, would require a separate nonzero condition.

Group terms to reveal a shared bracket

Some expressions have no useful factor common to all terms at once. Group x² + 3x + 2x + 6 as x(x + 3) + 2(x + 3). Now the whole bracket x + 3 is a common factor, giving (x + 2)(x + 3). Re-expansion returns all four original terms.

Grouping may require a different arrangement of terms or extraction of a negative factor. Keep each term’s sign when rearranging. A repeated bracket can be factored even when its value is zero: distributivity does not require division by that bracket.

A squared bracket includes two cross products

Squaring means multiplying the whole expression by itself. Expanding (a + b)(a + b) gives a² + ab + ba + b², so (a + b)² = a² + 2ab + b². Expanding (a − b)(a − b) gives a² − ab − ba + b², so (a − b)² = a² − 2ab + b².

The last term is +b² in both identities. In the squared difference, (−b)(−b) = b². The cross term has the appropriate sign and must not disappear. These identities hold for all real a and b, including negative and zero values; the sign of the numerical value of 2ab also depends on a and b. For example, (2 + (−3))² = 4 − 12 + 9 = 1.

The identities also factor perfect-square expressions: 4x² − 12x + 9 = (2x − 3)². The middle term must match −2(2x)(3) = −12x, as well as the end terms matching the squares of 2x and 3. Square terms at the two ends alone do not justify a squared bracket.

Opposite cross products cancel in a difference of squares

Expanding (a + b)(a − b) gives a² − ab + ab − b². The cross products cancel, leaving the identity a² − b² = (a − b)(a + b). For example, 9x² − 16 = (3x − 4)(3x + 4) because the two squared quantities are 3x and 4.

A squared difference and a difference of squares are different expressions: (a − b)² includes a cross term, while a² − b² factors into two different brackets. The difference-of-squares identity does not turn a² + b² into (a − b)(a + b).

Find factors by matching a sum and a product

For fixed numbers r and s, expansion gives (x + r)(x + s) = x² + (r + s)x + rs. Therefore, to factor x² + px + q in this form, seek fixed permitted coefficients with r + s = p and rs = q. For x² + 5x + 6, the numbers 2 and 3 have sum 5 and product 6, giving (x + 2)(x + 3).

State which coefficients are allowed: integer factors, rational factors and real factors are different requirements. Fractional coefficients can be legitimate; for example, x² − 1/4 = (x − 1/2)(x + 1/2). The method does not guarantee two real linear factors. For x² + 1, the conditions would be r + s = 0 and rs = 1. The first gives s = −r, so rs = −r² cannot be 1 for real r.

One nonmonic example is 2x² + 7x + 3. Split 7x into 6x + x, then group: 2x² + 6x + x + 3 = 2x(x + 3) + (x + 3) = (2x + 1)(x + 3). This factors an expression. It neither sets the product equal to zero nor asks for solutions of a quadratic equation.

Use a known algebraic form when comparing coefficients

After expansion and collection, expressions with the same coefficients of x², x and the constant term have the same value for every real x. This follows directly by substituting any real value into the identical simplified expression; it is not a conclusion drawn from a few successful numerical trials.

The converse for these quadratic forms can be justified without a general polynomial theorem. If their difference is Px² + Qx + R and it is zero for every real x, choosing x = 0 gives R = 0. Choosing x = 1 and x = −1 then gives P + Q = 0 and P − Q = 0. Adding gives 2P = 0, so P = 0 and Q = 0. Thus all three coefficient differences are zero. These particular checks work as part of a proof because the difference has already been shown to have this restricted form.

For an identity A(2x + 1) + B(x − 1) = 5x − 2, A and B are fixed unknown coefficients, while x ranges over all real values. Expand the left side to (2A + B)x + (A − B). Matching the coefficient of x and the constant gives the system 2A + B = 5 and A − B = −2. The second gives B = A + 2; substitution into the first gives 3A + 2 = 5, hence A = 1 and B = 3. With these constants the expanded expression is exactly 5x − 2, verifying every x. Matching at one chosen x alone would not determine both coefficients.

Separate positive geometric lengths from real algebraic values

An area picture for (a + b)² can use positive lengths a and b: the square contains areas a² and b² and two rectangles of area ab. For a difference-of-squares area picture, require a > b > 0, so both a − b and a + b are positive lengths. With a = 5 and b = 2, the area difference is 25 − 4 = 21, matching a rectangle with sides 7 and 3.

A negative number is not a physical side length. The algebraic identities nevertheless hold for every real a and b because their distributive derivations do not assume positive lengths. If a² − b² is negative, it is an algebraic difference, not the unsigned area of a rectangle. State the geometric domain when using an area interpretation.

Inspect symbolic coefficients and numerical checks separately

The model compares seven polynomial identities and one proposed false equality with an omitted cross term. Choose the proposal and numerical input independently. The symbolic expansions and matching coefficients establish equivalence for every real input; the numerical values show what happens at the selected input.

One counterexample disproves a proposed identity on its stated domain. A successful sample alone does not prove an identity: (x + 2)² = x² + 4 passes at x = 0 but fails at x = 1. Use the distributive derivation or justified coefficient argument to establish a claim for every permitted input, and keep the original expressions when checking samples.

Check the whole expression and state the scope

To check an expansion, account for every original product and sign. To check a factorisation, expand the complete product and collect like terms. State any restrictions from the original context. Polynomial rewriting here introduces no new excluded real values, including when a variable factor or bracket is zero.

Cubic factorisation, methods for solving quadratic equations, and cancellation of variable factors in algebraic fractions belong to later lessons. Fractions also require their original denominators to be nonzero. For now, preserve the expression’s value and distinguish that task from finding the solution set of an equation.

Worked examples

Try it yourself

1. Simplify 3(2x − 5) − 2(x + 1) by distributing to every term.

   Show the answer and reasoning

   4x − 17.

   Expansion gives 6x − 15 − 2x − 2. The variable terms combine to 4x and the constants to −17, for every real x.

2. Factor 4x² + 6x and explain whether x = 0 must be excluded.

   Show the answer and reasoning

   2x(2x + 3); zero is not excluded.

   Expanding the product gives 4x² + 6x. Both forms are zero at x = 0. Factoring by distributivity is valid there and does not divide by the variable factor.

3. Is (x + 4)² = x² + 16 an identity? Give a counterexample and the correct expansion.

   Show the answer and reasoning

   No. At x = 1, the sides are 25 and 17. The identity is (x + 4)² = x² + 8x + 16.

   The product (x + 4)(x + 4) contains the two cross products 4x and 4x. One failed permitted input disproves the proposed identity; full expansion justifies the corrected one.

4. Factor 25x² − 9 and check by expanding.

   Show the answer and reasoning

   (5x − 3)(5x + 3).

   These are the squares of 5x and 3. Expansion gives 25x² + 15x − 15x − 9; the cross products cancel. This is not the squared-difference formula.

5. Factor x² − x − 6 using integer coefficients.

   Show the answer and reasoning

   (x − 3)(x + 2).

   The fixed numbers −3 and 2 have sum −1 and product −6. Equivalently, group x² − 3x + 2x − 6 as x(x − 3) + 2(x − 3). Expanding verifies the identity for every real x.

6. Find fixed real A and B so that (x − 2)(x + 4) = x² + Ax + B is an identity for every real x.

   Show the answer and reasoning

   A = 2 and B = −8.

   Expansion gives x² + 4x − 2x − 8 = x² + 2x − 8. The coefficient argument identifies A = 2 and B = −8. Alternatively, x = 0 gives B = −8 and x = 1 gives A + B = −6; these two constraints determine the constants, and the expansion verifies every x.

7. Compare x(x + 1) = x² + x with x(x + 1) = x² + 1. Which is an identity?

   Show the answer and reasoning

   The first is an identity for every real x; the second is not.

   Distributivity proves the first equality. The second holds at x = 1, but at x = 0 its sides are 0 and 1. A successful value does not turn an equation into an identity.

What to carry forward

• An identity preserves an expression’s value for every permitted input; solving an equation asks which inputs satisfy an additional equality.
• Distribute to every term and retain signs; a squared bracket contains two cross products.
• Common factors and grouping use distributivity in reverse and do not exclude zero-valued variable factors.
• State the allowed coefficients when factoring, and check a proposed product by complete expansion.
• Use symbolic reasoning to justify every permitted input; a numerical counterexample disproves a claimed identity, while a successful sample alone does not prove it.
• Positive-length area interpretations have narrower domains than the corresponding real algebraic identities.