What Is G2 Mathematics | Why (x + 3)² Needs a 6x Term

To square a bracket, multiply the whole bracket by itself. For (a + b)², the four products are a², ab, ba and b². Since ab = ba for ordinary real numbers, the middle products combine to 2ab. Thus (a + b)² = a² + 2ab + b².

With a = x and b = 3, the middle term is 2 × x × 3 = 6x. With a = 2x and b = 3, it becomes 2 × 2x × 3 = 12x. The complete first term is also squared: (2x)² = 4x².

A numerical check can expose a missing middle term. At x = 1, (x + 3)² = 16, while x² + 9 = 10. One mismatch disproves the proposed identity. The distributive-law expansion explains why the correct expression works for every real x.

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In (x + 3)², the brackets show that the base being squared is the whole expression x + 3. The exponent does not attach only to x, only to three, or independently to each term. It means two factors of the entire bracket.

Compare x + 3², which means x + 9, with (x + 3)², which generally gives a different value. At x = 2, the first is eleven and the second is twenty-five. Brackets change which expression acts as the base of the exponent.

Similarly, 2(x + 3)² means two times the square of the bracket. It is not the same as [2(x + 3)]², which squares the outside two as well. The latter equals 4(x + 3)². Careful grouping prevents an apparently small notation change from altering a whole multiplier.

Before expanding, rewrite the square as a product of two identical brackets. This step makes the exponent’s scope visible and replaces a memory test about identities with a multiplication you can reconstruct from the distributive law.

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For real expressions, A(B + C) = AB + AC. The multiplier A acts on both terms in the bracket. When A itself is a sum, the same principle is applied again rather than replaced by a special exception.

For (x + 3)(x + 3), begin with x(x + 3) + 3(x + 3). Then expand each part: x² + 3x + 3x + 9. Every term has a source, and no term was created merely to fit a remembered pattern.

The middle products come from different selections: x from the first bracket with three from the second, and three from the first with x from the second. Although both simplify to 3x, they are two contributions and must both be counted.

That is the origin of the coefficient two in 2ab. It records two equal cross products. Once the learner understands the two selections, the identity becomes a compressed record of distribution. Without that understanding, a memorised formula can be applied to the wrong base or with the wrong sign.

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A double-bracket product initially gives four products because each of two terms pairs with each of two terms. Some of those products may be like terms and combine. In a squared binomial, the cross products are equal, so the usual simplified expansion has three terms.

For (x + 3)², the four products are x², 3x, 3x and 9. The two linear terms combine to 6x. They do not combine with x² because the variable parts differ, and they do not combine with nine because nine is constant.

Thus simplifying after expansion requires a second skill: collecting like terms. A learner may distribute correctly yet then combine 6x with x² incorrectly. Preserve the distinction between generating products and collecting the terms those products create.

There are also special inputs where the numerical middle contribution is zero, such as x = 0. That does not remove it from the general identity. An expression must remain equivalent for all permitted inputs, not only for one value where a term happens to vanish.

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Write (a + b)² = (a + b)(a + b). Distribution gives a² + ab + ba + b². Real-number multiplication is commutative, so ba = ab, and the middle sum becomes 2ab. Therefore the expansion is a² + 2ab + b².

The letters a and b can represent numbers or expressions. If a = 2x and b = 3, substitute those complete expressions into each part: a² = 4x², 2ab = 12x and b² = 9. The result is 4x² + 12x + 9.

Do not let the compact identity hide what must be squared. A common error is to turn (2x)² into 2x², forgetting that the coefficient two is also part of a. Since (2x)(2x) = 4x², both factors of two contribute.

You can always return to the two brackets if the substitution feels uncertain. Rebuilding a formula from its underlying multiplication is a practical safeguard, especially when coefficients, fractions or negative terms make the expression less familiar than the original x + 3 example.

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First rewrite the square as a product: (x + 3)(x + 3). Multiply x by both terms in the second bracket, giving x² + 3x. Then multiply three by both terms in that bracket, giving 3x + 9.

Combine all four products: x² + 3x + 3x + 9. Collect the like linear terms to obtain x² + 6x + 9. No further collection is possible because the remaining terms have variable parts x², x and one respectively.

Check at x = 2. The original is (2 + 3)² = 25. The expansion is 4 + 12 + 9 = 25. This numerical check is consistent with the identity established by distribution.

If the result was x² + 9, the missing amount at x = 2 would be twelve, exactly the value of the omitted 6x. A useful correction identifies that omitted contribution rather than simply saying that the formula was remembered incorrectly. The learner needs to see which two multiplications the short answer failed to include.

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A small table can record every pairing without relying on a remembered order of letters. Put the terms from one bracket across the top and those from the other down the side.

Adding the four cells gives x² + 6x + 9. The two off-diagonal cells are the cross products. Their equality does not make one redundant; each represents a separate part of the full product.

The table works for nonidentical brackets too. For (x + 2)(x + 3), the cross products are 3x and 2x, giving x² + 5x + 6. They combine because they are like terms, but their coefficients differ because the brackets are not identical.

Use the table when it clarifies the structure, then move to shorter working as confidence grows. The goal is complete distribution with a traceable source for every term. A mnemonic can help with order, but it should not replace the explanation of why those products belong.

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For x > 0, imagine a square with side length x + 3 units. Divide each side into a segment of length x and another of length three. The square separates into one x-by-x square, two x-by-three rectangles and one three-by-three square.

Their areas are x², 3x, 3x and nine square units. The total area is therefore x² + 6x + 9. Leaving out the middle term would leave out both rectangular regions, not merely change a convention about writing algebra.

The picture also explains why the two cross products are equal: the rectangles have swapped side orientations but the same area. Both regions are present in the larger square, so their areas add rather than cancel.

The positive-length assumption belongs to this geometric model. The algebraic identity holds for every real x because it follows from distribution. For negative inputs, use the algebraic argument rather than imagining a square with a negative physical side length. A helpful model should illuminate a valid case without silently narrowing the general expression’s domain.

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For (x − 3)², write (x − 3)(x − 3). The products are x², −3x, −3x and +9. The middle sum is −6x, while the final term is positive because a negative times a negative is positive.

Thus (x − 3)² = x² − 6x + 9. It is not x² − 9. Squaring the whole bracket produces two identical negative cross products and a positive constant, all of which follow directly from signed multiplication.

You can derive the general identity by replacing b with −b in the sum identity. Then (a − b)² = a² + 2a(−b) + (−b)² = a² − 2ab + b². The square on the last term removes its negative sign, but it does not remove the cross products.

At x = 1, the original gives (−2)² = 4. The expansion gives 1 − 6 + 9 = 4. This is a useful sign check. If the final constant is negative, inspect the multiplication of the two negative terms.

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Compare (x − 3)² with (x − 3)(x + 3). The first has identical brackets and expands to x² − 6x + 9. The second has opposite signs on the constants and expands to x² + 3x − 3x − 9 = x² − 9.

In the second product, the cross terms cancel because they have equal magnitudes and opposite signs. In the square, they reinforce because both are −3x. That sign relationship explains the difference between the two identities.

A learner who writes (x − 3)² = x² − 9 may be applying the difference-of-squares identity to the wrong bracket structure. The repair is to compare the two products explicitly, not merely to repeat the name of a formula.

The earlier Mathematics explainer directory includes a focused guide on the difference of two squares and the square of a difference. This article approaches the related issue through the origin of the middle term, so the two lessons support different entry points into the same algebraic structure.

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Expand (2x + 3)². The first square is (2x)² = 4x². The two cross products are 2x × 3 and 3 × 2x, giving 6x + 6x = 12x. The final square is nine. The answer is 4x² + 12x + 9.

The tempting answer 4x² + 6x + 9 includes only one cross product. Another tempting answer 2x² + 12x + 9 fails to square the coefficient in the first term. Those are separate errors and should be corrected separately.

Check at x = 1. The original bracket is five, whose square is twenty-five. The correct expansion gives 4 + 12 + 9 = 25. The single-cross-product version gives nineteen, while the unsquared-coefficient version gives twenty-three.

When using the identity, write a = 2x and b = 3 beside the working if helpful. This small substitution note ensures the complete term is used in a² and 2ab. It also makes later examples with larger coefficients feel like the same structure rather than a new rule.

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Expand (x + y)². Distribution gives x² + xy + yx + y². Since xy = yx, the middle contribution is 2xy. The result is x² + 2xy + y².

Now expand (2x − 5y)². The first square is 4x², the middle term is −20xy, and the final square is 25y². Both variables and both coefficients are part of the terms being multiplied.

The cross term is not 10xy because that counts only one pairing. It is not −20x²y² because multiplying x by y gives xy, not x²y². Exponents count repeated factors of the same variable; the middle product contains one factor of x and one of y.

A useful check uses x = 1 and y = 1. The original becomes (2 − 5)² = 9. The expansion becomes 4 − 20 + 25 = 9. Then use the distributive-law derivation to justify the identity generally. Numerical checks can catch mistakes, but the algebraic argument explains why every permitted input is preserved.

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Expand (x + 1/2)². The first term is x². The middle term is 2 × x × 1/2 = x. The final term is (1/2)² = 1/4. Therefore the result is x² + x + 1/4.

The coefficient two in the middle term can cancel a denominator in a coefficient, as it does here. This does not mean the middle contribution has disappeared; it has simplified to a single x.

For (3x − 1/2)², the expansion is 9x² − 3x + 1/4. Check the middle product carefully: 2 × 3x × (−1/2) = −3x. Squaring the negative one half gives positive one quarter.

If fractions make the identity feel less secure, expand the two brackets explicitly. The same four products appear. The new demand is fraction arithmetic, not a new algebraic principle. Separating those demands helps a learner retain confidence in the structure while practising the numerical coefficients that accompany it.

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Expand 2(x + 3)². First square the bracket to obtain x² + 6x + 9. Then distribute the outside two, giving 2x² + 12x + 18. The outside multiplier acts on every term of the expanded square.

Compare [2(x + 3)]². Here the entire product 2(x + 3) is squared, producing 4(x + 3)² = 4x² + 24x + 36. The brackets specify a different exponent base, so the result is different.

At x = 0, the first expression is eighteen and the second is thirty-six. This simple input exposes the grouping difference immediately. It is not enough to recognise the digits two and three; their structural positions determine which operations apply.

For a more involved example, 3(x − 2)² + x expands to 3x² − 12x + 12 + x, then 3x² − 11x + 12. Give expansion and collection separate lines so that the outside multiplier, the bracket square and the final like-term addition remain easy to trace.

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Simplify (x + 3)² − (x − 3)². Expand each square first: x² + 6x + 9 − (x² − 6x + 9). The subtraction applies to every term in the second expanded bracket.

Removing that bracket gives x² + 6x + 9 − x² + 6x − 9. The quadratic and constant terms cancel, leaving 12x. The middle terms add because subtracting negative six x contributes positive six x.

A compact alternative uses a difference of squares with A = x + 3 and B = x − 3. Then A² − B² = (A − B)(A + B) = 6 × 2x = 12x. This is valid because the complete expression really is a difference of two squares.

Both routes are useful once their structures are understood. If the expanded route produces zero, inspect whether the second bracket’s middle sign was handled correctly. Two expressions with similar shapes need not cancel term by term unless their actual signed terms match.

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For each expression, name a and b before expanding: (x + 4)², (3x + 2)² and (2x − y)². This step identifies the complete pieces that enter the square and cross product.

For the first, a = x and b = 4, giving x² + 8x + 16. For the second, a = 3x and b = 2, giving 9x² + 12x + 4. For the third, a = 2x and b = −y in the sum form, giving 4x² − 4xy + y².

Explain the middle term of each result. It is the sum of two equal cross products, not a number inserted by habit. In the second example each cross product is 6x, so their total is 12x.

If a result contains only the two square terms, rewrite the original as two identical brackets and list all four pairings. That method targets the omitted products directly. Once the pairings are secure, the shorter identity becomes a convenient summary rather than a fragile substitute for understanding.

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Expand (2x − 3)². The products are 4x², −6x, −6x and nine, giving 4x² − 12x + 9. Check at x = 2: the original is one squared, and the expansion gives 16 − 24 + 9 = 1.

Expand (−x + 3)². The first square is x², the middle term is −6x and the last term is nine. Thus it gives the same expansion as (x − 3)². This makes sense because −x + 3 = −(x − 3), and opposite values have equal squares.

Expand (−2x − 3)². Both terms inside the bracket are negative, so their cross products are positive. The result is 4x² + 12x + 9, matching (2x + 3)².

These examples show why reading only the visible subtraction signs can mislead. Multiply the complete signed terms. The middle sign comes from their product, while the two individual squares are nonnegative for real inputs. Keeping those roles separate makes the pattern dependable.

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Simplify (x + 2)² + 3x − 4. Expand the square as x² + 4x + 4, then add the remaining terms. Collecting gives x² + 7x because the constants cancel.

Now simplify 2(x − 1)² + (x + 1)². Expansion gives 2(x² − 2x + 1) + x² + 2x + 1. Distributing and collecting produces 3x² − 2x + 3.

Check the second result at x = −1. The original is 2(−2)² + 0² = 8. The expansion gives 3 + 2 + 3 = 8. A negative check value is helpful for detecting a wrongly signed linear contribution.

For a third task, expand (x + 5)² − 10x. The square is x² + 10x + 25, so the result is x² + 25. Here the middle term disappears only because the separate −10x cancels it. It was present in the expansion and then legitimately cancelled; it was not absent from the squared bracket.

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Factorising x² + 6x + 9 gives (x + 3)². The first and last terms are squares, and the middle term matches twice their bases: 2 × x × 3 = 6x. All three checks matter.

The expression x² + 9 has square first and last terms but no 6x middle term. It therefore cannot generally be written as (x + 3)². Recognising two squares alone is insufficient evidence for a perfect-square trinomial.

Similarly, 4x² − 12x + 9 = (2x − 3)². The square bases are 2x and three, and the negative middle term matches −2 × 2x × 3. Checking the middle coefficient prevents guessing a plausible bracket from the endpoints alone.

This is a useful bridge from expansion to factorisation. Expansion starts with the bracket and generates the coefficients; factorisation starts with the coefficients and tests a possible bracket structure. Re-expanding a proposed factorisation is a direct verification that all three terms, including the middle contribution, have been recovered.

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The identity explains why x² + 6x can be written as (x + 3)² − 9. Expanding the right side gives x² + 6x + 9 − 9, returning the original expression. The added and subtracted nine balance one another.

Thus x² + 6x + 5 = (x + 3)² − 4. This form can reveal a quadratic’s minimum or support later equation solving, where those topics are part of the learner’s course.

The three inside the bracket comes from half the coefficient six. That is because the cross term in (x + a)² is 2ax. Matching 2a with six gives a = 3, and squaring a creates the nine that must be accounted for.

This optional connection should follow a secure understanding of expansion. Memorising “halve and square” without seeing the middle term can become another unexplained routine. The identity gives the reason the procedure works and provides an expansion check whenever the rewritten form is uncertain.

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If (x + 3)² = 16, then x + 3 = 4 or x + 3 = −4. The solutions are x = 1 or x = −7. Both satisfy the original equation because both four and negative four square to sixteen.

Expanding gives x² + 6x + 9 = 16, or x² + 6x − 7 = 0. Factorising as (x + 7)(x − 1) = 0 gives the same two solutions. The two methods agree because the expansion preserves the expression for all real x.

If the middle term were omitted, the solver would instead obtain x² + 9 = 16, leading to x = ±√7. Those values solve a different equation. Correct later algebra cannot repair an earlier change in the expression being solved.

This is why expansion accuracy matters beyond a short exercise. A missing cross term changes roots, graph position and numerical predictions. The effect is structural, even when the wrong expression still looks recognisably quadratic and the subsequent calculations are carried out neatly.

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At x = 0, (x + 3)² and x² + 9 both equal nine. That input cannot detect the missing 6x because the omitted term is zero there. Agreement at zero is therefore a weak check for this particular error.

At x = 1 or x = 2, the omitted contribution is nonzero, so the wrong expansion is exposed. A negative input can also reveal a middle-sign error. Choose a permitted input that makes the suspected difference visible.

One counterexample disproves a proposed identity, but several successful substitutions do not prove an unrestricted identity. The proof comes from valid laws such as distribution and collection. Numerical checks are valuable error detectors alongside that reasoning.

When checking a long expression, use a small value that keeps arithmetic manageable. There is no benefit in choosing a complicated input that introduces new calculation mistakes. The purpose is to test whether the two forms agree, not to make the checking arithmetic more difficult than the original expansion.

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The expression (x − 3)² is always nonnegative for real x, yet its expansion contains −6x. There is no contradiction. The expanded expression’s total is nonnegative; its individual terms need not all be nonnegative at every input.

At x = 3, the terms are nine, negative eighteen and nine, summing to zero. At x = 1, they are one, negative six and nine, summing to four. The negative middle contribution is balanced by the other terms.

A learner may incorrectly change the middle sign to positive because “a square must be positive”. The correct statement is that a real square is nonnegative, including zero, and that property applies to the whole expression. It does not authorise changing individual coefficients.

The factorised form makes nonnegativity easy to see, while the expanded form makes coefficients easy to read. Different forms highlight different properties of the same expression. Keeping both available supports reasoning without requiring one form to display every feature equally clearly.

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If a learner writes (x + 3)² = x² + 9, ask them to write the square as two brackets. Then ask for x times each term, followed by three times each term. Record all four products before collecting anything.

This approach locates the missing contributions. Simply supplying the formula may produce the right answer on the next identical question without changing the belief that powers distribute over sums. The learner needs to see why that belief fails.

Follow with (x + 4)², then (2x + 3)². The first changes the constant; the second changes the complete first term. Ask for the middle contribution alone before requesting the full expansion. This tests whether both cross products and both coefficients are being considered.

Finally, compare (x + 3)² with x² + 3² and with (x − 3)(x + 3). The visible similarities make a useful sorting task when every answer must be justified by the operation. The goal is to recognise structure accurately, not to react to the appearance of two squares anywhere on the page.

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This guide concentrates on the square of a binomial and the origin of its cross term. The optional completing-square and equation examples should be selected according to the learner’s course and readiness. They explain consequences of the same identity without claiming identical assessment scope for all readers.

For official context, consult the 2027 G2 Mathematics syllabus. Use your school’s teaching sequence and question instructions for the required level of algebraic technique.

The existing Expanding Single and Double Brackets Without Losing Signs guide provides the broader distribution route. The algebraic structure guide develops the distinction between terms, factors and grouped operations.

Return to the Mathematics hub if the current difficulty concerns signed arithmetic, unlike terms or interpreting an exponent. Those foundations determine whether the identity can be used reliably. Choose the next lesson from the actual mistake rather than assuming every wrong expansion requires the same repair.

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Expand (x + 6)², (3x − 2)² and (x/2 + 1)². The answers are x² + 12x + 36; 9x² − 12x + 4; and x²/4 + x + 1. In each case, identify the two cross products that produced the middle term.

Now decide whether x² + 8x + 16 is a perfect square. It is (x + 4)² because the middle term is twice x times four. By contrast, x² + 16 is not generally that square because the required 8x contribution is absent.

Finally, simplify (x + 4)² − x². Expansion gives x² + 8x + 16 − x² = 8x + 16. The result is linear because the quadratic terms cancel, not because squaring a bracket originally produced only linear terms.

If these results are secure, try explaining one without a formula sheet. State what the exponent repeats, list the products and describe the collection. A clear explanation shows that the identity can be rebuilt when memory is uncertain.

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For another check of the structure, compare (x + 3)² with (x + 3)(x + 4). The first gives x² + 6x + 9 because the two cross products are equal. The second gives x² + 7x + 12 because its cross products are 4x and 3x. The general double-bracket rule applies to both; the special square identity is simply the case where the factors match. Recognising that relationship reduces the number of apparently separate formulas a learner has to remember.

You can also check a proposed square by inspecting its constant and linear coefficients together. Suppose someone claims x² + 10x + 16 = (x + 4)². The constant sixteen suggests a bracket constant of four in magnitude, but the required positive middle coefficient would then be eight, not ten. The proposed factorisation fails even before a numerical substitution. Matching only the first and last squares is insufficient; the middle term checks whether the endpoint choices fit the complete expression.

A numerical example makes the same idea accessible without letters. Calculate 23² as (20 + 3)². The expansion gives 400 + 120 + 9 = 529. Adding only 20² and 3² would give 409, omitting the two twenty-by-three contributions. Similarly, 19² = (20 − 1)² = 400 − 40 + 1 = 361. These are exact applications of the same identities, and they can support mental arithmetic when the chosen nearby number is convenient.

Do not turn that convenience into a demand to expand every numerical square mentally. A direct multiplication or calculator may be appropriate for the task. The value of the examples is that the middle term now has a visible numerical role: it contributes one hundred and twenty in the first calculation and negative forty in the second. Seeing its effect in ordinary numbers can help a learner recognise why it cannot simply disappear when the same multiplication is written using x.

When you see a squared bracket, identify the complete base and rewrite it as two identical factors. Multiply every term in one factor by every term in the other, keep the signs attached and collect only like terms.

For (x + 3)², that process gives x² + 3x + 3x + 9. The 6x comes from two equal products, each of which belongs to the full multiplication. Omitting it changes the expression rather than simplifying it.

Once that structure is understood, the identity a² + 2ab + b² becomes a useful shortcut. It records the first square, the two cross products and the last square. Apply it to complete expressions such as 2x or −3y, not merely to the most visible letter.

The checking question is, “Where did every term come from?” If each has a valid product source, and the collected form preserves the original values, the expansion is doing its job. A short expression is useful when it remains mathematically equivalent throughout the stated domain.

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The answers below address the common points where a squared bracket gets confused with a sum of squares, a difference of squares or an outside multiplier. Return to the two-bracket product whenever the notation feels ambiguous.

Why can I not square each added term separately?

Because squaring repeats the entire sum as a factor. Distribution produces cross products as well as the individual squares. For x + 3, those cross products are 3x and 3x, giving the additional 6x.

Why does a negative bracket constant give a positive final constant?

In (x − 3)², the last product is (−3)(−3) = 9. The cross products are negative, so the complete expansion is x² − 6x + 9. The signs follow from the individual multiplications.

Is (2x + 3)² equal to 4x² + 6x + 9?

No. Each cross product is 6x, and there are two of them. The correct middle term is 12x, giving 4x² + 12x + 9.

Can a squared-bracket expansion end with no middle term?

Special cases can make the cross term zero, and later terms in a larger expression can cancel it. For the general bracket (x + 3)², however, the cross term is 6x and must be included before any legitimate cancellation or substitution is performed.

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