Algebraic identities

The difference-of-squares identity is \(a^2-b^2=(a+b)(a-b)\). Therefore, \((a+b)(a-b)\) is the correct factor form. In contrast, \((a-b)^2=a^2-2ab+b^2\), so it is not equal to \(a^2-b^2\). Exam tip: whenever you see \(x^2-y^2\), factor it as \((x+y)(x-y)\).

The direct answer is B: the term \\(18x\\) does not appear. Multiply using the difference-of-squares identity \\( (a+b)(a-b)=a^2-b^2\\). Here, \\(a=x\\) and \\(b=9\\), so \\( (x+9)(x-9)=x^2-9^2=x^2-81\\). If expanded directly, the terms are \\(x^2-9x+9x-81\\); the middle terms cancel because \\(-9x+9x=0\\). Option A, \\(x^2\\), does appear. Option B, \\(18x\\), does not appear, so it is correct. Option C, \\(-81\\), appears as the constant term. Option D, \\(x^2-81\\), is the complete simplified result. A common mistake is to add the middle terms without noticing their opposite signs. Exam cue: conjugate brackets with opposite signs produce no middle term.

\((a+b)^2=(a+b)(a+b)=a^2+ab+ab+b^2=a^2+2ab+b^2\), so A is correct. In \((a-b)^2\), the middle term is \(-2ab\). Exam tip: always check the sign of the middle term.

Expanding, \((x-2)(x-7)=x^2-7x-2x+14=x^2-9x+14\). Therefore, the term without \(x\), the constant term, is \(14\). Note that \(-9\) is the coefficient of the middle term, not the constant term. Exam tip: the product of two negative numbers is positive.

\((x-4)(x+6)=x^2+6x-4x-24=x^2+2x-24\). Therefore, the coefficient of \(x\) is \(2\). The number \(-24\) is the constant term, not the coefficient of \(x\). Exam tip: simplify the expression by combining like terms before identifying a coefficient.

Expanding \((a+b)^2\) gives \(a^2+2ab+b^2\), so A is correct. In \((a-b)^2\), the middle term is \(-2ab\), not \(+2ab\). Exam tip: always check the sign of the middle term.

The expansion of \((a-b)^2\) is \(a^2-2ab+b^2\). Its middle term is always \(-2ab\); \(a^2+2ab+b^2\) is the identity for \((a+b)^2\). Exam tip: the last term is always the square of the second term.

In the square identity, the last term is (v^2) and remains positive. Exam tip: the minus sign affects the middle term.

The direct answer is option B: \((7+3)^2=7^2+2\cdot7\cdot3+3^2\). Use the identity \((a+b)^2=a^2+2ab+b^2\). Here \(a=7\) and \(b=3\), so substitute them in order: first square is \(7^2\), the middle term is positive \(2\cdot7\cdot3\), and the last square is \(3^2\). Numerically this is \(49+42+9=100\), matching \(10^2\). Option A has a negative middle term, which belongs to \((a-b)^2\), not a plus sign. Option B has the exact required pattern and is correct. Option C, \(7^2-3^2\), is a difference of squares and does not equal a square of a sum. Option D omits the cross term \(2ab\), so it is incomplete. Memory cue: plus inside the bracket means plus in the middle term; minus means minus.

The direct answer is option C. Let x=10 and y=3. The identity for a difference squared is (x-y)^2=x^2-2xy+y^2. Therefore (10-3)^2=10^2-2×10×3+3^2=100-60+9=49. Option A uses a plus middle term, so it represents (10+3)^2, not a difference. Option B is the difference of squares, 10^2-3^2, which equals (10-3)(10+3), not the square of 10-3. Option C has the correct negative middle term and positive final square. Option D omits the middle term entirely, so it is incomplete. Remember: in a squared subtraction, the middle term is negative, but the last square remains positive.

Here, \(104=100+4\) and \(96=100-4\). Thus the product is of the form \((100+4)(100-4)\), so \((a+b)(a-b)=a^2-b^2\) applies. Hence, \(104\times96=100^2-4^2=10000-16=9984\). The identity \((a-b)^2\) is used for squaring one binomial, not for multiplying conjugate binomials. Exam tip: when two numbers are equally distant from the same number, look for the difference-of-squares identity.

Option A is the correct square-of-a-sum identity: \((a+b)^2=a^2+2ab+b^2\). Option B wrongly omits the middle term \(2ab\). Exam tip: always check the sign and coefficient of the middle term.

The difference of squares factorises as \(a^2-b^2=(a+b)(a-b)\), so A is correct. Options B and D miss the required middle term \(2ab\). Exam tip: opposite signs in binomials usually indicate this identity.

Option A is the correct square-of-a-sum identity: the middle term is \(2ab\). In option B, one \(ab\) term is missing. Exam tip: multiply each term in the first bracket by both terms in the second bracket.

Use the identity \((a+b)^2-(a-b)^2=4ab\). Here, \(a=x\) and \(b=4\), so the expression becomes \(4\times x\times 4=16x\). The result \(8x\) would arise if \(b=2\). Exam tip: Recognise the difference-of-squares identity to simplify such expressions quickly instead of expanding both squares.

\(x^2-49=x^2-7^2\), so it is a difference of squares and factorises as \((x+7)(x-7)\). \(x^2+49\) is a sum of squares. In exams, check for two perfect squares separated by a minus sign.

\(x^2-49=x^2-7^2\), so it matches \(a^2-b^2=(a-b)(a+b)\) and becomes \((x-7)(x+7)\). \(x^2+49\) is a sum of squares. Exam tip: check for two perfect squares separated by a minus sign.

Writing \(1002=1000+2\) allows direct use of the identity \((a+b)^2=a^2+2ab+b^2\). Since \(1000\) is a convenient round number, the calculation becomes easy. \(1000-2=998\), so it does not represent the given number. Exam tip: for squaring, express a number using a nearby power of 10 whenever possible.

The direct answer is option B: \\(2mn\\). To see why, expand the square as multiplication: \\(m+n\\)^2=(m+n)(m+n)\\). The four products are \\(m^2\\), \\(mn\\), \\(nm\\), and \\(n^2\\). Since \\(mn=nm\\), the two middle products combine to \\(2mn\\). Thus \\(m^2+2mn+n^2\\) is obtained. Compared with only \\(m^2+n^2\\), the additional part is exactly \\(2mn\\). Option A is wrong because the expressions are generally different; they are equal only in special cases such as \\(mn=0\\), not as an identity. Option B is correct because it gives the missing middle term. Option C has the wrong sign and belongs to the square of a difference. Option D has only one \\(mn\\), although two cross-products occur. Exam cue: whenever a sum is squared, add twice the product of the two terms.

Using the identity \((m-n)^2=m^2-2mn+n^2\), the extra term compared with \(m^2+n^2\) is \(-2mn\). The term \(2mn\) occurs in \((m+n)^2\), so it is a close but incorrect distractor. Exam tip: the middle term in a binomial square is always \(\pm2mn\); use the negative sign for subtraction.

In the identity \((a+b)^2=a^2+2ab+b^2\), the last term is \(b^2\). Here, \(a=x\) and \(b=3\), so the last term is \(3^2=9\). The term \(2\cdot x\cdot3=6x\) gives the middle term, not 9. Exam tip: in a square identity, the square of the second term gives the last term.

Using the identity \((x-y)^2=x^2-2xy+y^2\), the middle term is \(-2xy\). Here, \(x=2a\) and \(y=b\), so the middle term is \(-2\times 2a\times b=-4ab\). The option \(-2ab\) misses the coefficient 2 in \(2a\). Exam tip: always multiply the coefficients of both terms while finding the middle term.

Use the identity \((a+b)(a-b)=a^2-b^2\), where \(a=x\) and \(b=8\). Thus, \((x+8)(x-8)=x^2-8^2=x^2-64\). Option A incorrectly uses a sum instead of a difference of squares. Exam tip: when two binomials have the same terms but opposite signs, apply the difference of squares identity.

\(x^2-49=x^2-7^2\) is a difference of squares, so it factorises as \((x-7)(x+7)\). \(x^2+49\) is a sum of squares. Exam tip: check whether the constant is a perfect square.

The expression has two binomials with the same first term and opposite second terms. This matches the identity \\(a+b)(a-b)=a^2-b^2\\). The middle terms cancel, so there is no term containing only one power of x. Here, the common term is \\(a=3x\\), and the other term is \\(b=2\\). Therefore, the expression becomes \\((3x)^2-2^2\\), which is \\(9x^2-4\\). Thus, option A is correct. Option B incorrectly changes the minus sign to a plus sign, while options C and D do not follow the identity.

To verify by direct multiplication, multiply each term: \\(3x\cdot3x=9x^2\\), \\(3x\cdot(-2)=-6x\\), \\(2\cdot3x=6x\\), and \\(2\cdot(-2)=-4\\). The terms \\(-6x\\) and \\(+6x\\) cancel, leaving \\(9x^2-4\\). This confirms the answer without assuming any particular value of x. The result is therefore exactly the first choice.