Algebraic identities

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.

In the plus square identity, (+2ab) appears. Exam tip: keep ( (a+b)^2 ) and ( (a-b)^2 ) separate.

In the minus square identity, the middle term is subtracted. Exam tip: the final (3^2) is still added.

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.

( (m+n)^2=m^2+2mn+n^2 ), so the extra term is (2mn). Exam tip: remember this common mistake.

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.