Algebraic identities

The square of a sum contains three parts: the square of the first term, twice the product of the two terms, and the square of the second term. In algebraic form, the identity is \\(a+b\\)^2=a^2+2ab+b^2. The middle term appears because the two cross-products, \(ab\) and \(ba\), are equal and together make \(2ab\).

Applying this identity gives the expansion \(a^2+2ab+b^2\), so option B is correct. For example, if both terms are added and the result is multiplied by itself, the products are \(a^2\), \(ab\), \(ba\), and \(b^2\). The two middle products combine to \(2ab\). Option A omits these products, while option C gives the wrong sign and option D is not a square expansion.

Expanding \((x-y)^2=(x-y)(x-y)\) gives \(x^2-xy-xy+y^2=x^2-2xy+y^2\). Therefore, option C is correct. Option A is the expansion of \((x+y)^2\), while option B equals \((x-y)(x+y)\). Exam tip: in a squared binomial, the middle term is twice the product of the terms; here it is \(-2xy\) because of the minus sign.

On multiplying \((p+q)(p-q)\), we get \(p^2-pq+pq-q^2\). The middle terms \(-pq\) and \(+pq\) cancel, leaving \(p^2-q^2\). \(p^2+q^2\) does not result from this cancellation. Exam tip: remember \((a+b)(a-b)=a^2-b^2\) as the difference of squares identity.

Option A is correct because multiplying \((a+b)(a-b)\) cancels the middle terms and gives \(a^2-b^2\). In option C, the left side is \((a-b)^2\). Exam tip: check the sign of the middle term carefully.

When \(a-b\) is squared, the middle term is negative, so \((a-b)^2=a^2-2ab+b^2\). Option B is the identity for the square of a sum. Exam tip: always check the sign of the middle term.

The expansion of \((a-b)^2\) is \(a^2-2ab+b^2\), so its middle term \(-2ab\) is negative. In option A, the middle term is \(+2ab\). Exam tip: the sign between the terms determines the sign of the middle term.

\(x^2-9=x^2-3^2\), so using \(a^2-b^2=(a-b)(a+b)\), it becomes \((x-3)(x+3)\). \(x^2+9\) is a sum of squares. Exam tip: check for two perfect squares separated by a minus sign.

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

A difference of squares, \(a^2-b^2\), factorises as \((a+b)(a-b)\). For \(x^2-25\), take \(a=x\) and \(b=5\) to get \((x+5)(x-5)\). Exam tip: check for subtraction between two perfect squares, not addition.

The identity \\(p+q)^2=p^2+2pq+q^2\\) shows that the middle term is twice the product of the two terms. Here, take \\(p=3a\\) and \\(q=2b\\). The middle term is therefore \\(2(3a)(2b)=12ab\\). The complete expansion is \\(9a^2+12ab+4b^2\\), so the central term is the term containing both variables.

Thus option C, \\(12ab\\), is correct. Option B is the square of the first term, \\(9a^2\\), and option D is the square of the second term, \\(4b^2\\); neither is the middle term. Option A would result from missing a factor of 2. Keeping the order of the identity clear prevents that common error.

In \((a-b)^2\), the middle term is negative, so the identity is \(a^2-2ab+b^2\). Option C gives the difference of squares, not a square. Exam tip: look for the \(-2ab\) middle term.

\((x+y)^2\) is the square of the binomial \(x+y\). Its expansion is \(x^2+2xy+y^2\). In contrast, \((x+y)(x-y)\) represents a difference of squares. Exam tip: look first for an outside power of 2.

Using the distributive property, \((x+2)(x+3)=x(x+3)+2(x+3)=x^2+3x+2x+6=x^2+5x+6\). Therefore, \(x^2+5x+6\) is correct. In option A, the constant term is incorrectly written as 5 instead of \(2\times3=6\). Exam tip: in \((x+a)(x+b)=x^2+(a+b)x+ab\), the middle coefficient is the sum of the constants and the final term is their product.

\(p^2-q^2\) is a difference of two squares and factorises as \((p+q)(p-q)\). Option B represents \((p+q)^2\), not a difference. Exam tip: check that both terms are perfect squares with a minus sign.

\((p-7)(p+7)\) matches \((a-b)(a+b)\), so it represents \(a^2-b^2\). In contrast, \((p+7)^2\) is a perfect-square form. Exam tip: look for identical terms with opposite signs.

The correct form is \((100-1)^2\), because 99 is just 1 less than 100. Using \((a-b)^2=a^2-2ab+b^2\), we get \((100-1)^2=10000-200+1=9801\). The expression \((99+1)(99-1)\) equals \(99^2-1\), so it is not equal to \(99^2\). Exam tip: for squaring a number, rewrite it using the nearest convenient multiple of 10 or 100.

\(x^2-9=x^2-3^2=(x-3)(x+3)\), so it is a difference of two squares. \(x^2+9\) is a sum of squares. Exam tip: look for \(a^2-b^2\).

Conjugate binomials have the same first term but opposite signs before the second term. Hence, \(x+y\) and \(x-y\) are conjugates. Their product is \((x+y)(x-y)=x^2-y^2\). Exam tip: check the sign of the second term first.

In the identity \((x+y)^2=x^2+2xy+y^2\), the first term is \(x^2\). Here, \(x=2a\), so the first term is \((2a)^2=4a^2\). \(10a\) is not the first term; the middle term is \(2\times 2a\times 5=20a\). Exam tip: while squaring a term, square its coefficient as well.

\(x^2-49=x^2-7^2=(x-7)(x+7)\), so it is a difference of squares. \(x^2+14x+49=(x+7)^2\) is a perfect square, not a difference. Exam tip: \(a^2-b^2\) has no middle term.