Algebraic identities

( (x+4)(x+6)=x^2+10x+24 ). Exam tip: the coefficient of (x) comes from the sum of constants.

The difference-of-squares identity is \((a+b)(a-b)=a^2-b^2\). On expansion, the middle terms cancel. Exam tip: a product of a sum and a difference gives a difference of squares.

The expression has the form
\((a+b)^2=a^2+2ab+b^2\). The middle term is the product term in the centre, formed by twice the product of the two terms being added. Here, those terms are \(2m\) and \(3\). Therefore, the middle term is \(2(2m)(3)=12m\). Thus, option B is correct. Option C, \(4m^2\), is the first square, and option D, \(9\), is the last square; neither is the middle term.

To solve similar questions, first identify \(a\) and \(b\), then use \(2ab\). Substituting gives \(2\times2m\times3=12m\). The coefficient is multiplied by both the variable term and the constant, so stopping at \(6m\) misses the required factor of 2. The supplied answer and explanation are mathematically correct and clearly identify the middle term.

This uses ( (a+b)^2-(a-b)^2=4ab ) where (a=4x) and (b=3y). Exam tip: take full terms as (a) and (b).

Using the sum identity gives (2(5a)^2+2(2b)^2=50a^2+8b^2). Exam tip: notice cancellation of middle terms.

The (pq) term is (2\cdot2p\cdot q=4pq) and the (pr) term is (4pr). Exam tip: use double product for every pair.

(9x^2=(3x)^2), (25y^2=(5y)^2), and the middle term is (30xy). Exam tip: check both square roots and the middle term.

Using \((A-B)^2=A^2-2AB+B^2\), take \(A=5x\) and \(B=3y\). This gives \(25x^2-30xy+9y^2\). Option C has a positive middle term, so it represents \((A+B)^2\). Exam tip: always check the sign of the middle term.

The sum of constants is (9+(-4)=5) and the product is (-36). Exam tip: check sum and product separately in mixed signs.

Using \((a+b)^2=a^2+2ab+b^2\), \((2x+3)^2=4x^2+12x+9\). Subtracting \(4x^2+9\) cancels the \(4x^2\) and \(9\) terms, leaving \(12x\). \(6x\) is a close distractor because the middle term is \(2\times 2x\times 3=12x\). Exam tip: always calculate the middle term of a binomial square as \(2ab\).

It matches \(a^2-2ab+b^2=(a-b)^2\), with \(a=p\) and \(b=5q\). Check the middle term: \(-2\times p\times5q=-10pq\). Perfect-square outer terms alone are not enough. In exams, always verify the \(\pm2ab\) term.

The difference of the first two squares is (4pq), and then (4pq) is subtracted. Exam tip: check remaining terms after applying identity.

\(p^2+16q^2=p^2+(4q)^2\) is a sum of squares, whereas the identity is \(A^2-B^2=(A+B)(A-B)\). In \(9a^2-b^2\), the terms are subtracted. Exam tip: check the sign between the two square terms first.

The first expansion is (x^2+10x+21) and the second is (x^2+10x+24). Exam tip: cancel like terms and subtract constants.

The pair terms in the square are (4ab), (2bc), and (4ac). Exam tip: take the double product of each pair.

Using \((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\), the term containing \(yz\) comes from \(2(2y)(5z)\). Thus, \(2\times2\times5\,yz=20yz\), so \(20yz\) is correct. \(10yz\) results from forgetting the factor \(2\) in the cross-product term. Exam tip: while squaring three terms, multiply every product of two unlike terms by \(2\).

Since \((a+b)^2=(a+b)(a+b)\), multiplication gives \(a^2+ab+ab+b^2=a^2+2ab+b^2\). Option B is the expansion of \((a-b)^2\). Exam tip: the middle term is always \(2ab\).

Write 997 as \(1000-3\). Using \((a-b)^2=a^2-2ab+b^2\), \((1000-3)^2=1000^2-2\times1000\times3+3^2=1000000-6000+9=994009\). Therefore, the correct answer is 994009. The value 997009 results from an incorrect middle-term calculation. Exam tip: use \((a-b)^2\) for squaring numbers close to 1000.

Using (a^2-b^2=(a+b)(a-b)), we get (2000\cdot8=16000). Exam tip: multiply the sum and difference.

Use the identity \((a+b)^2=a^2+2ab+b^2\) with \(205=200+5\). Thus, \((205)^2=200^2+2\times200\times5+5^2=40000+2000+25=42025\). In 41025, the middle term \(2ab\) is not calculated correctly. Exam tip: while squaring a binomial, always include the \(2ab\) term.