Algebraic identities

Using the distributive property, \((x-6)(x-2)=x^2-2x-6x+12=x^2-8x+12\). Therefore, option A is correct. In option C, both the middle term and the constant term are incorrect. Exam tip: use \((x-a)(x-b)=x^2-(a+b)x+ab\) for quick expansion.

( (z+1)(z+9)=z^2+10z+9 ). Exam tip: the coefficient of (z) comes from the sum of the constants.

In the square of three terms, mixed terms are (2xy), (2yz), and (2zx). Exam tip: include all pairs.

Each pair gives a double mixed term in three terms. Exam tip: do not miss (2ab), (2bc), and (2ca).

( (x+1)^2=x^2+2x+1 ), so the difference is (2x). Exam tip: do not forget the middle term while expanding.

Using \((a+b)^2=a^2+2ab+b^2\), we get \((5x+2)^2=25x^2+20x+4\). A term with no \(x\) is called the constant term, so the constant term is \(4\). \(20x\) is the middle term and is not constant. Exam tip: after expanding, identify the term without the variable.

The middle term is (-2\times4p\times1=-8p). Exam tip: the sign is very important in a subtraction square.

The number 1002 is very close to the simple number 1000, so the most useful rewrite is \\(1002=1000+2\\). Hence \\(1002^2=(1000+2)^2\\). Using \\((a+b)^2=a^2+2ab+b^2\\), this can be calculated as \\(1000^2+2(1000)(2)+2^2=1,000,000+4,000+4=1,004,004\\). The question asks for the useful form, not merely the final numerical value.

Therefore option A is correct. The form \\(1000-2\\) represents 998, not 1002, so option B is unsuitable. Similarly, \\(100+2\\) is 102 and \\(10+2\\) is 12; their squares are different numbers. The key strategy is to express a number near a convenient base and then apply the square identity. This reduces long multiplication and lowers the chance of calculation errors.

Write (999) as (1000-1) and use ( (a-b)^2 ). Exam tip: choosing the nearest simple number is faster.

Using the identity \((u+v)(u-v)=u^2-v^2\), multiplication gives \(uv\) and \(-uv\), which cancel each other. Therefore, the term \(uv\) does not appear in the final expression. The term \(-v^2\) does appear, so it is not the correct choice. Exam tip: recognise \((a+b)(a-b)\) directly as \(a^2-b^2\).

In \((a+b)(a-b)\), the middle terms \(+ab\) and \(-ab\) cancel, leaving \(a^2-b^2\). Option B is the identity for the square of a sum. Exam tip: look for conjugate binomials with opposite signs.

Using the identity \((a-b)^2=a^2-2ab+b^2\), we get \((n-11)^2=n^2-22n+121\). Therefore, the last term is \(121\). \(-22n\) is the middle term; the last term is \(11^2\), which is positive. Exam tip: in \((a-b)^2\), the last term is always \(b^2\).

\(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so it is a perfect-square trinomial. In option B, the constant would need to be \(25\). Exam tip: match the middle term with \(2ab\).

In ( (a-b)^2 ) the mixed term is negative. Exam tip: write ((3x)^2=9x^2) and ((2y)^2=4y^2).

This uses the identity \((a+b)(a-b)=a^2-b^2\). Here, \(a=5m\) and \(b=4n\), so \((5m+4n)(5m-4n)=(5m)^2-(4n)^2=25m^2-16n^2\). Option D resembles the expansion of \((5m-4n)^2\), but the given factors have opposite signs. Exam tip: in conjugate factors, the middle terms cancel out.

A perfect square trinomial has the form \\(x^2+2kx+k^2=(x+k)^2\\). In the given expression, the last term is \\(81=9^2\\). Twice the product of \\(x\\) and 9 is \\(2(x)(9)=18x\\), which matches the middle term. Therefore \\(x^2+18x+81=(x+9)^2\\). Substitution or expansion can be used to verify this result.

Option A is correct. Option B would expand to \\(x^2-18x+81\\), so its middle-term sign is wrong. Options C and D use numbers that do not produce both the constant term and the coefficient of \\(x\\). A reliable method is to take the square root of the constant term and then check whether twice that number gives the middle coefficient.

(100=10^2) and the middle term is (-20k=-2\times k\times10). Exam tip: identify the negative middle term.

Use the difference-of-squares identity \\(a^2-b^2=(a+b)(a-b)\\). The expression is \\(37^2-27^2\\), so set \\(a=37\\) and \\(b=27\\). Their sum is \\(37+27=64\\), and their difference is \\(37-27=10\\). Thus \\(37^2-27^2=64\\times10=640\\). This avoids separately finding 1369 and 729 and then subtracting them.

Therefore, option C, 640, is correct. A quick direct check is \\(37^2=1369\\) and \\(27^2=729\\); subtracting gives \\(1369-729=640\\). The other values are not equal to the difference. Recognising two squared numbers separated by a minus sign is the essential step in this problem.

The numbers 45 and 55 are equally distant from 50: they can be written as \\(50-5\\) and \\(50+5\\). Their product therefore has the form \\((a-b)(a+b)\\), which uses the identity \\((a+b)(a-b)=a^2-b^2\\). Taking \\(a=50\\) and \\(b=5\\), we get \\(45\\times55=(50-5)(50+5)=50^2-5^2\\). The identity named in option C is exactly this rule.

Thus option C is correct. If the value is required, it is \\(2500-25=2475\\). Options A and B are square identities but do not directly represent the product of two conjugate binomials. Option D is an identity for three terms and is unnecessary here. The useful pattern is two numbers equally placed around a common centre.

In the expansion of a three-term square, the square of each individual term appears. Here, \((2b)^2=4b^2\), so option B is correct. \(2ac\) is a product term involving \(a\) and \(c\), not the source of \(4b^2\). Exam tip: when squaring a term, square its numerical coefficient as well.