Algebraic identities

\((2x-1)(2x+7)=4x^2+12x-7\), while \((2x+1)(2x+5)=4x^2+8x+5\). Therefore, their difference is \((4x^2+12x-7)-(4x^2+8x+5)=4x-2\). The value \(-12\) results from expanding the middle term of the second product incorrectly. Exam tip: when subtracting an expression in brackets, change the sign of every term in the second bracket before combining like terms.

\(a^4-b^4=(a^2-b^2)(a^2+b^2)\), and then \(a^2-b^2=(a-b)(a+b)\). Thus, the difference-of-squares identity is used twice. \(a^4+b^4\) has a plus sign, so this identity does not apply directly. Exam tip: first identify perfect-square terms.

Using the identity \((x+y)^2=x^2+y^2+2xy\), we get \(x^2+y^2=(x+y)^2-2xy\). Therefore, \(x^2+y^2=81-2(14)=81-28=53\). Hence, 53 is the correct answer. A common error leading to 67 is subtracting only \(xy\) instead of \(2xy\). Exam tip: always account for the \(2xy\) term when finding the sum of squares.

\((a+b)^2=a^2+2ab+b^2\), so option A is the square of a sum. Check the middle term: \(2\times a\times b=2ab\). Option B expands \((a-b)^2\). Exam tip: always verify the coefficient of the middle term.

\(a^2-25b^2=a^2-(5b)^2\), so it is a difference of squares and factors as \((a-5b)(a+5b)\). Option C is a perfect-square trinomial. In exams, check for two squares separated by a minus sign.

The expansion is (25a^2-20ab+4b^2), so like terms cancel and (-20ab) remains. Exam tip: watch the negative middle term.

(144=12^2) and (2k=24), so (k=12). Exam tip: verify using both the last term and the middle term.

(225=15^2) and (-2kx=-30x) gives (k=15). Exam tip: the middle term of ((x-k)^2) is (-2kx).

Use the difference-of-squares identity \\( (a+b)(a-b)=a^2-b^2\\). In the first product, let \\(a=3x\\) and \\(b=4\\). Thus, \\((3x+4)(3x-4)=(3x)^2-4^2=9x^2-16\\). The full expression is then \\((9x^2-16)-(9x^2-20)\\). Distribute the minus sign across the second bracket: \\(9x^2-16-9x^2+20\\). The x-squared terms cancel and the constants give \\(-16+20=4\\). Therefore, option A is correct.

A common error is to subtract only the first term inside the second bracket and leave its constant unchanged. The minus sign applies to both terms, so \\(-(9x^2-20)=-9x^2+20\\). Directly simplifying the two brackets shows the same result: the first is \\(9x^2-16\\), and after subtraction the variable terms disappear. Since no value of x is specified, the result must be the constant 4, not an expression depending on x.

(2\times2y\times3z=12yz). Exam tip: in the square of three terms, take twice the product of every pair.

Here, \(9p^2=(3p)^2\) and \(16q^2=(4q)^2\). The middle term is \(-2\times3p\times4q=-24pq\), so A equals \((3p-4q)^2\). Exam tip: check whether the middle term is \(\pm2ab\).

By identity, the difference is (4xy), so (4xy=72). Exam tip: use the identity first to shorten the equation.

In \((p-q)^2\), the product of the terms is \(p(-q)=-pq\), so the cross term is \(2\times(-pq)=-2pq\). In \((p+q)^2\), it is \(+2pq\). Exam tip: always check the sign of the middle term.

(a^2-b^2=(a+b)(a-b)=2000\times10=20000). Exam tip: difference of squares is useful for large numbers.

Here, \(123=120+3\) and \(117=120-3\). Using \((a+b)(a-b)=a^2-b^2\), we get \((120+3)(120-3)=120^2-3^2=14400-9=14391\). \(14400\) is only \(120^2\); subtracting \(3^2\) is essential. Exam tip: when two numbers are equally spaced from a common number, use the difference-of-squares identity.

Write 998 as \(1000-2\). Using \((a-b)^2=a^2-2ab+b^2\), we get \((1000-2)^2=1000^2-2\times1000\times2+2^2=1000000-4000+4=996004\). Therefore, option A is correct. Option D misses the final \(+4\) term. Exam tip: the middle term in \((a-b)^2\) is always negative.

The difference-of-squares identity is \(p^2-q^2=(p+q)(p-q)\), so option C is correct. In option D, the sign is wrong: expanding gives cancellation of middle terms and leaves \(-q^2\). Exam tip: when you see a difference of squares, look for factors formed by the sum and difference.

The first is (x^2+11x+24) and the second is (x^2+11x+28). Exam tip: cancel like terms and check the constant difference.

To find the coefficient of x, multiply the terms that produce exactly one x. In \\( (2x+5)(2x-3)\\), the cross-products are \\(2x\cdot(-3)=-6x\\) and \\(5\cdot2x=10x\\). Adding them gives \\(4x\\), so the coefficient of x is 4. Therefore option A is correct. The term 4x is not itself the coefficient; its coefficient is the number 4.

Expanding the whole product confirms this: \\( (2x+5)(2x-3)=4x^2-6x+10x-15=4x^2+4x-15\\). The first product gives the x² term, the two cross-products give the x term, and the constants give 15. Option C incorrectly gives the entire middle term rather than its coefficient. Options B and D come from incomplete or incorrect multiplication. Combining signed middle terms carefully prevents the common error of treating 6x and 10x as both positive.

The constant term is ((-2)\times6=-12). Exam tip: multiply only the constants for the constant term.