Algebraic identities

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

The direct answer is C, 12. Compare the expression with \\((x+k)^2=x^2+2kx+k^2\\). The coefficient of x on the right is 24, so 2k=24 and k=12. The constant term also confirms this: k^2=144, and the positive value giving 2k=24 is k=12. Substitution checks it: \\((x+12)^2=x^2+24x+144\\). Option A, 6, would give middle coefficient 12 and constant 36. Option B, 10, would give 20x and 100. Option C is correct because it satisfies both the middle and final terms. Option D, 24, would give 48x and 576. A common mistake is to use only the last term and forget to check the middle term; always compare both parts.

Expand the left side:
\[
(x-k)^2=x^2-2kx+k^2.
\]
Compare with \(x^2-30x+225\). The coefficient of \(x\) gives
\[
-2k=-30\Rightarrow 2k=30\Rightarrow k=15.
\]
Check constant term: \(k^2=15^2=225\), which matches.
So \(k=15\).

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 \(\pm2ab\).

The direct answer is B, 18. Apply \\((x+y)^2-(x-y)^2=4xy\\). The question says the difference is 72, so \\(4xy=72\\). Dividing both sides by 4 gives \\(xy=18\\). Option A, 12, would make the difference 48, not 72. Option B is correct because 4 multiplied by 18 is 72. Option C, 24, would give 96. Option D, 36, would give 144. The important step is to simplify the whole difference before trying to find x and y separately; only their product is requested. This identity follows because the x-squared and y-squared terms cancel, while the two middle contributions add. Memory cue: plus-square minus minus-square equals four times the product.

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.

The direct answer is B: \\(1005^2-995^2=20000\\). Use the difference-of-squares identity \\(a^2-b^2=(a+b)(a-b)\\). Here, \\(a=1005\\) and \\(b=995\\). First calculate the sum: \\(1005+995=2000\\). Then calculate the difference: \\(1005-995=10\\). Multiply: \\(2000\times10=20000\\). Thus B is correct. Option A, 10000, is half the correct product and may result from forgetting a factor. Option C, 5000, is unrelated to the identity calculation. Option D, 1000, is also not the product of the sum and difference. The identity is useful because it avoids separately squaring two large numbers and subtracting them. Memory cue: difference of squares equals sum times difference.

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.

Multiply: \((2x+5)(2x-3)=(2x)(2x-3)+5(2x-3)=4x^2-6x+10x-15=4x^2+4x-15\). Hence the coefficient of \(x\) is \(4\).

The constant term is obtained by multiplying the constant parts of the two binomials.
In \((3x-2)(3x+6)\), the constants are \(-2\) and \(6\). So the constant term is
\[
(-2)\cdot 6=-12.
\]
Therefore the constant term is \(-12\).

The sum of constants is (1+2+3=6) and the sign is negative. Exam tip: stepwise multiplication also gives the same result.

The direct answer is A: 2b(3a²+b²). Use the cube expansions \\(a+b\\)^3=a^3+3a^2b+3ab^2+b^3 and \\(a-b\\)^3=a^3-3a^2b+3ab^2-b^3. Subtract the second from the first: a³-a³=0, 3a²b-(-3a²b)=6a²b, 3ab²-3ab²=0, and b³-(-b³)=2b³. The result is 6a²b+2b³. Taking out the common factor 2b gives 2b(3a²+b²), so option A is correct. Option B, 2a(a²+3b²), is the form associated with adding the two cubes rather than taking this difference; it has the wrong factor. Option C, 6ab, misses the cubic contribution 2b³ and has the wrong powers. Option D, a³-b³, is not the result of subtracting these two binomial cubes; it ignores the full expansions. A quick check with a = 1 and b = 1 gives 8 - 0 = 8, while option A gives 2(1)(3+1)=8. Memory cue: in this difference, odd b terms remain and the common factor is 2b.

On adding, terms with odd powers of (b) cancel and (2a^3+6ab^2) remains. Exam tip: use symmetry.

A difference of two cubes has the form \(a^3-b^3\) and factorises as \((a-b)(a^2+ab+b^2)\). \(a^3+b^3\) is a sum of cubes, so it uses a different identity. Exam tip: for a difference, the first binomial is \(a-b\).

For a perfect square, \(x^2+px+q=(x+\frac p2)^2\). Squaring gives the constant term \(\frac{p^2}{4}\), so \(q=\frac{p^2}{4}\). The choice \(q=p^2\) wrongly ignores halving the coefficient of \(x\). Exam tip: halve the middle coefficient, then square it.

By identity, (a^3+b^3+c^3-3abc=(a+b+c)(\cdots)). Exam tip: if (a+b+c=0), remember (a^3+b^3+c^3=3abc).

The direct answer is option D: 0. Use the identity \(x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\). The question gives \(x+y+z=0\). Substituting this into the first factor gives \(0\) multiplied by the second factor, so the whole expression is 0. Option D is correct. Option A, \(xyz\), is not the value because the expression contains several cube and product terms, not only \(xyz\). Option B, \(3xyz\), would ignore the cube terms and the given condition. Option C, \(x^2+y^2+z^2\), is only part of the factorisation and is not the whole expression. The important step is to notice the factor \(x+y+z\) before trying separate numerical calculations. Memory cue: whenever three terms satisfy sum zero, this identity makes the expression vanish.

\(a^2-b^2=(a-b)(a+b)\), so \(a-b\) is a factor for every value of the variables. \(a^2+b^2\) does not generally contain this factor. Exam tip: factorise using identities before testing divisibility.

The extra mixed terms in the square of three terms are (2xy+2yz+2zx). Exam tip: write all three mixed terms.

((2x+3)) is common, giving ((2x+3)[(2x+3)-(2x-3)]). Exam tip: taking common factors makes work easier.