Algebraic identities

In \((a+b+c)^2\), all three terms are positive, so the pairwise products \(2ab\), \(2bc\), and \(2ca\) are all positive. A negative sign would make the related cross-term negative. Exam tip: check signs carefully in squared expansions.

Twice the product of (-2b) and (c) is (2\cdot(-2b)\cdot c=-4bc). Match both sign and coefficient carefully.

\(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so it is a perfect square of a binomial. In option B, the constant term should be \(5^2=25\), not 20. Exam tip: match the middle term with \(2ab\).

\(4x^2-25=(2x)^2-5^2\), so it is a difference of two perfect squares and factors as \((2x-5)(2x+5)\). Options B and D are perfect-square trinomials. Exam tip: check for two square terms separated by a minus sign.

\((a+b)^2=a^2+2ab+b^2\), so its middle term is \(+2ab\). In contrast, \((a-b)^2\) has \(-2ab\) as its middle term. Exam tip: check the sign between the two terms before choosing a square identity.

Here \(16a^2=(4a)^2\) and \(25b^2=(5b)^2\), while the middle term is \(-2\times4a\times5b=-40ab\). Hence it is \((4a-5b)^2\). Exam tip: always verify the sign of the middle term.

\((1000+3)^2=1000^2+2(1000)(3)+3^2=1000000+6000+9=1006009\). The student omitted the middle term \(2ab\). In exams, write all three terms of \((a+b)^2\).

\(a^2+2ab+b^2=(a+b)^2\), so it is always a perfect-square trinomial. Option B has middle term \(ab\), but a square of a binomial requires \(2ab\). Exam tip: check the middle coefficient first.

By identity the left side is (4\cdot3m\cdot n=12mn). (12mn=48mn) gives no valid nonzero fixed (m); the given equality is not generally true.

Expand each product: \((x+4)(x-6)=x^2-2x-24\) and \((x-4)(x+6)=x^2+2x-24\). On adding, the \(-2x\) and \(+2x\) terms cancel, giving \(2x^2-48\). In option B, the sign of the constant term is incorrect. Exam tip: after expanding, check the signs carefully before combining like terms.

\((x+4)(x-6)=x^2-2x-24\), while \((x-4)(x+6)=x^2+2x-24\). On subtracting the second expression, \(x^2-2x-24-(x^2+2x-24)=-4x\). Therefore, the correct answer is \(-4x\). The option \(4x\) results from an incorrect sign while subtracting the bracketed expression. Exam tip: when subtracting a whole expression, change the sign of every term inside its bracket.

The direct answer is A: the coefficient of \\(x^2\\) is \\(-4\\). To see this clearly, first multiply \\((x-2)(x+5)=x^2+3x-10\\). Now multiply by \\((x-7)\\). The \\(x^2\\) terms come from \\(x^2(-7)=-7x^2\\) and \\(3x(x)=3x^2\\). Their sum is \\(-7+3=-4\\), so the coefficient is \\(-4\\). Equivalently, in three monic factors \\((x+a)(x+b)(x+c)\\), the coefficient of \\(x^2\\) is \\(a+b+c\\); here \\(-2+5-7=-4\\). Option A is correct. B has the wrong sign. C is the product of selected constants and is not the x-squared coefficient. D comes from an incorrect combination and has the wrong value. Do not confuse the coefficient of \\(x^2\\) with the constant term, which would be \\((-2)(5)(-7)\\).

The constant term of a product of polynomials is obtained by multiplying their constant terms. In \((x+1)(x-3)(x+8)\), the constant terms are \(1\), \(-3\), and \(8\). So the constant term is \(1\cdot(-3)\cdot8=-24\).

Here \(27=3^3\), so \(x^3+27=x^3+3^3=(x+3)(x^2-3x+9)\), which fits the sum-of-cubes identity. \(x^3-27\) is a difference of cubes. Exam tip: check whether the constant is a perfect cube.

\(a^2-b^2\) is the difference of two perfect squares, so it factorises as \((a+b)(a-b)\). Option B is the identity for the square of a binomial, not a difference of squares. Exam tip: identify both square terms first.

\((a+b)^2=(a+b)(a+b)\). Multiplying gives \(a^2+ab+ab+b^2=a^2+2ab+b^2\). Option B is the expansion of \((a-b)^2\). Exam tip: remember that the middle term in a square expansion has coefficient 2.

The correct identity is \((x+y)(x-y)=x^2-y^2\). On multiplying, the middle terms \(-xy\) and \(+xy\) cancel, leaving a difference of squares. Exam tip: look for binomials with opposite signs.

In \((a+b)(a-b)\), the middle terms \(-ab\) and \(+ab\) cancel, leaving \(a^2-b^2\). In \((a+b)^2\), a \(+2ab\) term appears. Exam tip: identify same binomials with opposite signs as a difference of squares.

The expression has the form \(a^2-b^2\), so the difference-of-squares identity is the quickest method: \\(a^2-b^2=(a+b)(a-b)\\). Take \(a=1004\) and \(b=996\). Their sum is 2000, and their difference is 8. Therefore the required value is \\(2000\times8=16000\\).

Hence option A is correct. Calculating both large squares separately is unnecessary and creates more opportunity for arithmetic mistakes. The value 8000 would result from losing a factor of 2, while 4000 and 12000 do not match the product of the sum and difference. The identity works because the middle terms cancel when the two squares are subtracted.

Here, 503 = 500 + 3 and 497 = 500 - 3. Using the identity \((a+b)^2+(a-b)^2=2(a^2+b^2)\), we get \((503^2+497^2)=2(500^2+3^2)=2(250000+9)=500018\). The value 500000 includes only \(2\times500^2\); the term \(2\times3^2\) must also be added. Exam tip: When two numbers are equally spaced from a common middle number, use this identity for quick calculation.

By difference of squares, ( (203+197)(203-197)=400\cdot6=2400 ). Exam tip: directly find sum and difference.

Compute \((x^2+1)^2\):
\[
(x^2+1)^2=x^4+2x^2+1.
\]
Now subtract \(x^4\):
\[
(x^4+2x^2+1)-x^4=2x^2+1.
\]
So the simplified result is \(2x^2+1\).

\(9p^2-24pq+16q^2=(3p)^2-2(3p)(4q)+(4q)^2=(3p-4q)^2\), so A is a perfect-square trinomial. In B, the last term is not \((4q)^2\). Exam tip: match the middle term with \(-2ab\).

The constant term comes from multiplying the constant parts in each factor.
\[
(x+1)(x+2)(x+3)
\]
Constant parts are \(1\), \(2\), and \(3\). So the constant term is
\[
1\cdot 2\cdot 3=6.
\]
Therefore, the constant term is \(6\).

To find the constant term, take the term without \(x\) from each factor. Thus, the constant term is \(1\times(-3)\times5=-15\). The closest distractor, \(15\), results from ignoring the negative sign. Exam tip: multiply only the constant terms and track signs carefully.