Algebraic identities

Using the sum identity gives (2(3a)^2+2(4b)^2=18a^2+32b^2). Exam tip: notice cancellation of middle terms.

After removing square terms from a three-term square, only double products remain. Exam tip: remember (2xy), (2yz), (2zx).

The direct answer is option A, 0. First recognize the perfect-square trinomial identity \((x+7)^2=x^2+2(x)(7)+7^2=x^2+14x+49\). Therefore the expression becomes \((x+7)^2-(x+7)^2\). Any quantity subtracted from itself is zero, so the value is 0 for every value of x. Option A is correct. Option B, \(14x\), is only the middle term inside the square; it is not left over because the entire identical square is subtracted. Option C, 49, is only the constant term, and it also cancels with the corresponding 49 in the second square. Option D, \(x^2\), is only the first square term and likewise cancels. A common mistake is to subtract selected terms while forgetting that all three terms belong to the same perfect square. Memory cue: identify the complete square before subtracting; identical expressions give zero.

Using the identity \((a+b)^2=a^2+2ab+b^2\), the expression becomes \(a^2+2ab+b^2-(a^2+b^2)\). The terms \(a^2\) and \(b^2\) cancel, leaving \(2ab\). It is not \(0\) because the middle term \(2ab\) does not cancel. Exam tip: expand the identity first, then cancel like terms carefully.

The sum of constants is (4+(-6)=-2) and the product is (-24). Exam tip: handle mixed signs carefully.

Expanding \((x+r)^2\) gives \(x^2+2rx+r^2\), so \(p=2r\) and \(q=r^2\). Hence \(p^2=4r^2=4q\). The other relations do not ensure a perfect square. Exam tip: square the middle coefficient and compare it with four times the constant term.

\(25p^2-36q^2=(5p)^2-(6q)^2\), which is a difference of two squares. Using \(a^2-b^2=(a+b)(a-b)\), we get \((5p+6q)(5p-6q)\). Options A and D are squares of a difference and a sum respectively; their expansions contain a \(pq\) term, which is absent in the given expression. Exam tip: identify the square roots first, then use \((a+b)(a-b)\) for a difference of squares.

Using \((a+b)^2=a^2+2ab+b^2\), take \(a=x\) and \(b=5y\). The middle term is \(2\times x\times5y=10xy\), and the last term is \(25y^2\). Option C represents \((x-5y)^2\). Exam tip: always check the sign of the middle term.

In \(x^2+12x+25\), the square roots of the first and last terms are \(x\) and \(5\), so the middle term should be \(2\times x\times5=10x\), not \(12x\). Exam tip: always check the middle term using \(2ab\).

The direct answer is option B: \(4bc+2ac\). To expand a square containing three terms, use the rule that every pair of different terms produces twice their product. Here the terms are \(a\), \(2b\), and \(c\). For the \(bc\) term, multiply the two terms \(2b\) and \(c\), then double: \(2(2b)(c)=4bc\). For the \(ac\) term, multiply \(a\) and \(c\), then double: \(2(a)(c)=2ac\). Adding these required terms gives \(4bc+2ac\). Option A, \(2bc+2ac\), is wrong because it forgets that the second term is \(2b\), not \(b\). Option B is correct because both pairwise products are calculated with the correct coefficient. Option C, \(2abc\), multiplies three variables, but the question asks for separate two-variable terms, not a three-variable product. Option D, \(4ab+2ac\), contains the \(ab\) pair instead of the requested \(bc\) pair. Memory cue: in a square, each unlike pair gives twice its product; carefully keep the coefficient 2 in \(2b\).

\(p^2+10pq+25q^2=p^2+2(p)(5q)+(5q)^2=(p+5q)^2\). Its middle term is twice the product of the square roots of the first and last terms. In exams, check this middle-term condition quickly.

The first expansion is (x^2+7x+10) and the second is (x^2+7x+12). Exam tip: after like terms cancel, subtract constants.

Expand: \((x-8)(x+2)=x(x+2)-8(x+2)=x^2+2x-8x-16=x^2-6x-16\). Hence coefficient of \(x\) is \(-6\) and constant term is \(-16\).

By difference of squares, ( (1005+995)(1005-995)=2000\cdot10=20000 ). Exam tip: verify arithmetic carefully.

Expanding \((a+b)(a+b)\) gives \(a^2+ab+ab+b^2=a^2+2ab+b^2\). Option A misses the cross-term \(2ab\). Exam tip: always check the sign of the middle term in square identities.

The sum-of-cubes identity is \(x^3+y^3=(x+y)(x^2-xy+y^2)\). Its middle term is negative; \((x-y)\) is used for the difference of cubes. Exam tip: for a sum, first look for the factor \((x+y)\).

This is (4ab) where (a=7a) and (b=1), so (28a) is obtained. Exam tip: distinguish symbols and full terms.

In \\((p+q)(p-q)\\), the cross terms \\(pq\\) and \\(-pq\\) cancel, leaving only \\(p^2-q^2\\). Option B is the identity for a perfect-square trinomial. Exam tip: opposite signs in matching binomials signal a difference of squares.

The identity \((x+y)^2=x^2+2xy+y^2\) includes the cross-term \(2xy\) along with both square terms. Option B represents \((x-y)^2\). Exam tip: use the sign of the middle term to identify the identity.

On expanding, \((x+y)^2=x^2+2xy+y^2\) and \((x-y)^2=x^2-2xy+y^2\). When these are added, \(2xy\) and \(-2xy\) cancel, giving \(2x^2+2y^2\). Subtracting \(2x^2\) leaves \(2y^2\). Hence, \(2y^2\) is correct; \(2xy\) does not remain because the middle terms cancel. Exam tip: expand both squares first and combine like terms carefully.

This expression uses the difference-of-squares identity. For any two quantities, \\(a+b\\)^2-\\(a-b\\)^2=4ab. Here, the two quantities are \\(a=3p\\) and \\(b=q\\). Therefore, the difference between the first two squared expressions is exactly \\(4(3p)(q)=12pq\\). The final term in the expression is also \\(12pq\\), so it cancels this difference completely. Thus, the value is zero, making option A correct. The result does not depend on particular values of \\(p\\) and \\(q\\); it follows algebraically for all values.

Expanding also verifies the result directly: \\( (3p+q)^2=9p^2+6pq+q^2\\) and \\( (3p-q)^2=9p^2-6pq+q^2\\). Subtracting the second from the first gives \\(12pq\\), because the \\(9p^2\\) and \\(q^2\\) terms cancel while the middle terms add. The remaining subtraction is \\(12pq-12pq=0\\). Option C is only the difference of the first two squares, not the value of the complete expression.

Using the identity \((a+b)^2=a^2+2ab+b^2\), with \(a=x\) and \(b=2y\), we get \((x+2y)^2=x^2+4xy+4y^2\). Subtracting \(x^2\) and \(4y^2\) leaves \(4xy\). \(2xy\) is incorrect because the middle term is \(2\times x\times 2y=4xy\). Exam tip: always calculate the middle term as \(2ab\) when expanding a binomial square.

Option A has constant term \(-18\). A negative product means the two constant terms must have opposite signs: \(x^2-7x-18=(x-9)(x+2)\). Exam tip: inspect the sign of the constant term first.

Notice \(16a^2=(4a)^2\) and \(25b^2=(5b)^2\), and the middle term is \(-40ab=-2(4a)(5b)\). So it matches \((a-b)^2=a^2-2ab+b^2\):
\(16a^2-40ab+25b^2=(4a-5b)^2\).

First multiply the outer factors: \((x+1)(x+3)=x^2+3x+x+3=x^2+4x+3\). Therefore, the required form is \(x^2+4x+3\). In \(x^2+2x+3\), the middle terms \(3x\) and \(x\) have not been added correctly. Exam tip: while multiplying two binomials, write all four products and then combine like terms.