Algebraic identities

The governing concept is the pair of square identities: (x+y)^2=x^2+2xy+y^2 and (x-y)^2=x^2-2xy+y^2. Subtract the second expansion from the first. The x^2 and y^2 terms cancel, while 2xy-(-2xy)=4xy. The complete expression is therefore 4xy-4xy=0. Hence option A is correct. A common error is to stop after evaluating only the difference of the first two squared expressions; that gives 4xy, which is option B, but the original expression still contains the final subtraction of 4xy. Option C incorrectly treats the result as half of the intermediate term, and option D is not produced by subtracting these two square expansions. The cancellation must be completed before choosing the answer.

The (bc) term is (2\cdot2b\cdot c=4bc) and the (ac) term is (2ac). Exam tip: use double product for each pair.

\(p^2+10pq+25q^2=p^2+2(p)(5q)+(5q)^2=(p+5q)^2\). 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.

For a product of two binomials, the coefficient of x is obtained by adding the two cross-products, while the constant term is obtained by multiplying the two constant terms. Here the cross-products are 2x and 8x, so their sum is 6x. The constants are 8 and 2, whose product is 16. Thus the coefficient of x is 6 and the constant term is 16, exactly as shown in option B.

Expanding verifies this carefully: \\( (x-8)(x+2)=x^2+2x-8x-16=x^2-6x-16 \\). The coefficient is read from the term 6x, and the final number is 16. Option A has the wrong sign for the coefficient, while C uses an incorrect sum and constant sign. Option D gives the wrong magnitude for the coefficient. Checking signs separately is especially important when one constant is negative.

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.

The expression has the form of a perfect-square identity: \\(a^2-2ab+b^2=(a-b)^2\\). The first term, 16a², is the square of 4a, and the last term, 25b², is the square of 5b. Their middle product is \\(2(4a)(5b)=40ab\\), and the given middle term is 40ab. Therefore the expression is \\( (4a-5b)^2\\), so option B is correct. The negative sign in the middle term is the key clue.

Verification by expansion gives \\( (4a-5b)^2=(4a)^2-2(4a)(5b)+(5b)^2=16a^2-40ab+25b^2\\). The plus-square option would produce a positive middle term, not a negative one. Option C does not use the square roots of the first and last terms correctly, and option D represents a difference of squares, which would remove the middle term instead of producing 40ab. Thus the perfect-square form is unambiguous.

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.

First, \((x+1)(x+2)=x^2+3x+2\). Multiplying this by \((x+3)\) gives \(x^3+3x^2+3x^2+9x+2x+6=x^3+6x^2+11x+6\). Hence, option A is correct. In option B, the coefficient of \(x^2\) is incorrectly written as 5 instead of 6. Exam tip: combine like terms carefully after multiplication.

( (a+b)^3 ) contains both (3a^2b) and (3ab^2). Exam tip: remember the four terms of the cube identity.

Expanding \((p+q)^2\) gives \(p^2+2pq+q^2\), so A is the square of a binomial sum. B represents \((p-q)^2\). Exam tip: check that the middle term is exactly \(2pq\).