Algebraic identities

Expanding \((a+b)^2\) gives \(a^2+2ab+b^2\), so its \(ab\)-term is \(2ab\). The mixed terms in \((b+c)^2\) and \((c+a)^2\) are \(2bc\) and \(2ca\), respectively, not \(ab\). Therefore, the total \(ab\)-term is \(2ab\). Exam tip: Expand each square separately and collect only the requested term.

For the first part, \((x+y)^2-(x-y)^2=4xy\). Similarly, \((x+2y)^2-(x-2y)^2=8xy\). Therefore, the total is \(4xy+8xy=12xy\). It is not \(16xy\), because the two differences add up to 12. Exam tip: apply the identity \((a+b)^2-(a-b)^2=4ab\) directly.

Use the identity \((a-b)(a+b)=a^2-b^2\). Here, \(a=2x\) and \(b=5\), so \((2x-5)(2x+5)=4x^2-25\). Adding \(25\) gives \(4x^2-25+25=4x^2\). The option \(4x^2+25\) incorrectly fails to cancel the \(-25\) with the outside \(+25\). Exam tip: When conjugate factors appear, apply the difference-of-squares identity first.

The expression matches the identity \(a^2+b^2-2ab=(a-b)^2\), where \(a=x+3\) and \(b=x+4\). Therefore, its value is \(\big((x+3)-(x+4)\big)^2=(-1)^2=1\). It would be 0 only if the two terms were equal; here they differ by 1. Exam tip: First identify \(a^2+b^2-2ab\) and rewrite it as \((a-b)^2\).

Use the identity \((x+y)^2-(x-y)^2=4xy\). Here, \(x=3a\) and \(y=2b\), so the expression equals \(4\times 3a\times 2b=24ab\). Hence, option B is correct. \(9a^2-4b^2\) is the product \((3a+2b)(3a-2b)\), whereas the question gives the difference of the squares of these two binomials. Exam tip: for an expression of the form \((x+y)^2-(x-y)^2\), apply \(4xy\) directly.

\(x^2+4x+4=(x+2)^2\). The square of every real number is never negative, so this polynomial is always non-negative. In contrast, \(x^2-4\) is negative at \(x=0\). Exam tip: check whether the middle term is twice the product of the terms in the binomial.

In conjugate binomials, one term is the same and the other has an opposite sign. Thus, \((5x+2y)(5x-2y)=25x^2-4y^2\). In option B both binomials are identical, so it uses a square identity instead. Exam tip: check for matching terms and opposite signs.

The expression \\(x+y+z)^2\\) means that the whole sum of the three numbers is multiplied by itself. An algebraic identity gives the same result in expanded form: \\(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx\\). The important point is that the square applies to the complete sum, not separately to only one term. Thus, the expression has one definite numerical value after the given values are substituted.

Substitute \\(x=1\\), \\(y=2\\), and \\(z=3\\). First find the sum: \\(x+y+z=1+2+3=6\\). Now square it: \\(6^2=36\\). The identity confirms this: \\(1^2+2^2+3^2+2(1)(2)+2(2)(3)+2(3)(1)=1+4+9+4+12+6=36\\). Therefore, option B is correct. Option A, 25, would result from an incorrect sum or operation.

\(p^2+2pq+q^2=(p+q)^2\). The square of every real number is zero or positive, so this expression cannot be negative. Exam tip: check whether the middle term is \(2pq\) to spot a perfect square.

For three linear factors, the coefficient of (x^2) is (1+2+3=6). Exam tip: you can also check by expanding the first two factors.

In the product of three linear factors, the coefficient of \(x^2\) equals the sum of the constant terms. Thus, \((-2)+3+5=6\). Therefore, the correct answer is 6. Option 5 results from adding only the positive constants \(3+5\) and ignoring the negative sign of \(-2\). Exam tip: always include signs while adding constant terms.

Using \((a+b)^2=a^2+2ab+b^2\), with \(a=x\) and \(b=2y\), we get \((x+2y)^2=x^2+4xy+4y^2\). Subtracting \(x^2+4y^2\) cancels the like terms, leaving \(4xy\). \(2xy\) is a common error because the middle term is \(2\times x\times 2y=4xy\). Exam tip: calculate the coefficient of the middle term carefully while expanding a square.

Using \((a+b)^2=a^2+2ab+b^2\), take \(a=2x\) and \(b=3y\). The middle term is \(2(2x)(3y)=12xy\), so A is correct. B has 6xy instead. Exam tip: verify a perfect square by checking twice the product of the square roots of the end terms.

\(x^4-81y^4=(x^2)^2-(9y^2)^2\). Applying \(a^2-b^2=(a-b)(a+b)\) gives option A. Option C has a sum, not a difference. Exam tip: first rewrite both terms as perfect squares.

The first term is \(64x^2=(8x)^2\) and the last term is \(25=5^2\). Their middle term is \(2\times 8x\times 5=80x\), so \(64x^2+80x+25=(8x+5)^2\). Option B would give a middle term of \(-80x\). Exam tip: always verify the middle term using \(a^2+2ab+b^2=(a+b)^2\).

The first term is \(25m^2=(5m)^2\), and the last term is \(49n^2=(7n)^2\). Using \((a-b)^2=a^2-2ab+b^2\), with \(a=5m\) and \(b=7n\), the middle term becomes \(-2(5m)(7n)=-70mn\). Hence the expression is \((5m-7n)^2\). In contrast, \((5m+7n)^2\) would have the positive middle term \(+70mn\). Exam tip: while identifying a perfect square, always check the sign of the middle term and verify \(2ab\).

First use the identity \((x+a)(x-a)=x^2-a^2\). The expression becomes \((x^2-a^2)(x^2+a^2)\). Applying \((p-q)(p+q)=p^2-q^2\), with \(p=x^2\) and \(q=a^2\), gives \(x^4-a^4\). Option D is the expansion of \((x^2-a^2)^2\), so it is not applicable here. Exam tip: identify conjugate factors step by step and use the difference-of-squares identity.

Use the identity \((a+b)^2+(a-b)^2=2a^2+2b^2\). Here, \(a=2x\) and \(b=3y\), so \((2x+3y)^2+(2x-3y)^2=2(2x)^2+2(3y)^2=8x^2+18y^2\). Subtracting \(8x^2\) gives \(18y^2\). The \(12xy\) and \(24xy\) options are incorrect because the mixed \(xy\) terms cancel when the two squares are added. Exam tip: When \((a+b)^2\) and \((a-b)^2\) occur together, look for cancellation of the mixed terms.

\((x+5)(x-3)=x^2+2x-15\) and \((x-5)(x+3)=x^2-2x-15\). On adding, \(+2x\) and \(-2x\) cancel, giving \(2x^2-30\). \(2x^2+30\) results from incorrectly adding the constant terms. Exam tip: expand each product separately before combining like terms.

Expanding the first product gives \((x+5)(x-3)=x^2+2x-15\). Similarly, \((x-5)(x+3)=x^2-2x-15\). Subtracting the second expression gives \(x^2+2x-15-(x^2-2x-15)=4x\). Hence, the correct answer is \(4x\). The option \(8x\) may result from handling the negative signs while subtracting incorrectly. Exam tip: when a minus sign precedes brackets, change the sign of every term inside the brackets.