Algebraic identities

The difference-of-squares identity is \(p^2-q^2=(p+q)(p-q)\), so A is correct. Option C is \((p-q)^2\), not a sum–difference product. Exam tip: look for two square terms separated by a minus sign.

The expression has the form \(a^2+b^2-2ab=(a-b)^2\), where \(a=2x+7\) and \(b=2x-1\). Thus, \(a-b=(2x+7)-(2x-1)=8\), so the value is \(8^2=64\). Option 8 is only the difference \(a-b\), not its square. Exam tip: first identify the \(a^2+b^2-2ab\) pattern, then square the difference.

Use the expansion of a square of three terms: \\( (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\\). The expression subtracts \(a^2+b^2+c^2+2ab\) from this complete expansion. The three square terms cancel, and the \(2ab\) term also cancels.

The terms left are \(2bc+2ca\), so option A is correct. It is useful to keep the subtraction bracket together; every term inside it is removed from the expanded square. Option B incorrectly retains \(2ab\), option C omits the \(2bc\) term, and option D omits the \(2ca\) term. The answer does not require assigning numerical values to the variables.

The correct answer is A. Treat x, 2y, and z as the three terms in the square. Using (p+q+r)^2=p^2+q^2+r^2+2pq+2pr+2qr, we obtain (x+2y+z)^2=x^2+4y^2+z^2+4xy+2xz+4yz. The expression subtracts x^2+4y^2+z^2, so those three square terms cancel and only the pair-product terms remain: 4xy+2xz+4yz. Option B uses coefficients that are too small, as if the factor 2 had not been applied correctly to every pair. Options C and D assign the coefficients to the wrong pairs. The governing concept is the square of a sum of three terms and cancellation of like terms.

The (xy)-term is (2\cdot3x\cdot(-2y)=-12xy) and the (yz)-term is (2\cdot(-2y)\cdot z=-4yz). Their sum is (-12xy-4yz).

The coefficient of (x) is the sum of pairwise products of constants. (4(-1)+4(-6)+(-1)(-6)=-22), so check options carefully.

The coefficient of (x) is the sum of pairwise products of constants. (3(-4)+3(2)+(-4)(2)=-14), so be careful with options.

For (x), choose one (2x) and two constants. The coefficient is (2(1\cdot3+1\cdot(-5)+3\cdot(-5))=-34).

Write \(a^4-b^4\) as a difference of squares: \((a^2)^2-(b^2)^2=(a^2-b^2)(a^2+b^2)\). Hence \(a^2+b^2\) is a factor. Exam tip: for fourth powers, first check whether they can be treated as squares.

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

Binomials with the same terms but opposite signs give \((p+q)(p-q)=p^2-q^2\), so A is correct. \((p+q)^2\) is a perfect square and contains the middle term \(2pq\). Exam tip: first check for opposite signs.

In \((m+n)\) and \((m-n)\), only the middle sign changes, so they are conjugate binomials. Their product is \((m+n)(m-n)=m^2-n^2\). In exams, look for identical terms with opposite signs.

In \((a+b)^2=a^2+2ab+b^2\), taking \(a=x\) and \(b=5y\) gives the middle term \(2\cdot x\cdot5y=10xy\). Hence A is \((x+5y)^2\). Option C represents a square of a difference. Exam tip: always check the middle coefficient.

Using the identity a^2-2ab+b^2, (4x-5)^2=16x^2-40x+25. On subtracting (16x^2+25), both 16x^2 and 25 cancel, leaving -40x. Option D, -20x, may result from incorrectly taking half of the middle term 2(4x)(5)=40x. In exams, always check the sign and factor 2 in the middle term while expanding a square.

\((p+q)\) and \((p-q)\) are conjugate binomials. On multiplying, the middle terms cancel: \((p+q)(p-q)=p^2-q^2\). Options B and C give perfect squares. Exam tip: look for binomials differing only in sign.

Treat it as ( ((x+y)+z)^2-((x+y)-z)^2 ). The result is (4z(x+y)=20z), so (40z) is not correct.

\((a-b)^2\) is the square of a real number, so it is always zero or positive. In contrast, \(a^2-b^2\) can be negative; take \(a=1,b=2\). Exam tip: for “always” statements, look for a square form.

In \((3x+5)(3x-5)\), the two factors have the same terms with opposite signs. Thus take \(a=3x\) and \(b=5\) in the difference-of-squares identity. Exam tip: check for identical terms and opposite signs first.

\(4p^2-25q^2=(2p)^2-(5q)^2\), so it is a difference of squares and factorises as \((2p-5q)(2p+5q)\). The other options are perfect-square trinomials. Exam tip: first check whether both terms are perfect squares with a minus sign.

\((x+2)(x+6)=x^2+8x+12\) and \((x+3)(x+5)=x^2+8x+15\). Hence, their difference is \((x^2+8x+12)-(x^2+8x+15)=-3\). Option 3 is incorrect because the \(x^2\) and \(8x\) terms cancel, so no term containing x remains. Exam tip: while subtracting an expression in brackets, change the sign of every term in the second bracket.