Factorisation

First, write \(x^2+7x+12=(x+3)(x+4)\). Thus the expression becomes \((x+3)(x+4)-y^2\). Using \(a^2-b^2=(a-b)(a+b)\), it factorises as \((x+3-y)(x+4+y)\). In option B, the \(y^2\) term becomes positive, so it is incorrect. Exam tip: after factorising, expand once to check that the \(-y^2\) term is obtained.

This is the perfect square of (a), (b), and (-c). Exam tip: decide signs from (2ab), (-2ac), and (-2bc).

First write (x^4+2x^2y^2+y^4) as ((x^2+y^2)^2). Exam tip: treat larger expressions as one combined square.

(a^6-b^6) can be fully factored using both cube and square identities. Exam tip: factor both (a^3-b^3) and (a^3+b^3) further.

Expanding \((3x-4y)(x+2y)\) gives \(3x^2+6xy-4xy-8y^2=3x^2+2xy-8y^2\). Therefore, option A is correct. In option B, the middle term becomes \(-2xy\), not \(2xy\). Exam tip: expand the factors and check the coefficient of the middle term.

\((3a-4b)^2=9a^2-2(3a)(4b)+16b^2=9a^2-24ab+16b^2\), so A is a perfect-square trinomial. In B, the last term should be \(16b^2\). Exam tip: verify the middle term as \(-2xy\).

The expression (x^3+27y^3) has factor ((x+3y)). The inside becomes (x^2-12xy+9y^2), giving ((x-3y)^2).

Write (4a^2-4ab+b^2) as ((2a-b)^2). Exam tip: then use (25c^2=(5c)^2) for difference of squares.

It is ((3x^2)^2-2\cdot3x^2\cdot5+5^2). Exam tip: identify perfect squares involving (x^4) too.

Writing it as ((x+3)^2-(2y-3)^2) gives the simplified form ((x-2y+6)(x+2y)). Exam tip: simplify the final answer.

Split the middle term as \(5xy=8xy-3xy\), since \(8\times(-3)=-24=2\times(-12)\). Thus, \(2x^2+5xy-12y^2=2x^2+8xy-3xy-12y^2\), which factors by grouping to \((2x-3y)(x+4y)\). Option B gives a middle term of \(-5xy\). Exam tip: expand the factors briefly to check both the middle term and the constant term.

First write (x^4+4x^2y^2+4y^4) as ((x^2+2y^2)^2). Exam tip: arranging the terms correctly helps.

In option A, \(16x^2=(4x)^2\) and \(25y^2=(5y)^2\). Its middle term is \(-40xy=-2(4x)(5y)\), so \(16x^2-40xy+25y^2=(4x-5y)^2\). In option B, the last term is \(20y^2\), but \(25y^2\) is required for a perfect-square trinomial with this middle term. Exam tip: take the square roots of the first and last terms, then check whether the middle term equals \(\pm2ab\).

\((3a-b)(2a+3b)=6a^2+9ab-2ab-3b^2=6a^2+7ab-3b^2\), so A is correct. In B, the middle term becomes \(-7ab\). Exam tip: always verify the middle term after multiplication.

It is ((5m)^3+(6n)^3). Exam tip: in sum of cubes, the middle term of the second bracket is negative.

If \(x-2y\) is a factor, substituting \(x=2y\) must make the polynomial zero. For option A, \(3(2y)^2-2(2y)y-8y^2=12y^2-4y^2-8y^2=0\). Exam tip: test a linear factor by using its zero value.

The expression can be rearranged as \(x^2+2x+1-4y^2\). Here, \(x^2+2x+1=(x+1)^2\) and \(4y^2=(2y)^2\). Therefore, it is \((x+1)^2-(2y)^2\). Applying the identity \(a^2-b^2=(a-b)(a+b)\) gives \((x+1-2y)(x+1+2y)\). Option B would contain \((x-1)^2\), which gives a \(-2x\) term instead of the required \(+2x\) term. Exam tip: rearrange the expression to spot perfect squares before applying the difference of squares identity.

First write (49a^2-70ab+25b^2) as ((7a-5b)^2). Exam tip: then apply difference of squares with (c^2).

First write (x^4-2x^2y^2+y^4) as ((x^2-y^2)^2). Exam tip: treat the compound part as one unit.

Write the expression as \(a^2-(b^2-2bc+c^2)\). Since \(b^2-2bc+c^2=(b-c)^2\), it becomes \(a^2-(b-c)^2\). Applying \(x^2-y^2=(x-y)(x+y)\) gives \((a-b+c)(a+b-c)\). Option B represents \(a^2-(b+c)^2\), which produces a negative \(2bc\) term. Exam tip: identify the perfect-square trinomial first and then use the difference-of-squares identity.