Factorisation

Expanding and grouping the terms gives (-(a-b)(b-c)(c-a)). Exam tip: handle signs carefully in cyclic expressions.

\((3x-2)(2x+1)=6x^2+3x-4x-2=6x^2-x-2\), so B is correct. The student's factors give a middle term of \(+x\), not \(-x\). Exam tip: expand factors to verify the middle term.

Since \(49p^2=(7p)^2\) and \(64q^2=(8q)^2\), option A has the form \(a^2-b^2\), the difference of squares. Option B is a sum of squares, not this identity. Exam tip: check for two perfect squares separated by a minus sign.

Here, \(8p^3=(2p)^3\) and \(q^3=q^3\). Applying the sum-of-cubes identity gives the stated factorisation. Exam tip: in \(a^3+b^3\), the middle term of the second factor has a negative sign.

The direct answer is option A: \(5x-2y-6)(5x-2y+6)\). First, \(25x^2-20xy+4y^2=(5x-2y)^2\), because the square of \(5x-2y\) is \(25x^2-20xy+4y^2\). Therefore the expression is \( (5x-2y)^2-6^2\). Using \(A^2-B^2=(A-B)(A+B)\), we obtain option A. Option B uses \(5x+2y\), whose square has a positive \(20xy\), not a negative one. Option C does not reproduce the original quadratic and is not the required identity. Option D is only \( (5x-2y+6)^2\), which has no subtraction of \(36\) and produces extra terms. Memory cue: a perfect square minus a square factors as difference and sum.

In option A, grouping gives 3x(x+2)+5(x+2). The common binomial is (x+2), so it factorises as (3x+5)(x+2). Exam tip: check that both groups contain the same binomial.

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 expression matches the identity \((p+q+r)^2=p^2+q^2+r^2+2pq+2pr+2qr\), with the three parts chosen as \(a\), \(b\), and \(-c\). Therefore it becomes \((a+b-c)^2\). The signs are determined by the cross terms: \(+2ab\), \(-2ac\), and \(-2bc\). Hence option D is correct.

Expanding verifies it: \((a+b-c)^2=a^2+b^2+c^2+2ab-2ac-2bc\). Option B would make all cross terms positive. Options A and C place a negative sign on a different variable, changing the signs of the corresponding mixed terms. Matching every cross term is the safest way to identify a perfect-square identity.

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).

Direct answer: Option B,
(2a-b-5c)(2a-b+5c). Start by grouping the first three terms: 4a²-4ab+b²=(2a-b)², because (2a-b)²=4a²-4ab+b². Also, 25c²=(5c)². Therefore the whole expression is (2a-b)²-(5c)². Using u²-v²=(u-v)(u+v), it becomes (2a-b-5c)(2a-b+5c). Option A uses 2a+b; its square gives +4ab, not -4ab. Option B has the required negative middle term and is correct. Option C does not follow a valid identity and generally expands to different terms. Option D is only one square and has no subtraction of 25c². Memory cue: a negative middle term in the first trinomial means (2a-b)², then use difference of squares.

Notice it matches a perfect square: \(9x^4-30x^2+25\) has first term \((3x^2)^2\), last term \(5^2\), and middle term \(2\cdot(3x^2)\cdot(-5)=-30x^2\). Therefore \(9x^4-30x^2+25=(3x^2-5)^2\).

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.

The direct mathematical answer is C: \((5m+6n)(25m^2-30mn+36n^2)\), but the supplied key should be checked carefully because this is a sum of cubes. We have \(125m^3=(5m)^3\) and \(216n^3=(6n)^3\). Using \(a^3+b^3=(a+b)(a^2-ab+b^2)\), we obtain \((5m+6n)(25m^2-30mn+36n^2)\), so C is correct. Option A has \(5m-6n\), which belongs to a difference pattern and is wrong for a sum. Option B has the correct first factor but a positive middle term; that would produce the wrong mixed terms. Option C has the correct first factor, negative middle term, and both square terms. Option D uses 125 and 216 directly instead of their cube roots and is not the identity. Memory cue: sum of cubes means plus in the first bracket and minus in the middle of the second.

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.

The direct answer is B: \(7a-5b-c)(7a-5b+c)\). The first three terms form \(49a^2-70ab+25b^2=(7a-5b)^2\), since \(2(7a)(-5b)=-70ab\). Also \(c^2=c^2\), so the whole expression is \((7a-5b)^2-c^2\). Using \(u^2-v^2=(u-v)(u+v)\) gives option B. Option A has \(7a+5b\), which creates a positive \(70ab\). Option C does not reproduce the terms through the required identity. Option D is only a square and does not subtract \(c^2\). Thus B has both the correct inner square and the correct difference-of-squares signs. Check the middle sign carefully.

The direct answer is C: \((x^2-y^2-z)(x^2-y^2+z)\). First recognize \(x^4-2x^2y^2+y^4\) as \((x^2-y^2)^2\), using \(A^2-2AB+B^2=(A-B)^2\) with \(A=x^2\), \(B=y^2\). The expression becomes \((x^2-y^2)^2-z^2\). Now apply the difference-of-squares identity \(P^2-Q^2=(P-Q)(P+Q)\), where \(P=x^2-y^2\) and \(Q=z\). This gives C. Option A uses \(x^2+y^2\), which changes the original middle sign. Option B has only linear x and y terms and cannot reproduce fourth-degree terms. Option C follows both identities exactly. Option D is a square, but the original expression is a difference of two squares and generally is not one square. Memory cue: first make the four-term part a square, then use difference of squares.