Factorisation

First treat (x^2+6x+9) as ((x+3)^2). In exams, identify the perfect square and then use difference of squares.

It is the perfect square of the difference of (2a) and (3b). In exams, confirm with the middle term (-2\cdot2a\cdot3b).

The direct answer is option A: \(3p-4q)(3p+4q)\). Here \(9p^2=(3p)^2\) and \(16q^2=(4q)^2\), so the expression is \( (3p)^2-(4q)^2\). Apply \(A^2-B^2=(A-B)(A+B)\) to get \( (3p-4q)(3p+4q)\). Option A expands to \(9p^2-12pq+12pq-16q^2=9p^2-16q^2\), so the middle terms cancel. Option B expands to \(9p^2-7pq-16q^2\), adding an unwanted term. Option C gives \(9p^2-24pq+16q^2\), not a difference of squares. Option D has leading coefficient \(81p^2\), not \(9p^2\). Memory cue: when two squares have a minus sign, use conjugate factors.

\(81p^2-25q^2=(9p)^2-(5q)^2\). Therefore, using \(a^2-b^2=(a-b)(a+b)\), it factorises as \((9p-5q)(9p+5q)\). Option C is \((9p-5q)^2\), a perfect-square trinomial rather than a difference of two squares. Exam tip: a difference of squares has two square terms separated by a minus sign.

Here, \(49p^2=(7p)^2\) and \(64q^2=(8q)^2\), so the expression is a difference of perfect squares. \(49p^2+64q^2\) is a sum, not a difference. Exam tip: check both terms for perfect squares first.

\(16a^2-25b^2=(4a)^2-(5b)^2\) is a difference of two squares. Using \(p^2-q^2=(p+q)(p-q)\), it factorises as \((4a+5b)(4a-5b)\), a pair of conjugate binomials. Option C factorises as \((4a-5b)^2\); its two binomials are identical, not conjugates. Exam tip: look for a minus sign between two perfect-square terms when identifying conjugate factors.

For \(x^2-10x+25\), the discriminant is \(b^2-4ac=(-10)^2-4(1)(25)=0\), so it has equal linear factors: \((x-5)^2\). Option B has a positive discriminant and therefore two distinct factors. Exam tip: a zero discriminant indicates a repeated factor.

A perfect-square trinomial has the form \(a^2+2ab+b^2=(a+b)^2\). Here, \(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so option A is correct. In option B, the constant term \(20\) is not \(5^2\); in option C, the middle term should be \(2\cdot x\cdot5=10x\), not \(5x\). Exam tip: take the square roots of the first and last terms, then check whether the middle term is twice their product.

A perfect-square trinomial has the form \(a^2-2ab+b^2=(a-b)^2\). In option A, \(x^2=(x)^2\), \(36y^2=(6y)^2\), and the middle term is \(-2\cdot x\cdot 6y=-12xy\). Thus, \(x^2-12xy+36y^2=(x-6y)^2\). In option D, the final term is \((6y)^2\), but its middle term should be \(-12xy\), not \(-6xy\). Exam tip: take the square roots of the first and last terms, then check whether the middle term is \(\pm2ab\).

\(9a^2-25b^2=(3a)^2-(5b)^2\), so it factors as \((3a-5b)(3a+5b)\). Option C is a perfect-square trinomial, not a difference of squares. Exam tip: check for two squares separated by a minus sign.

\(25m^2-n^2=(5m)^2-n^2\) is a difference of two squares, so it factorises as \((5m-n)(5m+n)\). Options A and B are perfect-square trinomials, while D is a sum of squares. Exam tip: first rewrite each term as a square.

For \(x^2+4x+7\), two integers must have product 7 and sum 4. The pair 1 and 7 has sum 8, so no such integer pair exists. In contrast, \(x^2+5x+6=(x+2)(x+3)\). Exam tip: check the required product and sum first.

First take (2) common and then apply difference of squares to (4p^2-q^2). In exams, factorise the final answer completely.

First take (3y) common and then factorise (4x^2-9y^2). In exams, keep checking after taking the common factor.

First (x^4-81=(x^2-9)(x^2+9)), and (x^2-9) factors further. In exams, factor further wherever possible.

\(8x^3+27y^3=(2x)^3+(3y)^3\) is a sum of cubes. Using \(a^3+b^3=(a+b)(a^2-ab+b^2)\), one factor is \(2x+3y\). In contrast, \(8x^3-27y^3\) gives \(2x-3y\). Exam tip: first rewrite terms as perfect cubes.

The direct answer is option A: \(x+y)(x+y-4)\). Notice that the first three terms are \(x^2+2xy+y^2=(x+y)^2\). The remaining terms are \(-4x-4y=-4(x+y)\). Let \(t=x+y\). The expression becomes \(t^2-4t=t(t-4)\). Replacing \(t\) gives \( (x+y)(x+y-4)\). Option A matches this exact reasoning. Option B contains \(x-y\), but the original square has \(x+y\), so it changes the signs of the cross terms. Option C is \( (x+y-2)^2\), which also contains an extra constant \(4\) and is not equal. Option D has the wrong second factor and does not produce \(-4x-4y\). Memory cue: take a repeated binomial as one temporary variable.

The first three terms form a perfect square: \(m^2+4mn+4n^2=(m+2n)^2\). The whole expression is therefore \((m+2n)^2-25\). Since \(25=5^2\), apply \(u^2-v^2=(u-v)(u+v)\), with \(u=m+2n\) and \(v=5\). This gives \((m+2n-5)(m+2n+5)\), so option A is correct.

Expanding the factors confirms the result because their product is \((m+2n)^2-5^2=m^2+4mn+4n^2-25\). Option B uses \(m-2n\), which changes the sign of the middle term. C is a sum rather than a difference of squares, and D does not factor the expression in the required way. First identify the hidden square, then apply the appropriate identity.

For \(x^2-2x-1\), the discriminant is \(b^2-4ac=(-2)^2-4(1)(-1)=8\), which is not a perfect square. Hence its roots are not integers, so it has no linear factorisation with integer coefficients. Exam tip: check whether the discriminant is a perfect square first.

Expanding ((5y-3)(y-4)) gives (5y^2-23y+12), so it is not correct. The correct option is ((5y-4)(y-3)).

In \(p^2+2pq+q^2\), the first and last terms are \(p^2\) and \(q^2\), while the middle term is \(2pq\). Hence it equals \((p+q)^2\). Exam tip: always check both the sign and coefficient of the middle term.

In A, grouping gives \(p(x+y)+q(x+y)\), so \(x+y\) is the common binomial in both groups. In B, the binomials are \(x+y\) and \(x-y\). Exam tip: factor each pair first and then compare the remaining binomials.

In \(x^2-10x+25\), \(25=(-5)^2\) and the middle term is \(2\times x\times(-5)=-10x\). Hence it is \((x-5)^2\). \(x^2-25\) is a difference of squares, not a perfect-square trinomial. Exam tip: verify the middle term using \(2ab\).

Grouping the terms gives \(x^2-3x-xy+3y=x(x-3)-y(x-3)\). Since \((x-3)\) is the common binomial factor, the expression is \((x-3)(x-y)\), or equivalently \((x-y)(x-3)\). Option B produces \(+xy\) on expansion, but the given expression has \(-xy\). Exam tip: in grouping, factor each pair first and then identify the common binomial factor.

First take (x) common and then factorise (x^2+3x+2). In exams, do not leave factorisation incomplete.