Factorisation

\(25m^2-n^2=(5m)^2-n^2\) 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.

First (x^2+2xy+y^2=(x+y)^2) and the rest is (-4(x+y)). In exams, treat the common binomial as a new term.

The first three terms are ((m+2n)^2), and (25=5^2) is being subtracted. In exams, identify the hidden square.

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.

First (x) is common and (2x^2+7x+3=(2x+1)(x+3)). In exams, factor the trinomial after taking the common factor.

First take (a) common and then factor (4a^2-9b^2) by difference of squares. In exams, factorise completely.

Let \(x^2+px+q=(x+a)^2\). Comparing coefficients gives \(p=2a\) and \(q=a^2\), hence \(p^2=(2a)^2=4q\). The condition \(p^2=q\) does not match the middle-term coefficient. Exam tip: square the coefficient of the middle term and compare it with four times the constant term.

The expression can be written as \((a+b)^2-4c^2=(a+b)^2-(2c)^2\). Applying \(x^2-y^2=(x-y)(x+y)\), where \(x=a+b\) and \(y=2c\), gives \((a+b-2c)(a+b+2c)\). Option B would give \((4c)^2=16c^2\), so it is incorrect. Exam tip: treat a complete bracket such as \((a+b)\) as one term when applying an identity.

Write it as (x^2-4y^2+2(x+2y)), so ((x+2y)) is common. In exams, group after difference of squares.