Factorisation

\(x^2-16=x^2-4^2\) is a difference of squares, so it factorises as \((x-4)(x+4)\). \((x-4)^2=x^2-8x+16\), not the given expression. Exam tip: spot \(a^2-b^2\).

\(a^2-b^2\) is a difference of two squares, so it factorises as \((a-b)(a+b)\). Option C is the identity for the square of a difference, not a difference of squares. Exam tip: first identify both terms as squares.

This is a difference of squares because \(9a^2=(3a)^2\) and \(25b^2=(5b)^2\). Applying \(x^2-y^2=(x-y)(x+y)\) with \(x=3a\) and \(y=5b\) gives \((3a-5b)(3a+5b)\). Option A expands to \(9a^2-30ab+25b^2\), so it does not match the given expression. Exam tip: for a difference of squares, write the difference and sum of the two square roots as factors.

(4x^2=(2x)^2), (9=3^2), and (12x=2\cdot2x\cdot3). In exams, check all three conditions of a perfect square.

The expression is a perfect-square trinomial: \(25r^2=(5r)^2\) and \(9=3^2\). Also, \(-30r=-2\times 5r\times 3\). Using \(a^2-2ab+b^2=(a-b)^2\), with \(a=5r\) and \(b=3\), we get \(25r^2-30r+9=(5r-3)^2\). In option A, the middle term would be \(+30r\). Exam tip: take the square roots of the first and last terms, then verify the middle term using \(\pm2ab\).

Group the terms: \(ax+ay+bx+by=a(x+y)+b(x+y)\). The binomial \((x+y)\) is common, so taking it out gives \((a+b)(x+y)\). Expanding option B gives \(ab+ay+bx+xy\), which is not the given expression. Exam tip: expand the factorised form once to verify that it reproduces the original expression.

It can be written as (x(p+q)+y(p+q)). In exams, group the first two and last two terms correctly.

A perfect-square trinomial has the form \(a^2+2ab+b^2\). Here, \(x^2+10x+25=x^2+2\cdot x\cdot5+5^2\), so it is \((x+5)^2\). Option B has the same middle term but its constant term is not \(25\). Exam tip: halve the middle coefficient and square it.

\(x^2-16=x^2-4^2\) is a difference of squares. Using \(a^2-b^2=(a-b)(a+b)\), it becomes \((x-4)(x+4)\). Expanding \((x-4)^2\) gives \(x^2-8x+16\), not the given expression. Exam tip: identify the identity before factorising.

(7-6=1) and (7\cdot(-6)=-42). In exams, use the middle-term sign to decide the larger number's sign.

\(x^2-12x+36=(x-6)^2=(x-6)(x-6)\), so its two linear factors are identical. In option B, the constant term is not \(36\). Exam tip: in a perfect-square trinomial, the middle term is \(\pm2ab\).

The greatest common factor is (6xy). In exams, take the common part of both numbers and variables.

Both terms have (5ab) common. In exams, after taking the greatest common factor, check the bracket.

In \(10a^2b\) and \(15ab^2\), 5, \(a\), and \(b\) are common, so the HCF is \(5ab\). Option D has HCF \(6ab\). Exam tip: use the smallest power of every common variable.

\((x-3y)^2=x^2-6xy+9y^2\), since the middle term is \(-2\cdot x\cdot3y\) and the last term is \((3y)^2\). Option A has a positive middle term. Exam tip: check \(a^2-2ab+b^2\).

This is the perfect square of the difference of (x) and (y). In exams, identify ((x-y)^2) from the negative middle term.

Taking \(a=x\) and \(b=5\) gives \(x^2+2(x)(5)+5^2=x^2+10x+25\). Hence it factorises as \((x+5)^2\). The expression \(x^2-10x+25\) has a negative middle term. Exam tip: always check the sign of the middle term.

(49x^2=(7x)^2) and the middle term is (2\cdot7x\cdot y). In exams, check square roots of first and last terms.

(81a^2=(9a)^2) and the middle term is (-2\cdot9a\cdot b). In exams, keep the negative sign carefully.

Expanding \((2x+1)(x+2)\) gives \(2x^2+4x+x+2=2x^2+5x+2\), so option A is correct. Option B expands to \(2x^2+4x+2\), which does not have the middle term \(5x\). Exam tip: Multiply the factors and check both the leading and middle terms.