Factorisation

The greatest common factor of 6x and 9 is 3, since 6x = 3(2x) and 9 = 3(3). Therefore, (6x+9) = 3(2x+3). The number 6 is not a factor of 9, 9 is not a factor of 6x, and x is not a factor of 9. Exam tip: check common numerical and variable factors in every term separately.

(4) is common in both terms and (4a+8=4(a+2)). Exam tip: multiply back the factor taken out to check.

The terms are \(xy\) and \(xz\). The factor \(x\) occurs in both terms, so factoring it out gives \(xy+xz=x(y+z)\). \(y\) occurs only in the first term and \(z\) only in the second; \(yz\) is not a factor of either term. Exam tip: Identify the number or variable that appears in every term before factorising.

(5) is common in both terms and (15\div5=3). Exam tip: keep the negative sign correctly inside the bracket.

In option A, \(6m^2=3m\times2m\) and \(9mn=3m\times3n\). Thus, \(3m\) is a common factor of both terms. In option B, \(6n\) does not contain \(m\), so it is not divisible by \(3m\). Exam tip: for a common factor, both the numerical factor and the variable must occur in every term.

\(x^2-49=x^2-7^2\), so it is of the form \(a^2-b^2=(a-b)(a+b)\). Hence, its factorisation is \((x-7)(x+7)\). \(x^2+49\) is a sum of squares, whereas \(x^2+14x+49=(x+7)^2\) is a perfect-square trinomial. Exam tip: look for two perfect-square terms separated by a minus sign before applying the difference-of-squares identity.

(x^2-9=x^2-3^2), so the difference of squares identity is used. Exam tip: identify both terms as squares.

(16=4^2), so (x^2-16=(x-4)(x+4)). Exam tip: in difference of squares, write both sum and difference factors.

This is a perfect square trinomial because the middle term is (2ab). Exam tip: check the roots of the first and last terms.

\(x^2-25=x^2-5^2\), so it is a difference of two perfect squares and factorises as \((x+5)(x-5)\). \(x^2+25\) has a plus sign, not a difference. Exam tip: check for perfect squares separated by a minus sign.

\(a^2+2ab+b^2=(a+b)^2\), so it is a perfect-square binomial expansion. In option C, the last term is negative, so it does not match this identity. Exam tip: check for the middle term \(2ab\).

This is the difference-of-squares identity: \(a^2-b^2=(a-b)(a+b)\). Therefore, \(p^2-q^2=(p-q)(p+q)\), so option A is correct. The closest distractor, \((p-q)^2\), expands to \(p^2-2pq+q^2\), which has a middle term and a different sign for \(q^2\). Exam tip: Factor a difference of two squares as ‘difference × sum’.

The common factor in both \(3x\) and \(3y\) is 3. Taking 3 outside gives \(3x+3y=3(x+y)\). The option \(3(x-y)\) is incorrect because expanding it gives \(3x-3y\). Exam tip: Expand your factorised expression once to check whether it reproduces the original expression.

The greatest common factor of \(7a^2\) and \(-14a\) is \(7a\). Hence, \(7a^2-14a=7a(a-2)\). So, after taking \(7a\) outside, \(a-2\) remains in the bracket. \(7(a-2)\) is incorrect because the common factor \(a\) has not been taken out. Exam tip: Verify factorisation by multiplying the factors to recover the original expression.

(2+3=5) and (2\cdot3=6). Exam tip: find two numbers whose sum gives the middle term and product gives the last term.

\(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so it is a perfect-square trinomial. In option B, the constant term should be \(5^2=25\), not 20. Exam tip: match the middle term with \(2ab\).

\(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so it is a perfect-square trinomial. In the other options, the square of half the middle-term coefficient does not equal the constant term. Exam tip: check \(a^2+2ab+b^2\).

The identity for a difference of squares is a² − b² = (a − b)(a + b). Here, x² − 25 is x² − 5², so it fits this identity. x² + 25 is a sum of squares. Exam tip: check for two perfect squares separated by a minus sign.

\(x^2-25=x^2-5^2\), which is a difference of two squares. Applying \(a^2-b^2=(a-b)(a+b)\) with \(a=x\) and \(b=5\) gives \((x-5)(x+5)\). Expanding option B gives \(x^2+10x+25\), while option D gives \(x^2-10x+25\), so neither is correct. Exam tip: take the square root of the constant term and write the difference and sum as the two factors.

The expression is a difference of squares: \(9x^2-1=(3x)^2-1^2\). Applying \(a^2-b^2=(a-b)(a+b)\) gives \((3x-1)(3x+1)\). Options A and C expand to expressions containing middle terms \(-6x\) and \(+6x\), respectively, which are absent here. Exam tip: first check whether both terms can be written as perfect squares.