Factorisation

Both \(m^2\) and \(-6m\) have \(m\) as a common factor. Factoring out \(m\) gives \(m^2-6m=m(m-6)\). Therefore, option B is correct. Expanding option A gives \(m^2+6m\), so the sign of the second term is incorrect. Exam tip: Verify a factorisation by expanding it and comparing it with the original expression.

The greatest common factor of \(6xy\) and \(9xz\) is \(3x\). Taking \(3x\) outside gives \(6xy\div 3x=2y\) and \(9xz\div 3x=3z\). Therefore, \(6xy+9xz=3x(2y+3z)\). Option D expands to \(18xy+9xz\), so it is incorrect. Exam tip: expand the brackets after factorising to verify the original expression.

\(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so it is a perfect-square trinomial. In option B, the middle term is 10x, but the constant term should be \(5^2=25\). Exam tip: match the form \(a^2+2ab+b^2\).

\(x^2+12x+36=x^2+2\times x\times6+6^2\) is a perfect-square trinomial. Using \(a^2+2ab+b^2=(a+b)^2\), with \(a=x\) and \(b=6\), we get \((x+6)^2\). The close distractor \((x-6)^2\) expands to a middle term of \(-12x\), so it is incorrect. Exam tip: take the square root of the constant term and check whether twice its product with the first term’s root equals the middle term.

\(9p^2-16q^2=(3p)^2-(4q)^2\), so it is a difference of two perfect squares. \(9p^2+16q^2\) is a sum, not a difference. Exam tip: first identify whether both terms are perfect squares.

This expression is a difference of squares: \(25x^2=(5x)^2\) and \(36=6^2\). Using \(a^2-b^2=(a-b)(a+b)\), with \(a=5x\) and \(b=6\), gives \(25x^2-36=(5x-6)(5x+6)\). If \((5x-6)^2\) were used, it would produce a middle term of \(-60x\), which is not present. Exam tip: in a difference of squares, the two factors have the same terms but opposite signs between them.

Both terms, 7p and 7q, contain the common factor 7, so 7p + 7q = 7(p + q). Although a² − b² can also be factorised, it uses the difference-of-squares identity. In exams, first check the common factor in every term.

\(p^2-q^2\) is a difference of two squares, so \(p^2-q^2=(p-q)(p+q)\). Options C and D are perfect-square trinomials. In exams, check the sign between the squares first.

\(p^2-25=p^2-5^2\), so it is a difference of squares and factorises as \((p+5)(p-5)\). \(p^2+25\) is a sum of squares. In exams, check whether the constant is a perfect square.

The expression is a perfect-square trinomial: \(4x^2=(2x)^2\) and \(9=3^2\). Also, \(-2\times 2x\times 3=-12x\), so \(4x^2-12x+9=(2x-3)^2\). Option A gives a middle term of \(+12x\), so it is not correct. Exam tip: In \(a^2-2ab+b^2=(a-b)^2\), always check the sign of the middle term.

The greatest common factor of \(6x\) and \(9\) is \(3\). Taking \(3\) outside gives \(6x+9=3(2x+3)\), so option A is correct. Expanding the close distractor, \(3(2x+9)\), gives \(6x+27\), not the given expression. Exam tip: verify factorisation by expanding the brackets.

In \(x^2+5x\), \(x\) is the common factor of both terms. Taking it out gives \(x(x+5)\), since \(x(x+5)=x^2+5x\). Option \(5(x+1)\) expands to \(5x+5\), so it is not correct. Exam tip: multiply the factors to check whether you get the original expression.

\(a^2-b^2\) is a difference of two squares. Using \(x^2-y^2=(x-y)(x+y)\), with \(x=a\) and \(y=b\), gives \((a-b)(a+b)\). The expansion of \((a-b)^2\) is \(a^2-2ab+b^2\), so it is not correct. Exam tip: for a difference of squares, write the difference factor first and the sum factor next.

A perfect-square trinomial has the form \((x+k)^2=x^2+2kx+k^2\). Thus \(b=2k\), so \(k=\frac{b}{2}\) and \(c=k^2=\left(\frac{b}{2}\right)^2\). Exam tip: halve the middle coefficient and square it to check the constant term.

This is a perfect-square trinomial: \(p^2-12p+36=p^2-2\times p\times6+6^2\). Therefore, \(p^2-12p+36=(p-6)^2\), so option B is correct. Expanding \((p+6)^2\) gives a middle term of \(+12p\), so it is not correct. Exam tip: use the identity \(a^2-2ab+b^2=(a-b)^2\).

For x^2+5x+6, we need two integers whose product is 6 and sum is 5. The integers 2 and 3 satisfy this, so it factorises as (x+2)(x+3). Exam tip: check the required sum and product for the middle and constant terms.

\(x^2-49=x^2-7^2\), so it matches \(a^2-b^2=(a-b)(a+b)\). Options B and D are perfect-square trinomials, while C is a sum of squares. Exam tip: check for two squares separated by a minus sign.

\(x^2+10x+25=x^2+2\cdot x\cdot5+5^2\), so it equals \((x+5)^2\). Option D is also a square pattern, but its negative middle term gives \((x-5)^2\). Exam tip: always check the sign of the middle term.

(-3-4=-7) and ((-3)(-4)=12). In exams, for positive constant and negative middle term, use both negative signs.

\(4x^2-25=(2x)^2-5^2\), so it is a difference of squares and factorises as \((2x-5)(2x+5)\). Option C is a perfect-square trinomial, not a difference. Exam tip: check for two squared terms separated by subtraction.