Factorisation

Use the identity \(a^2+2ab+b^2=(a+b)^2\) with \(a=2m\) and \(b=n\). Then \(a^2=4m^2\), \(2ab=4mn\), and \(b^2=n^2\). Hence, \(4m^2+4mn+n^2=(2m+n)^2\). The close distractor \((2m-n)^2\) expands to give a middle term of \(-4mn\), so it is incorrect. Exam tip: for a perfect square, check both the sign and coefficient of the middle term.

The greatest common factor of \(7x^2\) and \(14xy\) is \(7x\). Factoring it out gives \(7x^2+14xy=7x(x+2y)\). The closest distractor, \(14x(x+y)\), expands to \(14x^2+14xy\), so its first term is different. Exam tip: expand the brackets to check whether the original expression is obtained.

Group the expression as \(ab-ac+db-dc=a(b-c)+d(b-c)\). The common binomial factor is \((b-c)\), so the factorised form is \((b-c)(a+d)=(a+d)(b-c)\). Expanding option B gives \(ab+ac-db-dc\), which is different from the given expression. Exam tip: expand the factorised form once to check the signs.

\((x+a)^2=x^2+2ax+a^2\). Taking \(a=5\) gives the middle term \(2\times x\times5=10x\) and constant term \(25\). Hence A is a perfect square; B has 20 instead of 25. Exam tip: use the square root of the constant to check the middle term.

(-6+3=-3) and ((-6)(3)=-18). In exams choose the sign of the larger magnitude according to the middle term.

\(p^2-q^2\) is a difference of two squares, so \(p^2-q^2=(p-q)(p+q)\). Options B and C expand squares, not a difference of squares. Exam tip: first check whether both terms are perfect squares.

The governing concept is factorisation by taking out the greatest common factor. For 12x²y and 18xy², the greatest common numerical factor is 6. Both terms contain x and y, so the greatest common monomial factor is 6xy. Dividing the first term by 6xy gives 2x, and dividing the second term by 6xy gives 3y. The original minus sign must remain, so the result is 6xy(2x−3y). Option B is algebraically equivalent after further factorisation, but it has not taken out the greatest common factor. Option C also leaves a common factor inside, while D changes subtraction into addition. Hence option A is the simplest correct form.

\(a^2-16b^2=a^2-(4b)^2\), so it matches \(x^2-y^2=(x-y)(x+y)\). Options C and D are perfect-square trinomials instead. In exams, check for two squares separated by a minus sign.

First take out (5), then (x^2-4) is a difference of squares. In exams keep checking after taking the common factor.

\(a^2-b^2\) is a difference of two perfect squares, so \(a^2-b^2=(a-b)(a+b)\). Option B is a perfect-square trinomial, \((a+b)^2\), not a difference of squares. Exam tip: check for a minus sign between square terms first.

The governing method is factorisation by grouping. Group terms that share useful common factors: 2xy + 6y + 5x + 15 = (2xy+6y) + (5x+15). Factoring each group gives 2y(x+3) + 5(x+3). Both terms now contain the common binomial x+3, so taking it outside produces (2y+5)(x+3). Therefore option A is correct. Expansion verifies the result: (2y+5)(x+3) = 2xy + 6y + 5x + 15. Option B changes the signs of the 5x and constant terms. Option C pairs the variables incorrectly and does not recreate the original expression, while option D gives unrelated products. Grouping is effective because it creates the same binomial factor in both groups.

In \(x^2+10x+25\), \(25=5^2\) and the middle term is \(10x=2\times x\times5\). Hence it has the form \((x+5)^2\). Option B has constant term 20, so it does not fit the identity. Exam tip: compare the middle term with \(2ab\).

(4x^2=(2x)^2), (9=3^2), and the middle is (12x). In exams check all three conditions of the identity.

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

Expanding \((3x+1)(2x+3)\) gives \(6x^2+9x+2x+3=6x^2+11x+3\). Therefore, option A is correct. Option B has the constant term \(3\), but its middle term is \(19x\), so it is not the correct factorisation. Exam tip: verify a factorisation by expanding the binomials and checking both the middle term and the constant term.

A difference of squares has the form \(a^2-b^2\). Using \(a^2-b^2=(a-b)(a+b)\), its factors are binomials with opposite signs. Exam tip: first check that both terms are perfect squares.

(-9+5=-4) and ((-9)(5)=-45). In exams decide the sign of the larger magnitude from the middle term.

The grouping gives (5p(2q+3)-2(2q+3)). In exams you may need to take a negative common factor from the second group.

In \((a-b)^2=a^2-2ab+b^2\), a middle term occurs and the last term is positive. Here, use \(a^2-b^2=(a-b)(a+b)\): \((7x-5y)(7x+5y)\). Exam tip: check signs before choosing an identity.

A perfect-square trinomial has the form \(p^2-2pq+q^2\). Here, \(-10a=-2\times a\times5\) and \(25=5^2\), so \(a^2-10a+25=(a-5)^2\). Exam tip: check whether the middle term equals \(2pq\).