Factorisation

\(9a^2=(3a)^2\) and \(16b^2=(4b)^2\), so \(9a^2-16b^2\) is a difference of perfect squares. \(9a^2+16b^2\) has a plus sign instead. Exam tip: check that both terms are squares and are separated by subtraction.

The expression \(y^2+12y+36\) matches the identity \(a^2+2ab+b^2=(a+b)^2\). Taking \(a=y\) and \(b=6\), we get \(2ab=2\times y\times6=12y\) and \(b^2=36\). Hence, its factorised form is \((y+6)^2\). Expanding \((y-6)^2\) gives a middle term of \(-12y\), so it is not correct. Exam tip: for a perfect-square trinomial, use the square root of the constant term and choose the bracket sign from the middle term.

The expression has the standard perfect-square form \(a^2-2ab+b^2\). The first term \(m^2\) is the square of m, while the last term 25 is the square of 5. To check the middle term, calculate \(-2(m)(5)=-10m\). This agrees exactly with the given expression. Therefore, the two quantities are m and 5, with a minus sign between them.

Applying the identity gives \(m^2-10m+25=(m-5)^2\). On expanding, \((m-5)^2=m^2-5m-5m+25=m^2-10m+25\), so the factorisation is verified. Hence option B is correct. Option A would give a positive middle term, \(+10m\), and options C and D do not reproduce the original three terms when expanded.

Here (4+(-5)=-1) and (4\cdot(-5)=-20). In exams use opposite signs for a negative constant term.

First it becomes (a(x+y)+b(x+y)), then ((x+y)) is taken common. In exams look for a common binomial.

\(9a^2-16b^2=(3a)^2-(4b)^2\), so it factorises as \((3a+4b)(3a-4b)\). Option D is a perfect square, \((3a-4b)^2\), not a difference of squares. Exam tip: identify the signs first.

(4x^2=(2x)^2) and (9=3^2), so it factors as a difference of squares. In exams identify both squares.

The greatest common factor in both terms is (3x). In exams it is better to take out the greatest common factor.

Since \(x^2-25=x^2-5^2\), use \(a^2-b^2=(a-b)(a+b)\). \(x^2+25\) is a sum, not a difference. Exam tip: look for a minus sign between square terms.

The governing identity is the square of a difference: a² − 2ab + b² = (a − b)². Match the given expression with this pattern by taking a = r and b = s. The first term r² matches a², the middle term −2rs matches −2ab, and the final term s² matches b². Therefore r² − 2rs + s² = (r−s)², so option B is correct. Expanding the selected factorisation gives r² − rs − rs + s² = r² − 2rs + s², confirming it exactly. Option A has a positive middle term, option C gives the difference of squares r²−s², and option D produces different coefficients and terms. The signs and the middle coefficient are decisive in this factorisation.

(16a^2=(4a)^2), so the difference of squares applies. In exams take the square roots correctly.

\(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so it is a perfect-square trinomial. Option B has 24 instead of the required constant term 25. Exam tip: compare the middle term with \(2ab\).

(-6+(-7)=-13) and ((-6)(-7)=42). In exams use two negative signs when the constant is positive and the middle term is negative.

Group the expression as \((5xy-10x)+(3y-6)\). This gives \(5x(y-2)+3(y-2)\). Taking the common binomial \((y-2)\) gives \((5x+3)(y-2)\). On expansion, option B gives incorrect signs for the \(x\)- and \(y\)-terms. Exam tip: expand your final factors once to check the signs of the middle and constant terms.

First take common (2), then (a^2-4b^2) is a difference of squares. In exams apply the identity after taking the common factor.

(9x^2=(3x)^2), (16y^2=(4y)^2), and the middle is (24xy). In exams identify the two square terms first.

(25p^2=(5p)^2), (9q^2=(3q)^2), and the middle term is negative. In exams choose the correct square by the sign.

\(x^2-9=x^2-3^2\) is a difference of squares. Using \(a^2-b^2=(a-b)(a+b)\), it becomes \((x-3)(x+3)\). In contrast, \((x-3)^2=x^2-6x+9\). Exam tip: check the middle term by expanding.

Here (5+(-3)=2) and (5\cdot(-3)=-15). In exams check opposite signs when the constant is negative.

Area = length × breadth. Since x² + 9x + 20 = (x + 4)(x + 5), dividing by the given side x + 4 gives x + 5. The factor x + 16 would not produce the required middle term. Exam tip: check that the two numbers add to 9 and multiply to 20.