Factorisation

First take out common (3), then factorise (a^2-9b^2). In exams stop only after complete factorisation.

In x^2 + 10x + 25, the first and last terms are x^2 and 5^2, while the middle term is 2 × x × 5 = 10x. Hence it is (x + 5)^2. In option B, the constant should be 25, not 20. Exam tip: match the middle term with 2ab.

(4x^2=(2x)^2), (9y^2=(3y)^2), and the middle term is negative. In exams the sign of the middle term is decisive.

Here \(49a^2=(7a)^2\) and \(64b^2=(8b)^2\). Using \(x^2-y^2=(x-y)(x+y)\) gives \((7a-8b)(7a+8b)\). Squaring \((7a-8b)\) would also produce \(-112ab\), which is absent. Exam tip: for a difference of squares, use conjugate factors.

Group the terms: \(xy+2x+3y+6=x(y+2)+3(y+2)\). The common binomial is \((y+2)\), so the factorised form is \((y+2)(x+3)=(x+3)(y+2)\). Expanding option B gives \(xy+3x+2y+6\), so its middle terms do not match the original expression. Exam tip: expand the chosen factors once to verify the original expression.

(64a^2=(8a)^2), (25b^2=(5b)^2), and (80ab=2\cdot8a\cdot5b). In exams match all three conditions of a perfect square.

Expanding gives ( (2x+5)(2x-3)=4x^2+4x-15 ). Exam tip: expand the option and check both the middle term and constant term.

First take out common (3), then factorise (x^2-4) using difference of squares. In exams apply the identity after taking the common factor.

Here (-7+(-8)=-15) and ((-7)(-8)=56). In exams use two negative signs when the constant is positive and the middle term is negative.

The grouping gives (4a(b+2)+5(b+2)), and ((b+2)) is common. In exams take out the common bracket when it appears.

On grouping, \(ax+ay+bx+by=a(x+y)+b(x+y)\). Since \((x+y)\) is common to both terms, taking it out gives \((a+b)(x+y)\). Expanding option B gives \(ax+ab+xy+by\), which is not the given expression. Exam tip: while grouping, pair terms so that the same binomial appears as a common factor.

Group the terms as \(px-py+qx-qy=p(x-y)+q(x-y)\). Since \((x-y)\) is common to both terms, taking it out gives \((p+q)(x-y)\). Option A expands to \(px+py-qx-qy\), which is different from the given expression. Exam tip: expand your factorised form once to verify the signs.

First take (3) common and split (a^2-4b^2) by difference of squares. Exam tip: check further factorisation after taking common factor.

The identity is \(A^2-B^2=(A-B)(A+B)\), so the expression must be a subtraction of two perfect squares. \(A^2+B^2\) does not factorise by this identity. In exams, check the square terms and the minus sign first.

In A, \(49a^2=(7a)^2\) and \(16b^2=(4b)^2\), with a minus sign between them. Hence it has the form \(U^2-V^2\). Option B is a sum. Exam tip: check for two square terms separated by a minus sign.

In \((a-b)^2=a^2-2ab+b^2\), the middle term is negative. Here, \(2\times7p\times5q=70pq\), so option A is \((7p-5q)^2\). Option D has a positive middle term and matches \((a+b)^2\). Exam tip: always check the sign of the middle term.

Here \(4a^2=(2a)^2\) and \(9b^2=(3b)^2\), while the middle term is \(-2(2a)(3b)=-12ab\). Thus it is \((2a-3b)^2\). Exam tip: always check the sign of the middle term.

Check multiplication carefully: ((2x+5)(x+3)=2x^2+11x+15). So the correct factorisation is ((2x+5)(x+3)).

\(9x^2+24x+16=(3x+4)^2\), so it is a product of identical binomials. Check: \(24^2=4\times9\times16=576\). In option B, \(4ac=540\). Exam tip: test \(b^2=4ac\) for a perfect-square trinomial.

This matches \(x^2-2xy+y^2=(x-y)^2\), with \(x=7a\) and \(y=5b\). The middle term is \(-2(7a)(5b)=-70ab\). The student used a difference-of-squares product, which cancels the middle term. Exam tip: always verify the middle-term sign.