Factorisation

\(x^2-49=x^2-7^2\), which matches \(a^2-b^2=(a-b)(a+b)\). Therefore, it factorises as \((x-7)(x+7)\). Option C is a perfect-square trinomial, \((x+7)^2\), not a difference of squares. Exam tip: check that both terms are perfect squares with a minus sign between them.

(p^2-q^2=(p-q)(p+q)) and (-5p+5q=-5(p-q)). In exams, identify common ((p-q)) with signs.

The expression simplifies to (x^2+4x+3), so the factors are ((x+1)(x+3)). First simplify the given terms carefully.

For x^2+x+1, the discriminant is b^2-4ac = 1-4 = -3, not a perfect square, so it has no linear factors with integer coefficients. The others factor as a difference of squares, (x+2)(x+3), and (x-2)^2. Exam tip: check the discriminant first.

For \(x^2+7x+10\), the discriminant is \(b^2-4ac=49-40=9\), a perfect square. Its roots are \(-5\) and \(-2\), so it factors as \((x+5)(x+2)\). In exams, check whether the discriminant is a perfect square.

A perfect-square trinomial has the form \(a^2-2ab+b^2=(a-b)^2\). Here, \(9p^2=(3p)^2\), \(4q^2=(2q)^2\), and \(-12pq=-2(3p)(2q)\). Therefore, \(9p^2-12pq+4q^2=(3p-2q)^2\). In option B, the last term is \(2q^2\), not \(4q^2\), so it is not a perfect square. Exam tip: take the square roots of the first and last terms, then check whether the middle term is \(\pm2ab\).

First take (4) common and (3x^2-5xy+2y^2=(3x-2y)(x-y)). In exams, factor the trinomial after the common factor.

Treat it as ((x^2-y^2)+5(x-y)+6) and use (u=x-y). In exams, keep the remaining (x+y) term carefully.

In \(ax+ay+bx+by=a(x+y)+b(x+y)\), both groups contain the common binomial \((x+y)\), giving \((a+b)(x+y)\). In option B, the binomials are different. Exam tip: group terms in pairs and check for an identical factor.

Using difference of squares gives ((2y)(2x)=4xy). In exams, write ((A^2-B^2)) as ((A-B)(A+B)).

A perfect-square trinomial has the form \(a^2+2ab+b^2\). In option A, \(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so it is a perfect square. In option B, the middle term is \(10x\), but the constant term is \(20\) rather than \(25\), so it is not a perfect square. Exam tip: take the square roots of the first and last terms and check whether the middle term is twice their product.

Difference of squares directly gives (A-B) and (A+B). In exams, you may simplify it to ((2a-3b)(4a-b)).

First (x^2+4xy+4y^2=(x+2y)^2). In exams, rearrange terms and identify the perfect square.

The first three terms are ((4a-3b)^2), and (25c^2=(5c)^2). In exams, use two-step factorisation.

Writing it as ((x+y)^2+4(x+y)-4z^2) does not directly factor as option (D). Verify by expansion carefully.

In \(x^2+10x+25\), \(25=5^2\) and the middle term is \(2\cdot x\cdot5=10x\). Hence it is \((x+5)^2\). The expression ending in 20 does not fit this pattern. Exam tip: verify \(b^2=4ac\).

((3a+b)(a-2b)) gives (3a^2-6ab+ab-2b^2). In exams, check that the middle terms add to (-5ab).

In option A, \(49a^2=(7a)^2\) and \(64b^2=(8b)^2\). Thus, \(49a^2-64b^2=(7a)^2-(8b)^2\), which is a difference of two perfect squares. Its factorisation is \((7a-8b)(7a+8b)\). Option B is a sum of squares, while in options C and D the second term is not a perfect square. Exam tip: before using \(x^2-y^2=(x-y)(x+y)\), verify that both terms are perfect squares.

First (4x^2+4xy+y^2=(2x+y)^2), and the rest is (-3(2x+y)). In exams, take the common binomial outside.

First (x^2-2xy+y^2=(x-y)^2) and the remaining part is (-4(x-y)). In exams, treat the common binomial as one term.