Factorisation

The grouping gives (x^2(x-4)-1(x-4)), and (x^2-1) factors. In exams look for both common bracket and difference of squares.

The grouping gives (x^2(2x+3)-4(2x+3)), and (x^2-4) factors. In exams factorise the remaining binomial after grouping.

The grouping gives (x^2(3x-5)+4(3x-5)). In exams do not factor (x^2+4) into real linear factors.

Grouping ((a^3-b^3)+ab(a-b)) gives ((a-b)(a^2+ab+b^2+ab)). In exams check whether the final bracket forms a perfect square.

\((p+3q)^2=p^2+2(p)(3q)+(3q)^2=p^2+6pq+9q^2\), so A is correct. In C, the middle term is negative, so it matches \((a-b)^2\), not \((a+b)^2\). Exam tip: compare the middle term with \(2ab\).

In \(p^2+14p+49\), \(49=7^2\) and the middle term \(14p=2\times p\times7\). Hence it is \((p+7)^2\). In exams, check whether the middle term equals \(2ab\).

By the Factor Theorem, \(x+2\) is a factor only if the polynomial becomes 0 at \(x=-2\). For option A, \(-8+12-4=0\). Option B gives \(-4\), so it is not divisible by \(x+2\). Exam tip: substitute the zero of the linear factor.

First (4x^2-12xy+9y^2=(2x-3y)^2), then apply difference of squares. In exams take the square root of the group correctly.

First (9a^2+12ab+4b^2=(3a+2b)^2), then use (16c^2=(4c)^2). In exams apply difference of squares after forming the perfect square group.

(x^3+8y^3=(x+2y)(x^2-2xy+4y^2)) and (x^2y+2xy^2=xy(x+2y)). In exams simplify inside terms after a common binomial appears.

First treat (x^2+6x+9) as ((x+3)^2). In exams, identify the perfect square and then use difference of squares.

It is the perfect square of the difference of (2a) and (3b). In exams, confirm with the middle term (-2\cdot2a\cdot3b).

(9p^2=(3p)^2) and (16q^2=(4q)^2). In exams, take both square roots and write conjugate factors.

\(81p^2-25q^2=(9p)^2-(5q)^2\). Therefore, using \(a^2-b^2=(a-b)(a+b)\), it factorises as \((9p-5q)(9p+5q)\). Option C is \((9p-5q)^2\), a perfect-square trinomial rather than a difference of two squares. Exam tip: a difference of squares has two square terms separated by a minus sign.

Here, \(49p^2=(7p)^2\) and \(64q^2=(8q)^2\), so the expression is a difference of perfect squares. \(49p^2+64q^2\) is a sum, not a difference. Exam tip: check both terms for perfect squares first.

\(16a^2-25b^2=(4a)^2-(5b)^2\) is a difference of two squares. Using \(p^2-q^2=(p+q)(p-q)\), it factorises as \((4a+5b)(4a-5b)\), a pair of conjugate binomials. Option C factorises as \((4a-5b)^2\); its two binomials are identical, not conjugates. Exam tip: look for a minus sign between two perfect-square terms when identifying conjugate factors.

For \(x^2-10x+25\), the discriminant is \(b^2-4ac=(-10)^2-4(1)(25)=0\), so it has equal linear factors: \((x-5)^2\). Option B has a positive discriminant and therefore two distinct factors. Exam tip: a zero discriminant indicates a repeated factor.

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

A perfect-square trinomial has the form \(a^2-2ab+b^2=(a-b)^2\). In option A, \(x^2=(x)^2\), \(36y^2=(6y)^2\), and the middle term is \(-2\cdot x\cdot 6y=-12xy\). Thus, \(x^2-12xy+36y^2=(x-6y)^2\). In option D, the final term is \((6y)^2\), but its middle term should be \(-12xy\), not \(-6xy\). Exam tip: take the square roots of the first and last terms, then check whether the middle term is \(\pm2ab\).

\(9a^2-25b^2=(3a)^2-(5b)^2\), so it factors as \((3a-5b)(3a+5b)\). Option C is a perfect-square trinomial, not a difference of squares. Exam tip: check for two squares separated by a minus sign.