Factorisation

The correct answer is A: \((a+b-c)(a+b+c)\). First use the perfect-square identity \(a^2+2ab+b^2=(a+b)^2\). The expression therefore becomes \((a+b)^2-c^2\). Now apply the difference-of-squares identity \(X^2-Y^2=(X-Y)(X+Y)\), taking \(X=a+b\) and \(Y=c\). This gives \((a+b-c)(a+b+c)\). Option A is correct because expanding the conjugate factors returns \((a+b)^2-c^2\), which is the original expression. Option B uses \(a-b\) instead of \(a+b\), so its middle term would be \(-2ab\). Option C is incomplete: it has only one factor and contains \(c\) rather than \(c^2\). Option D is only another unfactorised arrangement; \((a+c)(a-c)=a^2-c^2\), so the full expression is not presented as the required product of factors. Memory cue: complete the perfect square first, then use difference of squares.

The correct answer is A: \((x+1-y)(x+1+y)\). Rearrange the terms so that \(x^2+2x+1\) is together. This is the perfect square \((x+1)^2\), so the expression becomes \((x+1)^2-y^2\). Applying \(X^2-Y^2=(X-Y)(X+Y)\), with \(X=x+1\) and \(Y=y\), gives \((x+1-y)(x+1+y)\). Option A expands to \((x+1)^2-y^2\) and hence reproduces \(x^2+2x+1-y^2\). Option B expands to \(x^2-y^2+1\) and misses the term \(2x\). Option C is the square of \(x+y+1\) and would contain positive \(2xy\) and \(2y\) terms, which are absent. Option D represents \((x-1)^2-y^2\) and would give \(-2x\) rather than \(+2x\). Memory cue: the sign of the middle term tells whether the square is \((x+1)^2\) or \((x-1)^2\).

The correct answer is A: \((2x+3-y)(2x+3+y)\). The first three terms form a perfect square because \(4x^2=(2x)^2\), \(9=3^2\), and \(12x=2(2x)(3)\). Thus \(4x^2+12x+9=(2x+3)^2\). The complete expression is \((2x+3)^2-y^2\). Using \(X^2-Y^2=(X-Y)(X+Y)\) gives \((2x+3-y)(2x+3+y)\). Option A is therefore correct, and expansion confirms it. Option B would come from \((2x-3)^2\), whose middle term is \(-12x\). Option C squares the wrong quantities and does not produce \(4x^2+12x+9\). Option D has \(y\) rather than \(y^2\) and is not the result of the difference-of-squares identity. Memory cue: check the middle term as twice the product of the two square roots.

The correct answer is A: (a+2)(2b+3). Use grouping: 2ab+3a+4b+6 can be written as a(2b+3)+2(2b+3). Both groups contain the common binomial (2b+3), so factor it out to obtain (a+2)(2b+3). Expanding option A gives 2ab+3a+4b+6, exactly the original expression. Option B expands to 2ab+2a+6b+6, so both the a and b coefficients are wrong. Option C expands to 2ab+6a+4b+12, giving incorrect a and constant terms. Option D expands to 2ab+4a+3b+6, so the middle coefficients are interchanged. A useful habit is to group the first two and last two terms, then check whether the binomial factors match. Finally expand the selected product to confirm it.

The correct answer is A: (3x+5)(y-2). Group the expression as (3xy-6x)+(5y-10). Take out the common factor from each group: 3x(y-2)+5(y-2). The binomial (y-2) is common, so the result is (3x+5)(y-2). Expanding option A gives 3xy-6x+5y-10, which matches every term. Option B expands to 3xy+6x-5y-10, changing the signs of the second and third terms. Option C expands to 3xy-15x+2y-10, so its coefficients do not match. Option D expands to 3xy+3x-10y-10, which is also different. The important idea is that the second group must have the same binomial as the first group after taking out its numerical or variable factor. Always expand once as a check.

The correct answer is A: \(4x^2-12x+9\). Use \(A^2-2AB+B^2=(A-B)^2\). Here \(4x^2=(2x)^2\), \(9=3^2\), and \(-12x=-2(2x)(3)\). Therefore the trinomial is \((2x-3)^2\), which is the product of two identical binomials: \((2x-3)(2x-3)\). Option B has a negative constant term, so it cannot be the sum of two squared binomial parts in this pattern. Option C has the wrong middle coefficient: \(-2(2x)(3)\) would be \(-12x\), not \(-6x\). Option D has a negative constant term and also the wrong sign relationship for a perfect square. Memory cue: take square roots of the first and last terms, then check whether the middle term is plus or minus twice their product.

The correct answer is A: \(125x^3+64y^3\). A sum of cubes must have two positive cube terms. Here \(125x^3=(5x)^3\) and \(64y^3=(4y)^3\), so option A is exactly \((5x)^3+(4y)^3\). It can be factorised as \((5x+4y)(25x^2-20xy+16y^2)\) using \(a^3+b^3=(a+b)(a^2-ab+b^2)\). Option B is a difference of cubes, not a sum, so it requires the difference-of-cubes identity. Option C contains squares rather than cubes, so the stated identity does not apply. Option D has \(16y^3\), but 16 is not a perfect cube, so it cannot be written as \((\text{integer times }y)^3\) in the required direct form. The memory cue is to inspect both the exponent and the numerical coefficient: both terms must be perfect cubes with a plus sign.

The correct answer is A: \(9p^2-24pq+16q^2\). The first term is \((3p)^2\) and the last term is \((4q)^2\). Twice their product is \(2(3p)(4q)=24pq\), and the middle term is negative. Therefore the expression matches \(A^2-2AB+B^2\) and equals \((3p-4q)^2\), or \((3p-4q)(3p-4q)\). Option B has \(14q^2\) instead of the required \(16q^2\), so the last term is not the square of \(4q\). Option C has the wrong middle coefficient and the wrong sign; it would need \(+24pq\) for \((3p+4q)^2\). Option D has \(8p^2\), not \((3p)^2\), so its first term is unsuitable. Memory cue: the end terms determine the binomial parts, while the middle term must be plus or minus twice their product.

The correct answer is A: \(x^2-25y^2\). Rewrite \(25y^2\) as \((5y)^2\), so the expression is \(x^2-(5y)^2\). It has the exact form \(A^2-B^2\), with \(A=x\) and \(B=5y\). Therefore it factorises as \((x-5y)(x+5y)\). Option B is a sum of squares, not a difference, so the given identity does not apply. Option C is the perfect square \((x+5y)^2\) and is better handled by the square identity. Option D is the perfect square \((x-5y)^2\), not a difference of two separate squares in the required two-factor pattern. Expanding option A gives \(x^2+5xy-5xy-25y^2=x^2-25y^2\). Memory cue: look for a minus sign between two complete squares; the resulting factors are conjugates with opposite signs.

The correct answer is A: \(25p^2-16q^2\). Both terms are perfect squares: \(25p^2=(5p)^2\) and \(16q^2=(4q)^2\). Because they are separated by a minus sign, the expression is \((5p)^2-(4q)^2\). Applying \(A^2-B^2=(A-B)(A+B)\) gives \((5p-4q)(5p+4q)\). Option B is a sum of squares, so the difference-of-squares identity does not apply. Option C is \((5p-4q)^2\), a perfect-square trinomial, because its middle term is \(-2(5p)(4q)=-40pq\). Option D is \((5p+4q)^2\), another perfect-square trinomial. Thus only A has the required structure of two separate squares with subtraction. Expanding its factors cancels the cross terms and returns the original expression. Memory cue: identify the square roots first, then check the sign between the squares.

\(a^3-b^3=(a-b)(a^2+ab+b^2)\). On expansion, the middle terms cancel, leaving \(a^3-b^3\). Option B has the wrong sign in the middle term. Exam tip: for a difference, the first factor is always \(a-b\).

Both terms have common factor (27), and (216x^3=27\cdot8x^3). Exam tip: taking out the common factor first makes it easier.

The difference-of-cubes identity is \(u^3-v^3=(u-v)(u^2+uv+v^2)\), so the required quadratic factor is \(u^2+uv+v^2\). The factor \(u^2-uv+v^2\) is used for a sum of cubes. Exam tip: for a difference, the middle term of the quadratic factor is positive.

Use the identity \(p^3+q^3=(p+q)(p^2-pq+q^2)\), so A is correct. For \(p^3-q^3\), the quadratic factor has \(+pq\), not \(-pq\). Exam tip: check the middle sign first.

\(9p^2=(3p)^2\) and \(16q^2=(4q)^2\), while the middle term is \(-2(3p)(4q)=-24pq\). Hence it is \((3p-4q)^2\). In exams, always match the sign of the middle term.

Since (x^2-4x+4=(x-2)^2), the expression becomes ((x-2)^2-y^2). Exam tip: use (+4) to form a perfect square.

First (a^4-16b^4=(a^2-4b^2)(a^2+4b^2)), then (a^2-4b^2) factors further. Exam tip: stop only after complete factorisation.

Expanding \((x+m)(x+n)\) gives \(x^2+(m+n)x+mn\). Comparing corresponding terms gives \(m+n=p\) and \(mn=q\). Option B wrongly interchanges the sum and product. Exam tip: match the coefficient of \(x\) with the sum first.

((5x+3)(2x-1)) gives (10x^2-5x+6x-3). Exam tip: identify the small difference for the middle term (x).

((5x+2)(3x-4)) gives (15x^2-20x+6x-8). Exam tip: choose the pair that gives (-2x), not (-14x).

\(x^2+2x+2=(x+1)^2+1\), so it has no linear factors with integer coefficients. In contrast, option A is \((x+2)(x+3)\). Exam tip: test factor pairs of the constant term systematically.

In \(x^3-4x\), take out the common factor \(x\): \(x(x^2-4)\). Now \(x^2-4=x^2-2^2\), a difference of squares. In \(x^3+4x\), the remaining expression is a sum. Exam tip: check the HCF first.

The identity for a difference of cubes is \(p^3-q^3=(p-q)(p^2+pq+q^2)\), so A is the quadratic factor. On expansion, the middle terms cancel. Exam tip: in the quadratic factor for a difference of cubes, the middle term is positive.

We need two factors of \(20\) whose sum is \(-9\): \(-5\) and \(-4\). Thus, \((x-5)(x-4)=x^2-9x+20\). The student's factors give constant term \(-20\), not \(+20\). Exam tip: verify both the middle and constant terms.

It is of the form ((3a)^3+3(3a)^2(2b)+3(3a)(2b)^2+(2b)^3). Exam tip: match the (3AB) terms carefully in cube identities.