Factorisation

The direct answer is A: \((p-5q)(p+5q)\). Rewrite the second term as a square: \(25q^2=(5q)^2\). The expression is therefore \(p^2-(5q)^2\), which matches \(a^2-b^2=(a-b)(a+b)\), with \(a=p\) and \(b=5q\). Hence the factorisation is \((p-5q)(p+5q)\). Option B does not use the correct square root of \(25q^2\) and expands to unrelated terms. Option C is a square of a sum; its expansion contains a positive middle term \(10pq\). Option D is a square of a difference; its expansion contains a negative middle term \(-10pq\). The original expression has no \(pq\) term, because the cross terms must cancel. Memory cue: in a difference of squares, keep the same terms but use opposite signs in the two brackets.

The direct answer is B: \((x-2)(x-5)\). We need two numbers whose product is the constant term 10 and whose sum is the coefficient of \(x\), namely \(-7\). The numbers \(-2\) and \(-5\) satisfy both conditions. Thus \(x^2-7x+10=(x-2)(x-5)\). On expansion, \(x^2-5x-2x+10=x^2-7x+10\). Option A gives factors 1 and 10, whose sum is 11, so it produces \(-11x\), not \(-7x\). Option C gives a positive middle term \(+7x\), so its sign is wrong. Option D uses 3 and 4; it gives \(-7x\) but its product is 12, not 10. B is the only option satisfying both tests. Memory cue: the constant term checks the product, while the middle coefficient checks the signed sum.

The direct answer is A: (2r − 9)(2r + 9). First identify the two perfect squares: 4r² = (2r)² and 81 = 9². Thus, 4r² − 81 = (2r)² − 9². Apply the difference-of-squares identity a² − b² = (a − b)(a + b), giving (2r − 9)(2r + 9). Option A can be checked by multiplication: the cross terms are +18r and −18r, so they cancel, leaving 4r² − 81. Option B is incorrect because (2r − 9)² = 4r² − 36r + 81; it has a middle term and a positive constant. Option C does not reproduce the original expression; multiplying it gives 4r² + 27r − 81. Option D expands to 4r² + 36r + 81, so both the middle term and the sign of the constant are wrong. The important distinction is that a difference of two squares produces two conjugate factors, one with subtraction and one with addition. Memory cue: minus between perfect squares means “minus factor, plus factor.”

The direct answer is A: \(a^2+8a+16=(a+4)^2\). Use the identity \((u+v)^2=u^2+2uv+v^2\). The first term is \(a^2\), so the first part is \(a\); the last term is \(16=4^2\), so the second part is 4. The expected middle term is \(2\cdot a\cdot4=8a\), exactly as given. Option B has the correct first two terms but its constant 15 is not \(4^2\), so it is not this perfect square. Option C has first and last terms that suggest \((a+4)^2\), but that would require middle term \(8a\), not \(6a\). Option D has constant 15 and therefore fails the square test; its negative middle term does not fix that. Memory cue: first and last terms must be squares, and the middle term must equal twice their square roots multiplied together.

The direct answer is A: \(3x(2x+5)\). A common factor must divide every term. The terms are \(6x^2\) and \(15x\). Their numerical greatest common factor is 3, and both contain at least one \(x\), so the common factor is \(3x\). Divide each term by \(3x\): \(6x^2\div3x=2x\) and \(15x
div3x=5\). Thus \(6x^2+15x=3x(2x+5)\). Multiplying back confirms the result. Option B omits the common factor \(x\), so it does not reproduce the original expression. Option C is algebraically equal to the original expression but has not meaningfully taken out the greatest common factor. Option D is invalid because 5 does not divide the coefficient 6. Memory cue: take out the greatest numerical common factor and the lowest power of every variable shared by all terms.

The direct answer is A: \((a+d)(b+c)\). Group the terms as \((ab+ac)+(db+dc)\). Take \(a\) from the first group and \(d\) from the second: \(a(b+c)+d(b+c)\). Both terms now contain the common binomial \((b+c)\), so factor it out: \((a+d)(b+c)\). Option A is confirmed by expansion: \(ab+ac+db+dc\). Option B would expand to \(ac+ad+bc+bd\), which has different pairings. Option C gives \(ab+ad+bc+cd\), also different. Option D produces signs unlike the original expression. The key skill is arranging terms so the same bracket appears twice. Memory cue: after grouping, look for an identical binomial factor; if it is not identical, regroup or check the signs.

The direct answer is A: \((x-7y)(x+7y)\). Rewrite \(49y^2\) as \((7y)^2\). Then the expression becomes \(x^2-(7y)^2\), a difference of squares. Applying \(u^2-v^2=(u-v)(u+v)\), with \(u=x\) and \(v=7y\), gives \((x-7y)(x+7y)\). Expanding option A gives \(x^2+7xy-7xy-49y^2=x^2-49y^2\); the cross terms cancel. Option B uses 49y rather than 7y as the square-root part and does not expand correctly. Option C is \((x+7y)^2\), which contains the extra term \(+14xy\). Option D is \((x-7y)^2\), which contains \(-14xy\). Since the original has no \(xy\) term, neither square is suitable. Memory cue: difference of squares gives conjugate brackets, not two identical brackets.

The direct answer is A: \((n-7)(n+4)\). For \(n^2+bn+c\), find two numbers whose product is \(c\) and whose sum is \(b\). Here the product must be \(-28\), and the sum must be \(-3\). The numbers \(-7\) and 4 satisfy these conditions: \((-7)(4)=-28\) and \(-7+4=-3\). Therefore, \(n^2-3n-28=(n-7)(n+4)\). Expanding gives \(n^2+4n-7n-28=n^2-3n-28\). Option B has product \(-28\) but sum 3, producing \(+3n\). Option C has product \(-28\) but sum \(-12\), not \(-3\). Option D has product 28, not \(-28\), and so cannot give the negative constant. Memory cue: a negative constant means the two numbers have opposite signs; use the sign of the middle coefficient to decide which number is negative.

The direct answer is A: (3a + 4)². Begin with the identity p² + 2pq + q² = (p + q)². Here, 9a² = (3a)² and 16 = 4². The middle term produced by these two parts is 2 × 3a × 4 = 24a, exactly the middle term given. Therefore, 9a² + 24a + 16 = (3a)² + 2(3a)(4) + 4² = (3a + 4)². Option A is correct. Option B expands to 9a² − 24a + 16, so its middle sign is wrong. Option C gives 9a² + 40a + 16, not 24a. Option D gives 9a² + 30a + 16, also incorrect. Memory cue: for a perfect-square trinomial, check both end squares and the middle term 2ab.

The direct answer is A: (2x + 5)(y + 3). Use grouping. Put the first two terms together and the last two together: 2xy + 6x + 5y + 15 = 2x(y + 3) + 5(y + 3). Both groups contain the same bracket, (y + 3). Taking it common gives (y + 3)(2x + 5), which is the same as option A. Verify by multiplication: 2x·y + 2x·3 + 5·y + 5·3 = 2xy + 6x + 5y + 15. Option B expands to 2xy + 10x + 3y + 15, so coefficients do not match. Option C gives 2xy + 3x + 10y + 15. Option D gives negative middle terms. Memory cue: in grouping, the repeated bracket is the key signal.

The direct answer is B: (4u − 3v)². Use (a − b)² = a² − 2ab + b². Choose a = 4u and b = 3v. Then a² = 16u², −2ab = −2(4u)(3v) = −24uv, and b² = 9v². Thus (4u − 3v)² expands exactly to the given expression. Option A has a plus sign and therefore produces +24uv. Option C begins with 256u², not 16u², and its cross term is also wrong. Option D expands to 16u² − 32uv − 9v², so both the middle and final terms differ. A negative middle term is the important clue: the two square-root parts must be subtracted. Check the cross term rather than relying only on the first and last terms.

The direct answer is B: (s − 7)(s + 5). For an expression s² + bs + c, find two numbers whose product is c and whose sum is b. Here the product must be −35 and the sum must be −2. The numbers −7 and +5 satisfy both conditions: (−7)(5) = −35 and −7 + 5 = −2. Therefore s² − 2s − 35 = (s − 7)(s + 5). Expanding option A gives s² + 2s − 35, so its middle sign is wrong. Option C gives s² − 36s + 35. Option D gives s² + 34s − 35. The constant term being negative means the two numbers have opposite signs; their negative sum tells us that the negative number has the larger magnitude.

The direct answer is A: (x + y − 5)(x + y + 5). First recognise the initial three terms: x² + 2xy + y² = (x + y)². Also, 25 = 5². Thus the whole expression becomes (x + y)² − 5². Apply the difference-of-squares identity P² − Q² = (P − Q)(P + Q), with P = x + y and Q = 5. This gives (x + y − 5)(x + y + 5). Option B uses (x − y)², whose middle term would be −2xy. Option C does not have factors whose product gives the original terms. Option D changes subtraction into addition and is not a factorised product. Always identify the perfect square before applying the second identity.

The direct answer is A: (2x + 3 − y)(2x + 3 + y). First group the first three terms: 4x² + 12x + 9 = (2x + 3)², because (2x)² = 4x², 2(2x)(3) = 12x, and 3² = 9. The expression therefore becomes (2x + 3)² − y². Now use P² − Q² = (P − Q)(P + Q), taking P = 2x + 3 and Q = y. This gives option A. Option B corresponds to (2x − 3)² − y² and would have −12x. Option C does not reproduce the original middle and constant terms. Option D is a square of a sum and would contain positive y² and additional mixed terms. Memory cue: make the perfect square first, then use difference of squares.

The direct answer is A: (x − 6)(x − 7). For a monic quadratic x² + bx + c, find two numbers whose product is c and whose sum is b. Here their product must be 42 and their sum must be −13. The numbers −6 and −7 work because (−6)(−7) = 42 and −6 + (−7) = −13. Hence x² − 13x + 42 = (x − 6)(x − 7). Option B gives a positive middle term, x² + 13x + 42. Option C uses −3 and −14, whose sum is −17, not −13. Option D uses −2 and −21, whose sum is −23. Memory cue: when the constant is positive and the middle coefficient is negative, both factor signs are often negative; confirm by sum and product.

The direct answer is A: (5m + 3)(n − 2). Factor by grouping. Pair the first two terms and the last two terms: 5mn − 10m + 3n − 6 = (5mn − 10m) + (3n − 6). Take 5m common from the first group, giving 5m(n − 2). Take 3 common from the second group, giving 3(n − 2). Now the identical binomial (n − 2) is common, so the expression becomes (n − 2)(5m + 3), which can be written as (5m + 3)(n − 2). Option B expands to 5mn + 10m − 3n − 6, so its middle signs do not match. Option C expands to 5mn − 2m + 15n − 6, with incorrect coefficients. Option D expands to 5mn − 5m + 6n − 6, also different. Always expand your selected factors as a final check. Memory cue: after grouping, the remaining bracket from both groups must be exactly identical, including its signs.

The direct answer is A. Start with the identity (a − b)² = a² − 2ab + b². Here a = 7p and b = 5q. Therefore, (7p − 5q)² = (7p)² − 2(7p)(5q) + (5q)² = 49p² − 70pq + 25q², exactly the given trinomial. The student incorrectly used the difference-of-squares pattern (a − b)(a + b). Expanding that product gives a² − b², because the two cross terms cancel: (7p − 5q)(7p + 5q) = 49p² − 25q². It has no middle term, so it cannot equal the given expression. Option B is wrong because the proposed product omits −70pq. Option C changes the sign of the middle term; (7p + 5q)² would contain +70pq. Option D is wrong because 25q² is already a square, namely (5q)². Memory cue: in a perfect square trinomial, the middle term is ±2 times the two square roots.

The direct answer is A: (q + 8)(q − 7). For a quadratic of the form q² + bq + c, look for two numbers whose product is c and whose sum is b. Here the product must be −56 and the sum must be 1. The numbers 8 and −7 satisfy both conditions: 8 × (−7) = −56 and 8 + (−7) = 1. Hence q² + q − 56 = q² + 8q − 7q − 56. Grouping gives q(q + 8) − 7(q + 8) = (q + 8)(q − 7). Option A is therefore correct. Option B uses −8 and +7; their product is −56, but their sum is −1, so it would produce q² − q − 56. Option C uses 4 and −14; their product is −56, but their sum is −10. Option D uses −1 and 56; their product is −56, but their sum is 55. The safest check is to multiply the two proposed binomials and compare both the middle coefficient and the constant term. Memory cue: product gives the last term; sum gives the coefficient of the middle term.

The direct answer is A: (a + 4)(a² − 4a + 16). Since 64 = 4³, the expression is a³ + 4³, a sum of cubes. Use the identity u³ + v³ = (u + v)(u² − uv + v²). Taking u = a and v = 4 gives a³ + 4³ = (a + 4)(a² − 4a + 16). To verify, multiply: a(a² − 4a + 16) + 4(a² − 4a + 16) = a³ − 4a² + 16a + 4a² − 16a + 64 = a³ + 64. Option B is the difference-of-cubes pattern and would represent a³ − 64. Option C has the wrong signs in the second bracket; its expansion contains extra terms rather than cancelling them. Option D uses 8, but 8³ is 512, not 64, so it does not match the expression. Memory cue: for a sum of cubes, the signs are plus in the first bracket, then minus and plus in the second bracket.

The direct answer is A: (b − 3)(b² + 3b + 9). Since 27 = 3³, the expression is b³ − 3³, a difference of cubes. Apply u³ − v³ = (u − v)(u² + uv + v²), with u = b and v = 3. This gives (b − 3)(b² + 3b + 9). Expanding confirms it: b(b² + 3b + 9) − 3(b² + 3b + 9) = b³ + 3b² + 9b − 3b² − 9b − 27 = b³ − 27. Option B is the sum-of-cubes pattern and would begin with b³ + 27, not b³ − 27. Option C incorrectly uses 9 instead of the cube root 3; 9³ is not 27. Option D has the correct first factor but wrong signs in the quadratic factor, so its expansion contains an unwanted −6b term. Memory cue: in u³ − v³, the first bracket has a minus, while all three terms of the second bracket are positive.

The direct answer is A: (x + 2)(x + 4). For x² + bx + c, find two numbers whose sum is b and whose product is c. Here we need a sum of 6 and a product of 8. The numbers 2 and 4 work because 2 + 4 = 6 and 2 × 4 = 8. Therefore x² + 6x + 8 = x² + 2x + 4x + 8 = x(x + 2) + 4(x + 2) = (x + 2)(x + 4). Expanding option A gives x² + 4x + 2x + 8 = x² + 6x + 8. Option B gives x² − 6x + 8, so the middle sign is wrong. Option C gives x² + 9x + 8, so its middle coefficient is 9 rather than 6. Option D gives x² + 6x + 9, so its constant term is wrong. A useful habit is to compare both the sum and the product, not just one of them. Memory cue: for a monic quadratic, the factor constants add to the middle coefficient and multiply to the last term.

The direct answer is A: 7xy(x − 3y + 2). Find the greatest common factor of all three terms. The numerical coefficients 7, 21 and 14 have greatest common factor 7. Each term also contains x and y: 7x²y ÷ 7xy = x, −21xy² ÷ 7xy = −3y, and 14xy ÷ 7xy = 2. Therefore 7x²y − 21xy² + 14xy = 7xy(x − 3y + 2). Expanding option A gives 7x²y − 21xy² + 14xy exactly. Option B is algebraically equivalent to A because it has only taken out 7x; it is not the best or greatest-common-factor form requested. Option C is also algebraically equivalent to A after distributing the factor, so it has not extracted the greatest common factor completely. Option D expands to 14x²y − 14xy² + 14xy, which changes the coefficients and is incorrect. The word “best” signals that the greatest common factor should be removed. Memory cue: compare every term and take the largest numerical and variable factor common to all of them.

The direct answer is A: (t − 8)(t + 8). Recognise that 64 = 8², so t² − 64 = t² − 8². This is a difference of squares, and the identity is u² − v² = (u − v)(u + v). Taking u = t and v = 8 gives (t − 8)(t + 8). Expanding confirms the result: t² + 8t − 8t − 64 = t² − 64; the middle terms cancel. Option B has factors whose product is t² − 30t − 64, not the given expression. Option C expands to t² + 16t + 64, which is a perfect square with a positive constant. Option D expands to t² − 16t + 64, also not the given expression and has a positive 64. The minus sign between two squares is essential: it produces two different conjugate factors. Memory cue: difference of squares means “subtract the roots, multiply by their sum”; it is not the same as squaring one binomial.

The direct answer is A: (r + 2s − 3)(r + 2s + 3). First inspect the first three terms. They form a perfect square: r² + 4rs + 4s² = (r + 2s)², because squaring gives r² + 2(r)(2s) + (2s)² = r² + 4rs + 4s². The complete expression is therefore (r + 2s)² − 9 = (r + 2s)² − 3². Apply the difference-of-squares identity u² − v² = (u − v)(u + v), with u = r + 2s and v = 3. This gives option A. Option B uses r − 2s; its square would contain −4rs, not +4rs. Option C does not arise from either matching identity, and multiplication does not reproduce the original terms. Option D is a single square; it would end with +9 rather than −9 and would not represent the given expression. Memory cue: when three terms form a square and a number is subtracted, treat the whole square as one object before using difference of squares.

The direct answer is A: (x + 4)(y − 2). Use grouping. Put the first two terms together and the last two terms together: xy − 2x + 4y − 8 = (xy − 2x) + (4y − 8). Take x common from the first group, giving x(y − 2). Take 4 common from the second group, giving 4(y − 2). Since both terms contain the identical factor (y − 2), take it out: x(y − 2) + 4(y − 2) = (x + 4)(y − 2). Expanding option A gives xy − 2x + 4y − 8, exactly the original expression. Option B expands to xy + 2x − 4y − 8, so both middle signs are wrong. Option C expands to xy − 4x + 2y − 8, with the coefficients attached to x and y interchanged. Option D expands to xy + x − 8y − 8, which also does not match. A common confusion is to choose factors only because their product has the correct first and last terms; the middle terms must also agree. Memory cue: in grouping, search for the same binomial, including its signs, in both groups.