Factorisation

The correct answer is D: (3x + 1)². Use the identity (a + b)² = a² + 2ab + b². Let a = 3x and b = 1. Then (3x + 1)² = (3x)² + 2(3x)(1) + 1² = 9x² + 6x + 1. Every term matches the given trinomial. Option A begins with 81x², not 9x², so its first term is wrong. Option B gives (3x)² − 2(3x)(1) + 1 = 9x² − 6x + 1, so its middle sign is wrong. Option C gives x² + 6x + 9, with the first and constant terms interchanged in value. Option D is therefore the only complete match. Memory cue: the middle term of (a + b)² must be twice the product of the two terms, namely 2ab.

The correct answer is A: x² + 6x + 9 = (x + 3)². A square of a binomial has the form a² + 2ab + b². Here the first term is x² = x², the last term is 9 = 3², and the required middle term is 2 × x × 3 = 6x. Thus all three terms fit perfectly. Option B has the same first two terms but its constant is 8 instead of 9, so it cannot be (x + 3)². Option C is a difference of squares, (x − 3)(x + 3), not the square of one binomial. Option D has no middle term and is not a binomial square. The safe check is to take the square roots of the first and last terms and double their product. If that does not equal the middle term, the trinomial is not a perfect square.

The correct answer is B: b = 12. Since 36 = 6², a perfect-square trinomial with first term x² and constant term 36 must have the form (x + 6)² or (x − 6)². Expanding (x + 6)² gives x² + 2(x)(6) + 6² = x² + 12x + 36, so b = 12. The positive value is appropriate because the expression is written with a positive middle coefficient. Option A is only the square root of 36, not twice the square root, so it gives the wrong middle term. Option C is too large and would not equal 2 × 1 × 6. Option D is the constant term itself, not the required middle coefficient. Memory cue: when the leading coefficient is 1, the middle coefficient of a perfect square is twice the square root of the constant term, with a sign matching the bracket.

The correct answer is A: 7x² + 21x. A quantity is a common factor only if it divides every term of the expression. Divide both terms in A by 7x: 7x² ÷ 7x = x and 21x ÷ 7x = 3. Therefore 7x² + 21x = 7x(x + 3), proving that 7x is common to both terms. In option B, 14x is divisible by 7x, but the constant term 7 is not divisible by x. In option C, 7x² is divisible by 7x, but 14 has no factor x. In option D, 21x² is divisible by 7x, but the constant 7 is not. Thus each wrong option fails because one term lacks x. Memory cue: test a proposed common factor against every term, not just the first one.

The correct answer is A: x² + 10x + 25 = (x + 5)². Compare the expression with a² + 2ab + b². The first term is x², so take a = x. The constant term is 25 = 5², so take b = 5. The required middle term is 2ab = 2 × x × 5 = 10x, exactly as shown in option A. Option B has the wrong constant term: the matching value should be 25, not 20. Option C has the correct first and constant terms, but its middle term should be 10x rather than 8x. Option D fails because neither the middle coefficient nor the constant term forms the required pair. Thus A alone is a perfect square. Memory cue: check first square, last square, and middle term equal to twice the product of their square roots.

The correct answer is A: 5(3a − 4) = 15a − 20. Use the distributive property: multiply the outside factor 5 by every term inside the bracket. First, 5 × 3a = 15a. Next, 5 × (−4) = −20. Combining the results gives 15a − 20, exactly the original expression. Option A is therefore a correct verification. Option B multiplies the first term correctly but leaves the second term as −4 instead of multiplying it by 5. Option C treats −4 as though it were −5, producing −25. Option D adds 5 and 3a rather than multiplying 5 by 3a, so it does not follow distribution. Memory cue: when removing brackets, the outside factor must multiply every term inside, including negative terms.

The direct answer is D: (m + n)². Start with the identity a² + 2ab + b² = (a + b)². In the given expression, a = m and b = n. Therefore, m² + 2mn + n² = (m + n)². We can verify this by expanding: (m + n)² = (m + n)(m + n) = m² + mn + mn + n² = m² + 2mn + n². Option A, (m − n)², gives m² − 2mn + n², so its middle term has the wrong sign. Option B gives m² + 2mn, so it leaves out n². Option C is a difference of squares and gives m² − n², not the original trinomial. A useful memory cue is that a positive middle term 2mn indicates the plus-square identity; a negative middle term indicates the minus-square identity.

The direct answer is A: (c − 2)(c − 4). For a monic quadratic c² + bc + d, find two numbers whose product is d and whose sum is b. Here the constant term is 8 and the middle coefficient is −6. The numbers −2 and −4 have product (−2)(−4) = 8 and sum −2 + (−4) = −6. Hence c² − 6c + 8 = (c − 2)(c − 4). Expanding confirms it: c² − 4c − 2c + 8 = c² − 6c + 8. Option B has factors +2 and +4, producing c² + 6c + 8, so the middle sign is wrong. Option C has product 8 but sum −9, giving c² − 9c + 8. Option D has product −8, not +8, and gives c² − 2c − 8. The memory cue is: a positive constant with a negative middle term usually requires two negative factors.

The direct answer is B: 6y(3x − 4). To find the greatest common factor, compare the numerical and variable parts of both terms. The greatest common divisor of 18 and 24 is 6. Both terms contain y, while only the first contains x, so x cannot be part of the common factor. Thus the greatest common factor is 6y. Divide each term by 6y: 18xy ÷ 6y = 3x and 24y ÷ 6y = 4. Keeping the subtraction sign gives 6y(3x − 4). Option A is mathematically equal to the original expression, but 3y is not the greatest common factor because 6y can still be taken out; therefore it does not answer the question as asked. Option C expands to 18xy − 432y, and option D expands to 432xy − 24y. A good check is to multiply the factor outside the brackets back into every term.

The direct answer is C: (x − 1)(x + 1). This expression is a difference of two squares. The identity is A² − B² = (A − B)(A + B). Write x² − 1 as x² − 1². Here A = x and B = 1, so x² − 1² = (x − 1)(x + 1). Multiplication verifies the result: (x − 1)(x + 1) = x² + x − x − 1 = x² − 1. Option A, x(x − 1), expands to x² − x, so it has an unwanted middle term. Option B, (x − 1)², expands to x² − 2x + 1. Option D, (x + 1)², expands to x² + 2x + 1. The common confusion is to use a square identity without checking the sign. A difference of squares produces two conjugate factors with opposite signs, not two identical binomial factors.

The direct answer is A: x² + 14x + 49. The identity a² + 2ab + b² = (a + b)² requires the first and last terms to be perfect squares and the middle term to equal twice the product of their square roots. Here x² = x² and 49 = 7². Twice their product is 2 × x × 7 = 14x, exactly the middle term. Therefore x² + 14x + 49 = (x + 7)². Option B has 36 = 6², but 2 × x × 6 = 12x, not 14x. Option C has the correct last term 49, but its middle term is 7x instead of 14x. Option D has 42 as the constant term, which is not 7², and it also fails the perfect-square pattern. The safe method is to check all three parts: first square, twice-the-product middle term, and second square.

The direct answer is C: (q + 1)(q + 5). For q² + bq + c, find two numbers whose sum is b and product is c. The required numbers must have sum 6 and product 5. They are 1 and 5. Therefore q² + 6q + 5 = (q + 1)(q + 5). Expanding gives q² + 5q + q + 5 = q² + 6q + 5, confirming the answer. Option A uses 2 and 3; their product is 6 and their sum is 5, so it gives q² + 5q + 6. Option B gives q² − 6q + 5, with the wrong sign in the middle. Option D gives q² + 11q + 30, so both the middle and constant terms are wrong. The memory cue is to check two conditions, not one: the factor numbers must add to the middle coefficient and multiply to the constant term.

The direct answer is D: (6p − 7q)(6p + 7q). Recognise 36p² − 49q² as a difference of two squares. Since 36p² = (6p)² and 49q² = (7q)², apply A² − B² = (A − B)(A + B). Thus 36p² − 49q² = (6p)² − (7q)² = (6p − 7q)(6p + 7q). Multiplying the two factors gives (6p)² − (7q)², because the cross terms cancel. Option A uses 36p and 49q instead of their square roots, so it is much too large and expands incorrectly. Option B is a square of a difference and would create a middle term −84pq. Option C does not use matching square roots and cannot produce the given terms. Option D is correct because it has the same two square roots and opposite signs. A useful cue is: subtraction between two perfect squares leads to conjugate factors.

The direct answer is A: 5xy(3x + 4y). Factorisation by common factor begins by finding what every term contains. The coefficients 15 and 20 have greatest common factor 5. Both terms contain x and y, but the first has x² and the second has y², so only one x and one y can be taken out safely. The greatest common factor is therefore 5xy. Divide: 15x²y ÷ 5xy = 3x, and 20xy² ÷ 5xy = 4y. Hence 15x²y + 20xy² = 5xy(3x + 4y). Option B is algebraically equivalent to the original expression but does not take out the greatest common factor, because 5x leaves a common y inside both terms. Option C gives 15x²y + 300xy², and option D gives 300x²y + 20xy. Multiplying back is the reliable check for any factorisation.

The direct answer is B: (s − 6)². Use the identity (a − b)² = a² − 2ab + b². In this expression, s² is the first square, 36 = 6² is the last square, and the middle term is −12s = −2 × s × 6. Therefore s² − 12s + 36 = (s − 6)². Expanding option B gives s² − 6s − 6s + 36 = s² − 12s + 36. Option A, (s + 6)², gives s² + 12s + 36, so its middle sign is wrong. Option C expands to s² − 9s − 36, which has both an incorrect middle coefficient and an incorrect constant sign. Option D expands to s² − 15s + 36, so its middle coefficient is wrong. The important cue is the sign of the middle term: a negative middle term in a perfect-square trinomial requires subtraction inside the bracket.

The direct answer is C: 7 and 8. To factor x² + 15x + 56 as (x + a)(x + b), the two numbers must satisfy a + b = 15 and ab = 56. The pair 7 and 8 satisfies both conditions: 7 + 8 = 15 and 7 × 8 = 56. Therefore x² + 15x + 56 = (x + 7)(x + 8). Expanding gives x² + 8x + 7x + 56 = x² + 15x + 56. Option A, 6 and 9, has the correct sum but product 54. Option B, 4 and 14, has sum 18 and product 56, so its middle term would be 18x. Option D, 5 and 10, has sum 15 but product 50. The common mistake is checking only the sum; both the sum and product must match. For a monic quadratic, the pair’s sum gives the middle coefficient and its product gives the constant term.

The direct answer is D: \(9a^2(3a-1)\). To factor an expression, first find the greatest common factor of every term. The numerical coefficients 27 and 9 have greatest common factor 9. The common variable part is \(a^2\), because the smaller power among \(a^3\) and \(a^2\) is \(a^2\). Thus the greatest common factor is \(9a^2\). Dividing each term by it gives \(27a^3\div9a^2=3a\) and \(-9a^2\div9a^2=-1\). Therefore the expression is \(9a^2(3a-1)\). A is not the greatest factor and does not reproduce the original expression. B gives \(27a^3-27a^2\), so its second coefficient is wrong. C simplifies to \(27a^3-9a^3\), also incorrect. D expands to \(27a^3-9a^2\). Memory cue: take the numerical HCF and the lowest common power of each variable.

The direct answer is A: \((7u+v)^2\). Use the identity \((p+q)^2=p^2+2pq+q^2\). The first term \(49u^2\) is \((7u)^2\), and the last term \(v^2\) is \(v^2\). The middle term predicted by the identity is \(2(7u)(v)=14uv\), exactly the middle term in the question. Therefore \(49u^2+14uv+v^2=(7u+v)^2\). A is correct because its expansion gives all three terms with the required signs. B would begin with \(2401u^2\), not \(49u^2\). C expands to \(49u^2-14uv+v^2\), so its middle sign is wrong. D expands to \(u^2+14uv+49v^2\), so the first and last terms are interchanged. A useful check is that the middle term must be twice the product of the square roots of the first and last square terms.

The direct answer is B: \((y-3)(y-8)\). For a monic quadratic \(y^2+by+c\), find two numbers whose product is \(c\) and whose sum is \(b\). Here the product must be 24 and the sum must be -11. The required numbers are -3 and -8 because \((-3)(-8)=24\) and \(-3+(-8)=-11\). Hence \(y^2-11y+24=(y-3)(y-8)\). A gives \(y^2+11y+24\), so its middle sign is wrong. C gives \(y^2-10y+24\), not the required middle coefficient. D gives \(y^2+14y+24\), with both the sum and sign wrong. The sign pattern is helpful: a positive constant and a negative middle coefficient require two negative numbers. Always check both the product and the sum.

The direct answer is B: \(8mn(4m-5n)\). First find the greatest common factor. The numerical HCF of 32 and 40 is 8. Both terms contain m and n, and the lowest powers are m^1 and n^1, so the common factor is \(8mn\). Divide each term by it: \(32m^2n\div8mn=4m\), while \(-40mn^2\div8mn=-5n\). Thus \(32m^2n-40mn^2=8mn(4m-5n)\). A has the wrong sign and would produce a positive second term. C is an incomplete numerical factorisation; it expands to \(32m^2n-20mn^2\), not the original. D gives \(32m^2n-80mn^2\). B expands exactly to the given expression. A reliable habit is to multiply the outside factor by every term inside the bracket.

The direct answer is C: \((x+2y)^2\). Compare the expression with the identity \((p+q)^2=p^2+2pq+q^2\). Take \(p=x\) and \(q=2y\). Then \(p^2=x^2\), \(2pq=2(x)(2y)=4xy\), and \(q^2=4y^2\). These are exactly the three terms given, so the expression is \((x+2y)^2\). A would expand to \(x^2+8xy+16y^2\), so its middle and last terms are too large. B gives \(x^2-4xy+4y^2\), with the wrong middle sign. D uses the difference-of-squares identity and gives \(x^2-4y^2\), which has no middle term. The quick check is that the middle term must be twice the product of the two quantities being squared.

The direct answer is A: \(x^2-49\). The identity for a difference of squares applies when an expression has two square terms separated by subtraction. Since \(49=7^2\), we can write \(x^2-49=x^2-7^2\). Therefore \(x^2-49=(x+7)(x-7)\). A is correct because it has exactly the required form. B is a sum of squares, not a difference of squares, so this identity does not apply. C is a perfect-square trinomial: \(x^2+14x+49=(x+7)^2\), not a two-term difference. D contains \(x^3\), which is not the square of x, so it does not match the identity. Memory cue: look for two square terms and a minus sign; then use the sum and difference of their roots.

The direct answer is A: \((t+3)(t-2)\). For \(t^2+bt+c\), find two numbers whose product is the constant term and whose sum is the coefficient of t. Here the product must be -6 and the sum must be 1. The numbers 3 and -2 satisfy both conditions: \(3(-2)=-6\) and \(3+(-2)=1\). Hence \(t^2+t-6=(t+3)(t-2)\). A expands to \(t^2-2t+3t-6=t^2+t-6\). B uses -3 and 2, whose sum is -1, so it gives \(t^2-t-6\). C uses 1 and -6, whose product is -6 but sum is -5. D uses 6 and -1, whose sum is 5. Check both sum and product, not just one.

The direct answer is B: \((x-7)^2\). Use the identity \((p-q)^2=p^2-2pq+q^2\). The first term is \(x^2\), so take p=x. The constant term is \(49=7^2\), so take q=7. The middle term predicted by the identity is \(-2(x)(7)=-14x\), exactly the given middle term. Therefore \(x^2-14x+49=(x-7)^2\). A expands to \(x^2+14x+49\), so its middle sign is wrong. C expands to \(x^2+35x-686\), matching neither the middle nor constant term. D expands to \(x^2-98x+2401\), so both coefficients differ. A useful check is that a negative middle term requires the minus version of the perfect-square identity when the outer square roots are x and 7.

The direct answer is A: 3a(a + 4b). Find the greatest common factor of the terms 3a² and 12ab. Both contain 3, and both contain a, so 3a is common. Divide each term by 3a: 3a² ÷ 3a = a and 12ab ÷ 3a = 4b. Therefore 3a² + 12ab = 3a(a + 4b). Option B is mathematically equal to the original expression, but it does not take out the variable a, so it is not the complete greatest-common-factor form expected here. Option C is also equivalent but takes out only a, not the greatest common factor 3a. Option D expands to 6a² + 12ab, which is different. Common-factor check: multiply the outside factor by every term inside.