Factorisation

The direct answer is A: \(x^2+10x+25=(x+5)^2=(x+5)(x+5)\). Use the identity \(u^2+2uv+v^2=(u+v)^2\). Here, \(u=x\) and \(v=5\), so the first term is \(x^2\), the last term is \(5^2=25\), and the middle term is \(2(x)(5)=10x\). All three parts match exactly. In B, the last term is 20 rather than 25, so it is not this perfect square. C is a difference of squares, not a trinomial or a square of one binomial. D has the correct-looking first and last terms, but its middle term should be 10x, not 5x. A common confusion is checking only the first and last terms; the middle term must also equal twice the product of the square roots.

The direct answer is C: \(4m^2+12mn+9n^2=(2m+3n)^2\). Apply \(u^2+2uv+v^2=(u+v)^2\). The first term is \(4m^2=(2m)^2\), the last term is \(9n^2=(3n)^2\), and the middle term is \(2(2m)(3n)=12mn\). Thus every term agrees with the identity. Option A would expand to \(4m^2-12mn+9n^2\), because the minus sign makes the middle term negative. Option B would begin with \(16m^2\), not \(4m^2\), and would also have a different middle term. Option D expands to \(4m^2+15mn+9n^2\), so its middle coefficient is incorrect. The sign of the middle term is a useful clue: a positive middle term requires the plus version of the binomial square.

Correct answer: A, (x − 6)(x − 7). To factor x² − 13x + 42, find two numbers whose product is 42 and whose sum is −13. The numbers are −6 and −7 because (−6)(−7) = 42 and −6 + −7 = −13. Therefore x² − 13x + 42 = (x − 6)(x − 7). Checking by multiplication gives x² − 7x − 6x + 42 = x² − 13x + 42. Option A is correct because both required conditions match. Option B gives x² + 13x + 42, so its middle sign is wrong. Option C gives x² − 17x + 42, so the middle coefficient is wrong. Option D gives x² − 11x − 42, so both the middle and constant terms do not match. Memory cue: for a positive last term and a negative middle term, both factor numbers are negative.

Correct answer: A, x³ − 8y³. A difference of cubes has the form a³ − b³. Rewrite 8y³ as (2y)³. Thus x³ − 8y³ = x³ − (2y)³, where a = x and b = 2y. Using a³ − b³ = (a − b)(a² + ab + b²), the factorisation is (x − 2y)(x² + 2xy + 4y²). Option A has exactly the required subtraction and both terms are perfect cubes. Option B is a sum of cubes, not a difference; it would use a³ + b³. Option C contains squares, not cubes, and belongs to the difference-of-squares pattern. Option D has 4y³, but 4 is not a perfect cube, so it is not immediately a difference of two cubes in the required form. Memory cue: first check the sign, then check whether each coefficient is a perfect cube.

Correct answer: A, (4a − 5b)². Use the identity (u − v)² = u² − 2uv + v². The first term 16a² is (4a)², and the last term 25b² is (5b)². The middle term should therefore be −2(4a)(5b) = −40ab, which is exactly present. Hence 16a² − 40ab + 25b² = (4a − 5b)². Option A matches all three terms after expansion. Option B expands to 16a² + 40ab + 25b², so its middle sign is incorrect. Option C has first square (16a)² = 256a² and last square (25b)² = 625b², so it is much too large. Option D has first square (8a)² = 64a², not 16a². Memory cue: take square roots of the outer terms, then use the middle term to decide the sign.

Correct answer: A. The identity is (a − b)² = a² − 2ab + b². In option A, 9x² = (3x)² and 16y² = (4y)². The required middle term is −2(3x)(4y) = −24xy, exactly as given. Therefore 9x² − 24xy + 16y² = (3x − 4y)². Option B has the correct-looking outer coefficients but its final term is negative, whereas a square expansion has a positive final square term. Option C is a difference of squares, (3x − 4y)(3x + 4y), not a square of one binomial. Option D has 8y² instead of the required 16y², so its middle term does not fit the identity. Memory cue: for a square trinomial, check first square, last square, and middle term −2ab.

Correct answer: A. A perfect-square trinomial has the form a² ± 2ab + b². Here 16p² = (4p)² and 9q² = (3q)². The required middle term for the minus form is −2(4p)(3q) = −24pq. Therefore 16p² − 24pq + 9q² = (4p − 3q)², giving two identical factors, (4p − 3q)(4p − 3q). Option B has 6q² instead of 9q², so the last term is not the square needed. Option C is a difference of squares, not a perfect-square trinomial. Option D has a negative final square term, so it cannot be the square of a real binomial in this pattern; its middle sign also differs. Memory cue: the middle coefficient must equal twice the product of the outer square roots, with the appropriate sign.

Correct answer: A. Rewrite the expression as (3p)³ − (2q)³ because (3p)³ = 27p³ and (2q)³ = 8q³. The difference-of-cubes identity is a³ − b³ = (a − b)(a² + ab + b²). Substituting a = 3p and b = 2q gives (3p − 2q)(9p² + 6pq + 4q²). Option A follows the identity exactly. Option B has a plus in the first factor, which would correspond to a sum of cubes, and its signs do not reproduce the original expression. Option C does not use the cube roots as the first factor and does not expand correctly. Option D has the wrong sign in the middle term of the second factor; expansion would produce an incorrect cross term. Memory cue: for a³ − b³, signs are minus, plus, plus.

Correct answer: A. Since 64 = 4³, the expression is x³ + 4³. Use the sum-of-cubes identity a³ + b³ = (a + b)(a² − ab + b²). Taking a = x and b = 4 gives x³ + 64 = (x + 4)(x² − 4x + 16). Option A has the correct first factor and the required minus sign in the middle of the second factor. Option B begins with x − 4, so it represents the wrong cube relationship. Option C has a plus middle term; multiplying it gives x³ + 8x² + 32x + 64, not the original expression. Option D treats 64 as 8³, but 8³ = 512, so its first factor is incorrect. Memory cue: for a³ + b³, the signs are plus, minus, plus.

Correct answer: A, 5xy(x − 3y + 2). Look for the greatest common factor present in every term. The numerical factors 5, 15 and 10 have greatest common factor 5. Each term also contains x and y, so the greatest common monomial factor is 5xy. Divide each term by 5xy: 5x²y ÷ 5xy = x, −15xy² ÷ 5xy = −3y, and 10xy ÷ 5xy = 2. Therefore the factorised form is 5xy(x − 3y + 2). Option B is algebraically equivalent but leaves y inside every term and is not the simplest complete common-factor form. Option C is also equivalent but leaves 5 inside the bracket, so it is not fully extracted. Option D gives twice the original expression. Memory cue: take the greatest common numerical factor and the lowest power of every variable shared by all terms.

Correct answer: A. If the outer terms are a² and b², a perfect-square trinomial must have the form a² + 2ab + b² = (a + b)² or a² − 2ab + b² = (a − b)². Thus the middle term must be ±2ab: twice the product of the square roots of the first and last terms, with its sign determining whether the binomial has a plus or minus sign. Option B is wrong because the middle term is not generally the sum of the outer terms. Option C is wrong because multiplying the outer terms gives a⁴b⁴ in symbolic form, not the required cross term. Option D is wrong because the middle coefficient can be many values, depending on a and b; it is the relationship ±2ab that matters. Memory cue: outer squares alone are not enough—always test the middle term against ±2ab.

The correct answer is A: (3a + 5)(2b + 3). Arrange the four terms in two groups: (6ab + 9a) + (10b + 15). Take the common factor 3a from the first group and 5 from the second group. This gives 3a(2b + 3) + 5(2b + 3). Both terms now contain the identical binomial (2b + 3), so take it outside: (2b + 3)(3a + 5), which is the same as option A because multiplication can be written in either order. Option B would expand to 6ab + 15a + 10b + 25, so its coefficients do not match. Option C gives an incorrect coefficient for the a-term. Option D expands to 6ab + 18a + 5b + 15, also incorrect. A useful check is to multiply the final factors back and compare every term, not just the first and last terms.

The correct answer is A: (x + 3y)(x − 2y). Since the first term is x², begin with two binomials whose first terms are x and x. We need two numbers, multiplied as coefficients of y, whose product is −6 and whose sum is +1, the coefficient of xy. The numbers are +3 and −2. Therefore write (x + 3y)(x − 2y). Expanding confirms it: x·x = x², x(−2y) = −2xy, 3y·x = 3xy, and 3y(−2y) = −6y². Combining the middle terms gives −2xy + 3xy = xy, exactly the original expression. Option B gives −xy as the middle term. Option C gives 5xy and −6y², while option D gives −5xy and −6y². Memory cue: for a monic trinomial, the two inner coefficients must have the required sum and product.

The correct answer is A: a² − 14ab + 49b² = (a − 7b)². When a trinomial is the product of two identical binomials, it is the square of one binomial, so it must follow one of these identities: (p + q)² = p² + 2pq + q² or (p − q)² = p² − 2pq + q². Take p = a and q = 7b. Then p² = a², q² = 49b², and −2pq = −2(a)(7b) = −14ab. All three terms match option A exactly. Option B fails because the last term should be 49b², not 14b². Option C has the required positive middle term for (a + 7b)², but its last term is 14b² instead of 49b². Option D has 49b², but the middle coefficient is −7 rather than −14. Memory cue: in a perfect-square trinomial, the middle term is twice the product of the two square-root terms, with the appropriate sign.

The correct answer is A: a³ + b³ = (a + b)(a² − ab + b²). This is the sum-of-cubes identity, u³ + v³ = (u + v)(u² − uv + v²). Substitute u = a and v = b. The first factor must be a + b because the expression is a sum. In the second factor, the middle term has a minus sign. Verify by multiplication: a(a² − ab + b²) + b(a² − ab + b²) gives a³ − a²b + ab² + a²b − ab² + b³. The unlike middle terms cancel, leaving a³ + b³. Option B is the difference-of-cubes identity and gives a³ − b³. Option C leaves extra terms instead of cancelling them. Option D has both the wrong first sign and wrong middle sign. Memory cue: for a sum of cubes, signs are plus, minus, plus across the two factors: (a+b)(a²−ab+b²).

The correct answer is B: a³ − b³ = (a − b)(a² + ab + b²). This is the difference-of-cubes identity u³ − v³ = (u − v)(u² + uv + v²). Taking u = a and v = b gives the stated factorisation. To verify it, multiply: a(a² + ab + b²) − b(a² + ab + b²) = a³ + a²b + ab² − a²b − ab² − b³. The mixed terms cancel, leaving a³ − b³. Option A is the sum-of-cubes formula and produces a³ + b³. Option C has the correct first factor but the wrong sign in the second factor, so the mixed terms do not cancel correctly. Option D has the wrong first factor and therefore cannot represent a difference. Memory cue: difference of cubes uses minus in the first factor, but all plus signs inside the second factor.

The correct answer is A: (2m + 3n)(4m² − 6mn + 9n²). Recognise each term as a cube: 8m³ = (2m)³ and 27n³ = (3n)³. Apply u³ + v³ = (u + v)(u² − uv + v²), with u = 2m and v = 3n. The first factor is 2m + 3n. In the second factor, u² = 4m², −uv = −(2m)(3n) = −6mn, and v² = 9n². Thus the result is option A. Option B uses the difference-of-cubes pattern and has the wrong first sign. Option C does not use the correct cube roots and therefore has incorrect coefficients. Option D has a positive 6mn term; multiplying it would create additional positive mixed terms rather than the cancellation required for a sum of cubes. Memory cue: sum of cubes has + in the first factor and − between the squared and mixed terms in the second factor.

The correct answer is A: (5x − y)(25x² + 5xy + y²). First recognise 125x³ as (5x)³. The expression is therefore (5x)³ − y³, a difference of cubes. Use u³ − v³ = (u − v)(u² + uv + v²), taking u = 5x and v = y. Then u² = 25x², uv = 5xy, and v² = y², giving (5x − y)(25x² + 5xy + y²). Option B corresponds to a sum of cubes because its first factor is 5x + y. Option C uses 125x rather than the cube root 5x and has the wrong structure. Option D has a negative mixed term; that sign belongs to the sum-of-cubes identity, not the difference-of-cubes identity. Multiplication of option A shows the mixed terms cancel, leaving exactly 125x³ − y³.