Factorisation

The direct answer is A: \((6a-b)(6a+b)\). Begin by noticing that \(36a^2=(6a)^2\) and \(b^2=b^2\). Thus the expression has the form \(U^2-V^2\), called a difference of two squares. The identity is \(U^2-V^2=(U-V)(U+V)\). Taking \(U=6a\) and \(V=b\) gives \(36a^2-b^2=(6a-b)(6a+b)\). To check A, multiply: \((6a-b)(6a+b)=36a^2+6ab-6ab-b^2=36a^2-b^2\); the middle terms cancel. B expands to \(36a^2+36ab-ab-b^2\), which is not the original expression. C expands to \(36a^2+12ab+b^2\), so both the middle and last signs are wrong. D expands to \(36a^2-12ab+b^2\), also incorrect. A common confusion is to use a square such as \((6a-b)^2\); that creates a middle term. In a difference of squares, the two factors have opposite signs between their corresponding terms.

The direct answer is A: \((k-7)(k-8)\). For a monic quadratic \(k^2+pk+q\), find two numbers whose sum is \(p\) and whose product is \(q\). Here the required sum is \(-15\) and the required product is \(56\). The numbers \(-7\) and \(-8\) work because \((-7)+(-8)=-15\) and \((-7)(-8)=56\). Therefore \(k^2-15k+56=(k-7)(k-8)\). Multiplying A confirms it: \(k^2-8k-7k+56=k^2-15k+56\). B gives \(k^2+15k+56\), so its middle sign is wrong. C gives \(k^2-18k+56\), so the middle coefficient is -18 rather than -15. D gives \(k^2-57k+56\), so its middle coefficient is also wrong. The positive constant 56 means the two numbers have the same sign; because their sum is negative, both must be negative. Memory cue: for a monic quadratic, check the pair by both sum and product, not by product alone.

The direct answer is A: (x-y+1)(x+y-1). First group the y-terms: -y^2+2y-1=-(y^2-2y+1)=-(y-1)^2. The whole expression becomes x^2-(y-1)^2. Now apply U^2-V^2=(U-V)(U+V), with U=x and V=y-1. This gives [x-(y-1)][x+(y-1)]=(x-y+1)(x+y-1). Option A is therefore correct. Expanding it gives x^2+x(y-1)-y+1? More directly, it is x^2-(y-1)^2=x^2-y^2+2y-1, exactly the original. B and C both produce x^2-(y+1)^2=x^2-y^2-2y-1, with the wrong sign for 2y. D is (x+y-1)^2 and introduces a positive 2xy term, which is absent. The main warning is that the negative sign belongs to the entire square (y-1)^2, not just to y^2.

The direct answer is A: \((2x+1)(x+3)\). Multiply the proposed factors: \((2x+1)(x+3)=2x^2+6x+x+3=2x^2+7x+3\), so it matches exactly. A systematic method is to multiply the leading and constant coefficients, 2 times 3=6, and find two numbers with product 6 and sum 7: 6 and 1. Split the middle term as 6x+x, then group: \(2x^2+6x+x+3=2x(x+3)+1(x+3)=(2x+1)(x+3)\). B expands to \(2x^2+5x+3\), so its middle coefficient is wrong. C expands to \(2x^2+7x+6\), so its constant term is wrong. D expands to \(2x^2-7x+3\), so the middle sign is wrong. A useful habit is to expand every proposed pair; this catches both coefficient and sign errors quickly.

The direct answer is A: $(3x+4)(x+2)$. Use the split-middle-term method. Multiply the first and last coefficients: $3×8=24$. Two numbers with product 24 and sum 10 are 6 and 4. Thus $3x^2+10x+8=3x^2+6x+4x+8$. Group the terms: $3x(x+2)+4(x+2)$. The common factor is $x+2$, giving $(3x+4)(x+2)$. Expanding A returns $3x^2+6x+4x+8=3x^2+10x+8$. B expands to $3x^2+14x+8$, so its middle term is too large. C expands to $x^2+10x+24$, changing the leading and constant terms. D expands to $3x^2-10x+8$, changing the sign of the middle term. The memory cue is: for $ax^2+bx+c$, first inspect $ac$, then split $bx$ into two terms that can be grouped with a common binomial.

The direct answer is A: $3(2a-3b)(2a+3b)$. First take out the greatest common numerical factor, 3: $12a^2-27b^2=3(4a^2-9b^2)$. The expression inside is still factorable because $4a^2=(2a)^2$ and $9b^2=(3b)^2$. Apply $U^2-V^2=(U-V)(U+V)$: $4a^2-9b^2=(2a-3b)(2a+3b)$. Hence the complete form is $3(2a-3b)(2a+3b)$. Option C is only a partial factorisation; it leaves a difference of squares unfinished. Expanding A gives $3[4a^2-9b^2]=12a^2-27b^2$. B expands to $12a^2+12ab-27ab-27b^2$, not the original. D expands to $12a^2-9ab-18b^2$, also incorrect. The word complete means continue factoring until no common factor or standard identity remains.

The direct answer is A: $(a+b-c)(a+b+c)$. The first three terms form a perfect square: $a^2+2ab+b^2=(a+b)^2$. Therefore the expression becomes $(a+b)^2-c^2$. This is a difference of squares with $U=a+b$ and $V=c$. Applying $U^2-V^2=(U-V)(U+V)$ gives $[(a+b)-c][(a+b)+c]=(a+b-c)(a+b+c)$. Option A is correct. Expanding it as a difference of squares returns $(a+b)^2-c^2=a^2+2ab+b^2-c^2$. B is based on $(a-b)^2-c^2$ and would contain $-2ab$, not $+2ab$. C is a sum of squares, not the factorisation of a difference; it is not generally a product of the required two linear factors. D is based on $(a+c)^2-b^2$ and changes the roles of b and c, producing different terms. Memory cue: first combine three terms into a perfect square, then factor the remaining difference of squares.

The direct answer is A: (3p+2)(2q+3). Group the terms in pairs: (6pq+9p)+(4q+6). From the first group take out 3p, giving 3p(2q+3). From the second group take out 2, giving 2(2q+3). Both groups now contain the common binomial (2q+3), so factor it out: (3p+2)(2q+3). Expanding A gives 6pq+9p+4q+6, exactly the original expression. B expands to 6pq+18p+4q+12, so the p-term and constant are wrong. C expands to 6pq+18p+8q+24, so several coefficients are wrong. D expands to 6pq+4p+9q+6, which interchanges the required coefficients. The key idea in grouping is not merely to make pairs, but to make the same bracket appear in both groups. Always compare the brackets before taking out the final common factor.

The direct answer is A: $(4m+5n)(16m^2-20mn+25n^2)$. Recognize $64m^3=(4m)^3$ and $125n^3=(5n)^3$. Use the sum-of-cubes identity $u^3+v^3=(u+v)(u^2-uv+v^2)$, with $u=4m$ and $v=5n$. Substitution gives $(4m+5n)[(4m)^2-(4m)(5n)+(5n)^2]=(4m+5n)(16m^2-20mn+25n^2)$. Option A is correct. B uses $4m-5n$ in the first factor, which belongs to a difference pattern, not this sum. C has a positive middle term; multiplying it creates the wrong mixed terms. D incorrectly uses $64m+125n$ instead of the cube roots $4m$ and $5n$, and its second factor is not the correct identity factor. A useful memory cue is: for a sum of cubes, signs are plus, minus, plus in the two factors: $(u+v)(u^2-uv+v^2)$.

The direct answer is A. First recognise that \(125x^3=(5x)^3\) and \(216y^3=(6y)^3\). The expression is therefore a difference of cubes, \(u^3−v^3\), where \(u=5x\) and \(v=6y\). The identity is \(u^3−v^3=(u−v)(u^2+uv+v^2)\). Substituting gives \((5x−6y)[(5x)^2+(5x)(6y)+(6y)^2]=(5x−6y)(25x^2+30xy+36y^2)\). Thus A reproduces the original expression when expanded. B uses the sum factor and is the identity for a sum of cubes, not this difference. C has the wrong negative sign in the middle term; expanding it does not cancel correctly. D does not take the cube roots as the first factor and its second factor is incomplete, so it cannot produce the required cubic expression. Memory cue: for a difference of cubes, the signs are minus in the first factor and plus, plus in the second factor.

The direct answer is A. For \(2a^2−11a+15\), multiply the leading coefficient and constant: \(2×15=30\). We need two numbers whose product is 30 and whose sum is −11. Those numbers are −5 and −6. Split the middle term as \(−5a−6a\): \(2a^2−5a−6a+15\). Group the terms: \(a(2a−5)−3(2a−5)\). Taking the common binomial gives \((2a−5)(a−3)\), which is A. B expands to \(2a^2−13a+15\), so its middle coefficient is wrong. C expands to \(a^2−11a+30\), changing both the leading and constant terms. D gives positive middle and constant contributions, not the signs in the question. A useful check is to multiply the two binomials: the outer and inner products are \(−6a\) and \(−5a\), adding to \(−11a\). Memory cue: for \(ax^2+bx+c\), find factors of \(ac\) whose sum is \(b\).

The direct answer is A. Expand each factor in A: \((3y+1)(y−5)=3y^2−15y+y−5=3y^2−14y−5\), exactly the given expression. Another way is to use the ac method: \(3×(−5)=−15\), and the numbers \(−15\) and \(1\) have sum \(−14\). Split the middle term as \(−15y+y\), then group: \(3y(y−5)+1(y−5)=(3y+1)(y−5)\). B expands to \(3y^2+15y−y−5=3y^2+14y−5\), so the middle sign is wrong. C has leading coefficient \(3\) but its constant term is \(−15\), not \(−5\). D expands to \(3y^2−2y−5\), whose middle coefficient is different. Always verify a proposed factorisation by multiplying; this is especially useful when signs are mixed. Memory cue: the product of the constants must equal \(−5\), while the middle terms must combine to \(−14y\).

The direct answer is A. Rearrange the terms as \(9x^2+6xy+y^2−1\). The first three terms form a perfect square: \(9x^2+6xy+y^2=(3x+y)^2\), because \((3x+y)^2=9x^2+6xy+y^2\). Thus the expression becomes \((3x+y)^2−1^2\). Apply the difference-of-squares identity \(p^2−q^2=(p−q)(p+q)\), giving \((3x+y−1)(3x+y+1)\). Therefore A is correct. B would come from \((3x−y)^2−1\) and has \(−6xy\), not \(+6xy\). C does not preserve the correct leading and cross-term coefficients when expanded. D is only a square and expands to additional terms \(6x\), \(2y\) and a different constant, so it cannot equal the original expression. A common confusion is forgetting that the terms may be rearranged before identifying the identity. Memory cue: first make the perfect square, then subtract the remaining square.

The direct answer is A. Start with the difference of squares: \(a^2−b^2=(a−b)(a+b)\). The remaining terms are \(−a+b\), which equal \(−(a−b)\), not \(−(a+b)\). So the whole expression becomes \((a−b)(a+b)−(a−b)\). Taking \((a−b)\) common gives \((a−b)[a+b−1]=(a−b)(a+b−1)\). Option B expands to \(a^2−b^2−a−b\), so the sign of b is wrong. Option C expands to \(a^2−b^2+a−b\), so the signs of the remaining linear terms are wrong. Option D is a square and produces different cross terms and coefficients. A common mistake is to mishandle \(−a+b\); carefully factor it as \(−(a−b)\). Memory cue: first use \(a^2−b^2\), then look for the same bracket in the leftover linear terms.

The direct answer is A: $(p+4)(p-5)$. For a monic quadratic $p^2+bp+c$, find two numbers whose product is c and whose sum is b. Here the product must be -20 and the sum must be -1. The numbers 4 and -5 satisfy both conditions: $4×(-5)=-20$ and $4+(-5)=-1$. Therefore $p^2-p-20=(p+4)(p-5)$. Expanding confirms it: $p^2-5p+4p-20=p^2-p-20$. B has factors -4 and 5; their product is -20 but their sum is 1, giving $p^2+p-20$. C uses -2 and 10, whose sum is 8, so it gives $p^2+8p-20$. D uses 20 and -1, whose sum is 19, producing $p^2+19p-20$. Always check both the product and the sum, not just the constant term.

The direct answer is B: (a+b)(x+y). Group the four terms in pairs: ax+ay+bx+by=(ax+ay)+(bx+by). Take the common factor from each group: ax+ay=a(x+y), and bx+by=b(x+y). The expression becomes a(x+y)+b(x+y). Now the binomial (x+y) is common to both terms, so take it out: (a+b)(x+y). Expanding verifies the result: (a+b)(x+y)=ax+ay+bx+by. A is only an intermediate grouped form, not the fully factorised form. C expands to ab+ax+by+xy, which is different. D expands to abx+aby and loses the separate terms, so it cannot equal the original expression. The useful memory cue is: group terms so that the same bracket appears twice, then take that bracket common.

The direct answer is A: \(9a^2-16b^2\). The difference-of-squares identity applies when two perfect-square terms are separated by a minus sign: \(P^2-Q^2=(P+Q)(P-Q)\). Here \(9a^2=(3a)^2\) and \(16b^2=(4b)^2\), so \(9a^2-16b^2=(3a+4b)(3a-4b)\). Expanding gives \(9a^2-12ab+12ab-16b^2=9a^2-16b^2\). B is a sum of squares, not a difference, so this identity does not apply. C is \((3a+4b)^2\) because its middle term is \(+24ab\). D is \((3a-4b)^2\) because its middle term is \(-24ab\). Those are perfect squares, not expressions of the required difference-of-squares form. A safe memory cue is to check two things: both terms must be squares, and the sign between them must be minus.

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, the product must be 42 and the sum must be −13. The numbers −6 and −7 satisfy both conditions: (−6)(−7) = 42 and −6 + (−7) = −13. Therefore x² − 13x + 42 = (x − 6)(x − 7). Option A is correct because expansion gives x² − 7x − 6x + 42 = x² − 13x + 42. Option B gives a positive middle term, x² + 13x + 42, so its signs are wrong. Option C gives factors with sum −17, producing x² − 17x + 42. Option D gives factors with sum −43, producing x² − 43x + 42. Memory cue: when the constant is positive and the middle coefficient is negative, both factor numbers are usually negative.

The direct answer is A: (5x + 3)(y − 2). In grouping, arrange the four terms into two pairs that have the same binomial factor. Write 5xy − 10x as 5x(y − 2), because 5x is common. Write 3y − 6 as 3(y − 2), because 3 is common. The expression becomes 5x(y − 2) + 3(y − 2). Now take the common factor (y − 2): (5x + 3)(y − 2). Option A is confirmed by expansion: 5xy − 10x + 3y − 6. Option B expands to 5xy + 10x − 3y − 6, so both variable-term signs differ. Option C expands to 5xy − 15x + 2y − 6, which has wrong coefficients. Option D expands to 5xy + 15x − 2y − 6, also incorrect. Memory cue: after grouping, the bracket taken out must be identical in both groups.

The direct answer is A: 2(a − 2b)(a + 2b). First identify the greatest common factor of 2a² and 8b², which is 2. Taking it out gives 2(a² − 4b²). The expression inside is a difference of squares because 4b² = (2b)². Use a² − c² = (a − c)(a + c): a² − (2b)² = (a − 2b)(a + 2b). Thus the complete factorisation is 2(a − 2b)(a + 2b). Option A expands back to 2(a² − 4b²) = 2a² − 8b². Option B would give 2(a² − b²), whose second term is −2b², not −8b². Option C does not produce the required terms when multiplied. Option D represents a square, but the original expression is a difference of squares, not a repeated square. Memory cue: take out a common factor before applying an identity.

The direct answer is A: (3x + 4y)². Compare the expression with the identity u² + 2uv + v² = (u + v)². The first term 9x² is (3x)², and the last term 16y² is (4y)². Now check the middle term: 2(3x)(4y) = 24xy, exactly as required. Therefore 9x² + 24xy + 16y² = (3x + 4y)². Option A is correct because its expansion gives all three original terms. Option B gives −24xy, so only the middle sign is wrong. Option C uses 9x and 16y instead of their square roots, so its first and last terms are much too large. Option D gives a middle coefficient 6x² + 24xy + 8y² after multiplication, not the required trinomial. Memory cue: for a perfect square, the middle term must be twice the product of the square roots of the first and last terms.

The direct answer is A: (5p − 3q)². Use the identity (u − v)² = u² − 2uv + v². The first term 25p² is (5p)², and the last term 9q² is (3q)². Their doubled product is 2(5p)(3q) = 30pq, and the given middle term is negative. Hence the correct sign between the two terms is minus: (5p − 3q)². Expanding confirms 25p² − 30pq + 9q². Option B would produce +30pq, so its sign is wrong. Option C takes incorrect square roots and changes the coefficients drastically. Option D expands to 25p² − 50pq + 9q², so its middle coefficient is incorrect. A useful check is to take the square roots of the first and last terms first, then use the middle-term sign.

The direct answer is B: (x − 3)(x + 3). Rewrite x² − 9 as x² − 3². This is a difference of squares, so use u² − v² = (u − v)(u + v), giving (x − 3)(x + 3). Reema’s expression (x − 3)² is not equivalent: expanding it gives x² − 6x + 9, which contains a middle term and has a positive constant. Option B expands to x² + 3x − 3x − 9 = x² − 9, so it is correct. Option A is the original error and adds unwanted terms. Option C expands to x² − 8x − 9, not x² − 9. Option D expands to x² + x − 12, also incorrect. Memory cue: a difference of squares has two different brackets, one with minus and one with plus; it is not a square of one bracket.

The direct answer is A: (x + 5)(x − 3). For x² + 2x − 15, find two integers whose product is −15 and whose sum is +2. The required numbers are +5 and −3. Therefore x² + 2x − 15 = (x + 5)(x − 3). Expanding gives x² − 3x + 5x − 15 = x² + 2x − 15. Option A matches both the middle and constant terms. Option B uses −5 and +3, whose sum is −2, so it gives x² − 2x − 15. Option C gives a middle coefficient −14, not +2. Option D has product (+3)(+5) = +15, but the constant must be −15. Memory cue: a negative constant means the two factor numbers have opposite signs; the positive middle coefficient tells us that the larger absolute value must be positive.

The direct answer is A: x + 5. The area of a rectangle equals the product of its two sides. Therefore the unknown side is (x² + 9x + 20) ÷ (x + 4). Factor the numerator by finding two numbers whose product is 20 and whose sum is 9: they are 4 and 5. Thus x² + 9x + 20 = (x + 4)(x + 5). Cancelling the given side x + 4 leaves x + 5. Option A is correct, and multiplying the two sides gives the original area. Option B would produce x² + 20x + 64, not the given polynomial. Option C gives x² − x − 20, and option D gives x² + 8x + 16. Memory cue: for rectangle-area questions, factor the area polynomial and identify the known side as one factor.