Factorisation

The direct answer is A: (2m + n)². Use u² + 2uv + v² = (u + v)² with u = 2m and v = n. Then u² = 4m², 2uv = 2(2m)(n) = 4mn, and v² = n². Every term matches the given expression, so 4m² + 4mn + n² = (2m + n)². Option A expands correctly. Option B gives 4m² − 4mn + n², with the wrong sign in the middle. Option C gives 16m² + 8mn + n², so its first two coefficients are wrong. Option D expands to 4m² + 5mn + n², not the given middle coefficient. Memory cue: in a perfect-square trinomial, identify the square roots of the first and last terms, then verify that the middle term is twice their product.

Direct answer: A, \((a+d)(b-c)\). Start by grouping the first two and the last two terms: \(ab-ac+db-dc=a(b-c)+d(b-c)\). Both groups contain the same binomial factor \((b-c)\). Taking it out gives \((a+d)(b-c)\). To check, expand: \((a+d)(b-c)=ab-ac+db-dc\), exactly the original expression. Option A is correct because every sign and term matches. Option B expands to \(ab+ac-db-dc\), so the signs of \(ac\) and \(db\) are wrong. Option C expands to \(ab-bd+ac-cd\), which has a different arrangement of signs and products. Option D expands to \(ab+ac+db+dc\), making all terms positive. Memory cue: when grouping, look for an identical bracket after factoring each pair; always expand once to verify signs.

Direct answer: A, \((x-6)(x+3)\). For a monic quadratic \(x^2+bx+c\), find two numbers whose product is \(c\) and whose sum is \(b\). Here their product must be \(-18\), and their sum must be \(-3\). The numbers are \(-6\) and \(+3\). Therefore \(x^2-3x-18=x^2-6x+3x-18=x(x-6)+3(x-6)=(x-6)(x+3)\). Option A is correct. B uses +6 and -3: their sum is 3, giving middle term \(+3x\), not \(-3x\). C uses -9 and +2: the sum is -7 and the product is -18, so the middle term is wrong. D uses -18 and +1: the sum is -17, also wrong. Memory cue: the product controls the constant term and the sum controls the middle coefficient; check both, not just one.

The governing concept is factorisation by taking out the greatest common factor. For 12x²y and 18xy², the greatest common numerical factor is 6. Both terms contain x and y, so the greatest common monomial factor is 6xy. Dividing the first term by 6xy gives 2x, and dividing the second term by 6xy gives 3y. The original minus sign must remain, so the result is 6xy(2x−3y). Option B is algebraically equivalent after further factorisation, but it has not taken out the greatest common factor. Option C also leaves a common factor inside, while D changes subtraction into addition. Hence option A is the simplest correct form.

Direct answer: A, \(5(x-2)(x+2)\). First identify the greatest common factor of both terms: \(5x^2-20=5(x^2-4)\). Then recognise \(x^2-4=x^2-2^2\), a difference of squares. Applying \(u^2-v^2=(u-v)(u+v)\) gives \(x^2-4=(x-2)(x+2)\). Hence the complete factorisation is \(5(x-2)(x+2)\). Option A is correct. B only takes out 5 and stops before fully factorising; also its inside expression is incorrectly written as \(x^2-20\). C uses 4 instead of the square root 2, so it would produce \(x^2-16\). D expands to \(5x^2-15x-20\), not the given expression. Memory cue: remove a common factor first, then look again for an identity; factorisation may require more than one step.

The governing method is factorisation by grouping. Group terms that share useful common factors: 2xy + 6y + 5x + 15 = (2xy+6y) + (5x+15). Factoring each group gives 2y(x+3) + 5(x+3). Both terms now contain the common binomial x+3, so taking it outside produces (2y+5)(x+3). Therefore option A is correct. Expansion verifies the result: (2y+5)(x+3) = 2xy + 6y + 5x + 15. Option B changes the signs of the 5x and constant terms. Option C pairs the variables incorrectly and does not recreate the original expression, while option D gives unrelated products. Grouping is effective because it creates the same binomial factor in both groups.

Direct answer: A, \((2x+3)^2\). Use \(u^2+2uv+v^2=(u+v)^2\). The first term \(4x^2\) is \((2x)^2\), and the last term 9 is \(3^2\). Their double product is \(2(2x)(3)=12x\), exactly the middle term. Therefore \(4x^2+12x+9=(2x)^2+2(2x)(3)+3^2=(2x+3)^2\). Option B would expand to \(4x^2-12x+9\), so its middle sign is wrong. Option C would have leading term \(16x^2\) and constant 81, both incorrect. Option D expands to \(4x^2+20x+9\), so its middle coefficient is wrong. Memory cue: check all three parts—first square, twice the product, and last square—rather than matching only the first and last terms.

Direct answer: A, \(x^2+10x+25=(x+5)^2\). For a trinomial \(x^2+2kx+k^2\), the first and last terms are squares and the middle term is twice their roots multiplied. Here \(25=5^2\), and twice the product of x and 5 is \(10x\), so the identity gives \((x+5)^2\). Option A is correct. Option B has constant 20 instead of 25, so it cannot be the square of a binomial with first term x² and middle term 10x. Option C has the correct constant 25, but its middle term should be 10x, not 8x. Option D has a negative constant, so it cannot match \((x+5)^2\) or \((x-5)^2\), both of which have positive constant 25. Memory cue: for \(x^2+bx+c\), if c=25, test roots ±5; the middle coefficient must then be ±10.

Direct answer: A, (3x+1)(2x+3). To factor a quadratic, multiply the first and last coefficients: 6×3=18. The two numbers whose product is 18 and sum is 11 are 9 and 2. Thus 11x can be split as 9x+2x: 6x²+9x+2x+3=3x(2x+3)+1(2x+3)=(3x+1)(2x+3). Option A expands to 6x²+9x+2x+3, exactly the given expression. Option B gives 6x²+19x+3, so its middle coefficient is wrong. Option C gives 6x²+9x+6x+3=6x²+15x+3, also wrong. Option D gives 6x²−9x−2x+3=6x²−11x+3, with the wrong sign for the middle term. Memory cue: after factorising, always expand the answer to check the middle term and the constant term.

Direct answer: A, (x−9)(x+5). For x²+bx+c, find two numbers whose product is c and whose sum is b. Here their product must be −45 and their sum must be −4. The numbers −9 and +5 satisfy both conditions: (−9)(5)=−45 and −9+5=−4. Therefore x²−4x−45=(x−9)(x+5). Expanding A gives x²+5x−9x−45=x²−4x−45. Option B gives x²+4x−45 because the middle signs combine to +4x. Option C gives x²−12x−45, so the middle coefficient is wrong. Option D gives x²−44x−45, also incorrect. Since the constant is negative, the two numbers must have opposite signs; the negative number must have the larger absolute value because the sum is negative. Memory cue: check both product and sum, not just the constant term.

Direct answer: A, (5p−2)(2q+3). Group the first two and last two terms: (10pq+15p)+(−4q−6). Factor each group: 5p(2q+3)−2(2q+3). Now the common binomial is (2q+3), so take it out: (2q+3)(5p−2), which is the same as (5p−2)(2q+3). Option A expands to 10pq+15p−4q−6, exactly as required. Option B expands to 10pq−15p+4q−6, so several signs are wrong. Option C expands to 10pq+60p−4q−24, not the given expression. Option D expands to 10pq+3p−20q−6, so both the p and q coefficients are wrong. The key step is arranging groups so that the same binomial factor appears in both. Memory cue: if needed, take a negative common factor from the second group.

Direct answer: A. The expression is a difference of squares: 49x²−25y²=(7x)²−(5y)². Therefore use u²−v²=(u−v)(u+v), giving (7x−5y)(7x+5y). The student incorrectly used the square-of-a-difference identity. Indeed, (7x−5y)²=(7x)²−2(7x)(5y)+(5y)²=49x²−70xy+25y². It has a middle term and a positive last term, neither of which occurs in the given expression. Option B is false because 49 and 25 are perfect squares. Option C is false because the expression has the factorisation just shown. Option D is false for the same expansion reason. Memory cue: (u−v)² has three terms, but u²−v² has two conjugate factors and no middle term.

Direct answer: C, a²−10a+25. A perfect-square trinomial has the form u²−2uv+v² or u²+2uv+v². Here a²=(a)² and 25=5². The required middle term for (a−5)² is −2(a)(5)=−10a, so a²−10a+25=(a−5)². Option A has the correct leading term and middle coefficient for neither listed square with constant 20; 20 is not 5² and the matching constant would be 25. Option B also has constant 20, so it cannot be (a−5)² or (a+5)². Option D has 25=5², but its middle term should be −10a, not −8a. Expanding option C confirms a²−10a+25. Memory cue: take square roots of the first and last terms, then check whether the middle term is plus or minus twice their product.

Direct answer: A, 3(a−3b)(a+3b). First take out the common factor 3: 3a²−27b²=3(a²−9b²). Inside the brackets, 9b²=(3b)², so a²−9b² is a difference of squares. Apply u²−v²=(u−v)(u+v): a²−(3b)²=(a−3b)(a+3b). Hence the complete factorisation is 3(a−3b)(a+3b). Option B would expand to 3a²−3b², so it misses the factor 9 on b². Option C expands to 3a²+3ab−27ab−27b², which has unwanted middle terms. Option D represents a square and expands to 3a²−54ab+27b², with both a middle term and the wrong sign on the last term. Memory cue: remove the numerical common factor first, then look for a standard identity; do not stop at 3(a²−9b²).

Direct answer: A, (2x−3y)². Use the identity (u−v)²=u²−2uv+v². Let u=2x and v=3y. Then u²=4x², −2uv=−2(2x)(3y)=−12xy, and v²=9y². These are exactly the three terms in the given expression, so 4x²−12xy+9y²=(2x−3y)². Option B expands to 4x²+12xy+9y², so its middle sign is wrong. Option C has incorrect square roots: its leading term is 16x² and its last term is 81y², not the given terms. Option D expands to 4x²−20xy+9y², so the middle coefficient is wrong. The negative middle term tells us to use a minus inside the square. Memory cue: square the first and last parts, then check whether the middle term is plus or minus twice their product.

Direct answer: A. Rewrite the terms as squares: 49a²=(7a)² and 64b²=(8b)². The expression is therefore (7a)²−(8b)², a difference of squares. Using u²−v²=(u−v)(u+v), its factorisation is (7a−8b)(7a+8b). The student used (u−v)² incorrectly. Expanding that claimed form gives (7a−8b)²=49a²−112ab+64b², which has a middle term and a positive final term; the original expression has neither. Option B repeats the incorrect claim and confuses a difference with a square. Option C does not expand to the original expression. Option D is false because the difference-of-squares identity factorises it immediately. Memory cue: a²−b² uses conjugate factors, while (a−b)² creates a middle term.

Direct answer: A, (x+3)(y+2). To factorise by grouping, arrange the four terms in two pairs: xy+2x and 3y+6. Take the common factor from each pair: xy+2x=x(y+2), while 3y+6=3(y+2). Thus the expression becomes x(y+2)+3(y+2). The same bracket, (y+2), occurs in both terms, so take it out: (x+3)(y+2). Option A is correct because expansion gives xy+2x+3y+6. Option B expands to xy+3x+2y+6, so the coefficients of x and y are interchanged. Option C gives xy-2x-3y+6, with incorrect signs. Option D gives xy-2x+3y-6, also incorrect. Memory cue: pair terms so that one identical bracket appears twice, then take that bracket common.

Direct answer: A, \((8a+5b)^2\). Use the identity \((P+Q)^2=P^2+2PQ+Q^2\). Here choose \(P=8a\) and \(Q=5b\). Then \(P^2=64a^2\), \(Q^2=25b^2\), and \(2PQ=2(8a)(5b)=80ab\). These are exactly the three terms in the given expression, so it equals \((8a+5b)^2\). Option B uses a minus sign; its middle term would be \(-80ab\), not \(+80ab\). Option C squares the whole terms incorrectly and would have degree-four terms such as \(a^2b^2\) after expansion. Option D expands to \(64a^2+208ab+25b^2\), so its middle coefficient is wrong. A useful check is that the square roots of the first and last terms are \(8a\) and \(5b\); the middle term must be twice their product, with the same sign as the bracket.

Direct answer: B, \((2x+5)(2x-3)\). Multiply the first terms: \(2x·2x=4x^2\). Multiply the last terms: \(5·(-3)=-15\). The cross terms are \(2x·(-3)=-6x\) and \(5·2x=10x\); their sum is \(4x\). Therefore the product is \(4x^2+4x-15\), exactly the original quadratic. Option A gives cross terms \(6x-10x=-4x\), so it produces \(4x^2-4x-15\). Option C gives \(4x^2-12x+5x-15=4x^2-7x-15\). Option D gives \(4x^2-2x+30x-15=4x^2+28x-15\). For a quadratic, check the leading product, constant product, and sum of cross terms. Memory cue: the signs of the factors must create a negative constant and a positive middle term, so the positive factor must have the larger cross contribution.

Direct answer: A, \(3(x-2)(x+2)\). First inspect all terms and take out the common factor 3: \(3x^2-12=3(x^2-4)\). The expression inside is a difference of squares because \(x^2-4=x^2-2^2\). Apply \(A^2-B^2=(A-B)(A+B)\), giving \(x^2-4=(x-2)(x+2)\). Hence the complete factorisation is \(3(x-2)(x+2)\). Option B would represent \(3(x^2-16)=3x^2-48\), not the given expression. Option C stops after extracting 3, so it is not complete because \(x^2-4\) can still be factorised. Option D represents \(3(x^2-4x+4)\), which contains an extra \(-4x\) term. The important habit is to look for a common numerical factor first and then check whether the remaining expression is a familiar identity.

Direct answer: A, \((x-7)(x-8)\). For \(x^2+bx+c\), find two numbers whose product is \(c\) and whose sum is \(b\). Here the product must be 56 and the sum must be -15. The numbers -7 and -8 satisfy both conditions: \((-7)(-8)=56\) and \(-7+(-8)=-15\). Therefore \(x^2-15x+56=(x-7)(x-8)\). Option B has product 56 but sum +15, so it gives a positive middle term. Option C has product 56, but its sum is -18, giving \(-18x\). Option D has product 56, but its sum is -57, giving \(-57x\). A reliable check is to multiply the constant terms and add the numbers in the brackets; both tests must work, not just one. When the constant is positive and the middle coefficient is negative, both integer factors are negative.

Direct answer: A, (4a+5)(b+2). Group terms that share a useful factor: (4ab+8a)+(5b+10). From the first group take 4a: 4ab+8a=4a(b+2). From the second group take 5: 5b+10=5(b+2). The expression is therefore 4a(b+2)+5(b+2). Since (b+2) is common, factor it out to obtain (4a+5)(b+2). Option A expands to 4ab+8a+5b+10 exactly. Option B changes the signs of the terms involving 5 and gives -5b-10. Option C expands to 4ab+2a+5b+10, so the coefficient of a is wrong. Option D gives 4ab+20a+10b+50, which is clearly different. The key skill is not merely pairing terms; it is choosing pairs that produce the same bracket.

Answer: A. A difference of squares has the form \(U^2-V^2\): two perfect-square terms separated by a minus sign. In option A, \(49a^2=(7a)^2\) and \(16b^2=(4b)^2\), so the expression is \((7a)^2-(4b)^2\). It can therefore be factorised as \((7a-4b)(7a+4b)\). Option B is a sum of two squares, not a difference. Option C is a three-term expression and is actually \((7a-4b)^2\), because its middle term is \(-56ab\). Option D has a mixed term and its final term is negative, so it is not the required two-square form. Memory cue: difference of squares means “square minus square,” and the factors have opposite signs.

Answer: A. The identity is \((u-v)^2=u^2-2uv+v^2\). The first term of A is \(49p^2=(7p)^2\), and the last term is \(25q^2=(5q)^2\). The required middle term is \(-2(7p)(5q)=-70pq\), exactly as given. Therefore, A becomes \((7p-5q)^2\). Option B has the wrong middle coefficient: it would need \(-70pq\), not \(-60pq\). Option C has a negative final square term, so it cannot be a square of a real binomial in this identity. Option D has a positive middle term and matches \((7p+5q)^2\), not the requested subtraction form. Always check both outer squares and the middle term.

Answer: A. For two identical binomial factors, use \((u-v)^2=u^2-2uv+v^2\). Here \(4a^2=(2a)^2\) and \(9b^2=(3b)^2\). The middle term required for their square is \(-2(2a)(3b)=-12ab\). Thus option A is \((2a-3b)^2\), a product of the identical binomials \((2a-3b)(2a-3b)\). Option B has 6b² instead of the required 9b². Option C is a difference of squares and has only two terms, so its factors are different. Option D has a negative final term and cannot be this binomial square. A useful check is to square the proposed binomial and compare every term, not just the first and last terms.