Factorisation

The direct answer is C: one minus sign and one plus sign. The difference-of-squares identity is a²−b²=(a−b)(a+b). To understand why, multiply the two binomials: (a−b)(a+b)=a²+ab−ab−b². The middle terms +ab and −ab cancel, leaving a²−b². Thus one factor contains a minus sign and the other contains a plus sign. Option A is wrong because (a+b)(a+b)=(a+b)²=a²+2ab+b², which has a positive middle term. Option B gives (a−b)²=a²−2ab+b², which has a middle term and a positive b² term. Option D is not a pair of algebraic factors at all. Memory cue: opposite signs make the middle terms cancel.

The direct answer is C: (4a−5b)(4a+5b). First identify the two square terms: 16a²=(4a)² and 25b²=(5b)². Therefore, 16a²−25b² has the form A²−B², where A=4a and B=5b. Applying A²−B²=(A−B)(A+B) gives 16a²−25b²=(4a−5b)(4a+5b). Option A is the square of a binomial; expanding it gives 16a²−40ab+25b², which has the wrong signs and a middle term. Option B gives 16a²+40ab+25b², also incorrect. Option D does not complete the factorisation and, when expanded, simply retains the original structure. Check the answer by multiplying the two conjugate factors: the mixed terms cancel.

The direct answer is A: p²−q². This expression contains two perfect-square terms, p² and q², separated by a minus sign. It exactly matches the identity A²−B²=(A−B)(A+B), so p²−q²=(p−q)(p+q). Option B, p²+q², is a sum of squares and does not match this identity over ordinary real polynomial factorisation. Option C is a perfect-square trinomial because p²+2pq+q²=(p+q)². Option D is also a perfect-square trinomial because p²−2pq+q²=(p−q)². The important distinction is that the difference-of-squares form has only two square terms, while the perfect-square trinomials include a middle term ±2pq. Memory cue: minus between two squares gives conjugate factors, one with plus and one with minus.

The direct answer is B: (x−y)². Use the identity A²−2AB+B²=(A−B)². Here A=x and B=y, so x²−2xy+y²=(x−y)². A step-by-step check is useful: the square of x gives x², the square of y gives y², and the middle term in (x−y)² is −2xy, exactly as required. Option A is incorrect because (x+y)²=x²+2xy+y²; its middle sign is positive. Option C uses the difference-of-squares identity and expands to x²−y², so it lacks the −2xy term. Option D expands to x²−3xy+2y², which does not match the original expression. A common confusion is to choose the sign from the first and last terms only; always check the middle term, since its sign determines whether the binomial is x+y or x−y.

The direct answer is A: 3. To find a common factor, inspect every term: 3x², 6x, and 3. The number 3 divides all three coefficients, so 3 is a common factor. However, x is not present in the last term 3, so x and 3x cannot be common factors of all terms. Also, 6 does not divide the coefficient 3, so 6 is not common. Taking out 3 gives 3x²+6x+3=3(x²+2x+1), after which the bracket can be recognised as (x+1)². Thus removing 3 is the correct first step and makes the remaining factorisation simple. Option B fails because x cannot be divided out from the constant term. Option C fails numerically. Option D fails for both reasons. Memory cue: a common factor must divide every term without leaving fractions or missing variables.

The direct answer is A: 3(x+1)². First take the greatest common numerical factor, 3, from all terms: 3x²+6x+3=3(x²+2x+1). Now examine the trinomial inside the brackets. It has the form x²+2·x·1+1², so by A²+2AB+B²=(A+B)², it becomes (x+1)². Therefore the complete factorisation is 3(x+1)². Option B would expand as 3(x²+4x+4)=3x²+12x+12, so both the middle and constant terms are wrong. Option C expands to 9x²+6x+1 and has a different leading coefficient and constant term. Option D gives 3(x²−2x+1), whose middle term is negative, not positive. The phrase complete factorisation means that no further common factor or applicable identity remains. Always expand the final answer to verify it.

The direct answer is A: a(a+5). Look for the greatest common factor of the two terms a² and 5a. Both contain a, so take a outside. Dividing each term by a leaves a from a² and 5 from 5a; hence a²+5a=a(a+5). Checking by distribution, a(a+5)=a·a+a·5=a²+5a, exactly the original expression. Option B expands to 5a+25 and does not contain a². Option C expands to a³+5a², which has powers that are too large. Option D expands to a²+10a+25, adding terms that are not present. The key idea is that factorisation reverses the distributive law: instead of multiplying a through a bracket, we identify what was common and place it outside. Do not confuse a common factor with a term added inside a square.

The direct answer is B: m(m−6). Both terms, m² and −6m, contain m as a common factor. Taking m outside leaves m from m²÷m and −6 from −6m÷m, so m²−6m=m(m−6). Verification by expansion gives m·m+m·(−6)=m²−6m. Option A has the wrong sign: m(m+6) expands to m²+6m. Option C expands to 6m−6 and is not equal to the original quadratic expression. Option D is a square of a binomial; its expansion is m²−12m+36, which has an incorrect middle term and an extra constant term. A useful habit is to carry the sign of each term into the bracket when taking out a positive common factor. The minus sign belongs before 6 inside the bracket, not outside as an accidental change.

The direct answer is A: 3x(2y+3z). Find the greatest common factor of 6xy and 9xz. The numerical greatest common factor of 6 and 9 is 3, and x appears in both terms, so the common factor is 3x. Divide each term by 3x: 6xy÷3x=2y and 9xz÷3x=3z. Therefore, 6xy+9xz=3x(2y+3z). Check by distribution: 3x·2y=6xy and 3x·3z=9xz, recovering both original terms. Option B would expand to 6xy+27xz, so the second coefficient is wrong. Option C expands to 18xy+9xz, so the first coefficient is wrong. Option D expands to 18xy+9xz as well, and 9 is not the greatest common numerical factor in the required bracket form. The safest method is divide every term by the factor taken outside.

Correct answer: A. A perfect-square trinomial follows \(a^2+2ab+b^2=(a+b)^2\) or \(a^2-2ab+b^2=(a-b)^2\). In A, \(x^2=x^2\), the constant 25 is \(5^2\), and the middle term is \(10x=2\cdot x hickspace\cdot5\). Therefore, \(x^2+10x+25=(x+5)^2\). A is correct because all three parts match the identity. B has the correct first and middle terms, but its constant should be 25, not 20. C has only two terms, so it is not a trinomial and it is not the displayed binomial square. D has a negative middle term but also a negative constant; a square such as \((x-5)^2\) would be \(x^2-10x+25\), not D. Memory cue: the last term is the square of half the middle-term coefficient when the first term is \(x^2\).

Correct answer: A. Use \(a^2+2ab+b^2=(a+b)^2\). Here the first term is \(x^2=x^2\), the constant term is \(36=6^2\), and twice the product of the two square roots is \(2\cdot x\cdot6=12x\), exactly the middle term. Thus \(x^2+12x+36=(x+6)^2\). A expands back to the original expression, so it is verified. B would expand to \(x^2+24x+144\), changing both the middle and constant terms. C would give \(x^2-12x+36\), so its sign is wrong. D expands to \(x^2+15x+36\), not \(x^2+12x+36\). A useful method is to take the square root of the first and last terms, then check the middle term. Warning: the sign inside the bracket controls the sign of the middle term.

Correct answer: A. The identity for a difference of two squares is \(a^2-b^2=(a-b)(a+b)\). In A, \(9p^2=(3p)^2\) and \(16q^2=(4q)^2\), and the sign between them is minus. Hence \(9p^2-16q^2=(3p)^2-(4q)^2\), exactly the required form; it can further factorise as \((3p-4q)(3p+4q)\). B contains a plus sign, so it is a sum of squares, not a difference. C and D contain three terms and include the mixed term \(24pq\); they are perfect-square trinomials, not two-square differences. In C, the identity is \((3p-4q)^2\), while D is \((3p+4q)^2\). Memory cue: difference of squares has exactly two square terms separated by a minus sign, and its factors have opposite signs.

Correct answer: B. First identify the expression as a difference of squares: \(25x^2=(5x)^2\) and \(36=6^2\). Apply \(a^2-b^2=(a-b)(a+b)\) with \(a=5x\) and \(b=6\). Therefore, \(25x^2-36=(5x-6)(5x+6)\). Expanding B gives \(25x^2+30x-30x-36=25x^2-36\), confirming it. A is \((5x-6)^2=25x^2-60x+36\), which has an unwanted middle term and the wrong constant sign. C expands to \(25x^2-11x-36\), so it is not equivalent. D gives \(25x^2+60x+36\), also incorrect. Common confusion: do not square each bracket separately; the difference-of-squares factors use opposite signs and no middle term remains after multiplication.

Correct answer: C. To take a common factor, look for a factor present in every term. In \(7p+7q\), both terms contain 7, so taking it outside gives \(7p+7q=7(p+q)\). Multiplying back confirms the result. A has no obvious common algebraic factor between \(x^2\) and 9; it is not generally factorised over the usual school-number system by simply taking a common factor. B can be factorised, but by the special identity \(a^2-b^2=(a-b)(a+b)\), not by extracting one common factor from both displayed terms. D is a trinomial and is handled by finding suitable factors of 6 whose sum is 5, not by one common factor shared by all terms. Memory cue: common-factor method means every term must contain the same factor; take it out once and divide each term by it.

Correct answer: A. The difference-of-squares identity is \(a^2-b^2=(a-b)(a+b)\). Option A has exactly two square terms, \(p^2\) and \(q^2\), separated by a minus sign. Therefore, \(p^2-q^2=(p-q)(p+q)\). B has a plus sign, so it is a sum of squares and does not match this identity. C matches \(p^2+2pq+q^2=(p+q)^2\), the square of a sum. D matches \(p^2-2pq+q^2=(p-q)^2\), the square of a difference. The middle terms in C and D are important clues that those are trinomials, not two-term differences. Memory cue: ‘minus between two squares’ means difference of squares; ‘minus twice the product’ means a square of a difference.

Correct answer: A. Rewrite the constant as a square: \(25=5^2\). Then option A becomes \(p^2-5^2\), which is exactly \(a^2-b^2\) with \(a=p\) and \(b=5\). Applying the identity gives \(p^2-25=(p-5)(p+5)\). B is a sum of squares, \(p^2+5^2\), so the difference-of-squares identity does not apply. C is a perfect-square trinomial because \(p^2+10p+25=(p+5)^2\), not a two-square difference. D does have a common factor, \(p^2+5p=p(p+5)\), but it is not presented as a difference of two perfect squares. Thus A is the only suitable choice. Memory cue: check both the minus sign and whether the constant is a perfect square before applying the identity.

Correct answer: C. Use \(a^2-2ab+b^2=(a-b)^2\). The first term is \(4x^2=(2x)^2\), and the last term is \(9=3^2\). The middle term predicted by the identity is \(-2\cdot(2x)\cdot3=-12x\), exactly the given middle term. Hence \(4x^2-12x+9=(2x-3)^2\). A has the same first and last square roots but a plus sign, so it expands to \(4x^2+12x+9\). B expands to \(16x^2-24x+9\), changing the leading term. D expands to \(4x^2-36x+81\), changing both the middle and constant terms. The safest check is to identify the square roots of the first and last terms and then verify the middle term and its sign.

Answer: A, (a − b)(a + b). The expression a² − b² is called a difference of two squares because a² and b² are squares joined by a minus sign. The identity for this pattern is x² − y² = (x − y)(x + y). Put x = a and y = b; therefore a² − b² = (a − b)(a + b). We can verify it by multiplication: (a − b)(a + b) = a² + ab − ab − b² = a² − b², because the middle terms cancel. Option A is correct. Option B gives (a − b)² = a² − 2ab + b², which has a middle term and a positive b². Option C gives a² + 2ab + b², so both the middle term and the sign pattern are different. Option D gives a² − ab, not the required expression. Memory cue: difference of squares means difference factor times sum factor; never square only one binomial.

Answer: B, (p − 6)². Compare the expression with the identity u² − 2uv + v² = (u − v)². Here u = p and v = 6. The middle term is −2 × p × 6 = −12p, and the last term is 6² = 36. Therefore p² − 12p + 36 = (p − 6)². Option A has the wrong sign: expanding it gives p² + 12p + 36. Option C expands to p² − 9p − 36, so neither the middle term nor the constant term matches. Option D expands to p² − 15p + 36, which also has the wrong middle coefficient. A useful check is to take the square root of the first and last terms: p and 6; then use the sign of the middle term. Since the middle term is negative, the binomial must contain subtraction. Memory cue: negative middle term means (first square root − second square root)² when the last term is positive.

Answer: A, x² − 49. The difference-of-squares identity is r² − s² = (r − s)(r + s). Since 49 = 7², x² − 49 can be written as x² − 7². Therefore x² − 49 = (x − 7)(x + 7). Option B is not a difference of squares; it is a perfect-square trinomial: x² + 14x + 49 = (x + 7)². Option C is a sum of squares, x² + 7², and the ordinary difference-of-squares identity does not apply to a plus sign. Option D is also a perfect-square trinomial: x² − 14x + 49 = (x − 7)². Only A has exactly two square terms separated by subtraction. Memory cue: identify two squares first, then check that the sign between them is minus; a plus sign changes the pattern completely.

Answer: A, (x − 4)(x + 4). Since 16 = 4², the expression is x² − 16 = x² − 4², a difference of two squares. The correct identity is r² − s² = (r − s)(r + s), so x² − 16 = (x − 4)(x + 4). Ravi treated the expression as the square of a difference, but that is a different identity. Option B expands to x² − 8x + 16, which contains an unwanted middle term and has a positive constant term. Option C expands to x² + 8x + 16, also not the given expression. Option D expands to x² − 15x − 16, so it does not match either. Multiplying option A verifies the result: x² + 4x − 4x − 16 = x² − 16. Memory cue: a² − b² produces two different conjugate binomials, (a − b) and (a + b); a squared binomial produces a middle term.

Answer: A, a² − b² = (a − b)(a + b). The left side contains two square terms, a² and b², separated by subtraction, so it is a difference of squares. Its factorisation identity is obtained by multiplying conjugate binomials: (a − b)(a + b) = a² + ab − ab − b² = a² − b². Option B is the identity for the square of a sum and expands to a² + 2ab + b², not a difference of squares. Option C is the identity for the square of a difference and includes the middle term −2ab. Option D is a correct identity for the difference of cubes, but it applies to a³ − b³, not a² − b². The exponent pattern must be checked before selecting a formula. Memory cue: square difference gives two conjugate factors; squared binomials give a middle term, and cube difference uses three-term factors.

The direct answer is A: \(9(x+y)\). First group terms that contain the same binomial: \(7x+7y=7(x+y)\) and \(2x+2y=2(x+y)\). Therefore the whole expression becomes \(7(x+y)+2(x+y)\). The common factor is \((x+y)\), so factor it out: \((7+2)(x+y)=9(x+y)\). Expanding A gives \(9x+9y\), which is the same as \(7x+7y+2x+2y\). B expands to \(7xy+7x+2y+2\), introducing terms not present. C also creates unrelated products and does not reproduce the expression. D expands only to \(7x+7y\), leaving out \(2x+2y\). Memory cue: when two groups contain the same bracket, take that entire bracket as the common factor.

The direct answer is A: \((3+x)(a+b)\). Arrange the four terms in two useful groups: \(3a+3b=3(a+b)\) and \(xa+xb=x(a+b)\). Thus the expression becomes \(3(a+b)+x(a+b)\). The common factor is the whole bracket \((a+b)\), so taking it out gives \((3+x)(a+b)\). Expanding A confirms this: \(3a+3b+xa+xb\). Option B would expand to \(3a-3b+xa-xb\), changing both signs involving \(b\). Option C would give \(3a+3b-xa-xb\), changing the signs of the \(x\)-terms. Option D gives \(3ax+3bx\), which has products not present in the original. Common-confusion warning: do not multiply the outside coefficients together; they are being added after the shared bracket is identified.

The direct answer is A: \((10a-b)(10a+b)\). Recognize a difference of squares: \(100a^2=(10a)^2\) and \(b^2=b^2\). Apply \(P^2-Q^2=(P-Q)(P+Q)\) with \(P=10a\) and \(Q=b\). This gives \(100a^2-b^2=(10a-b)(10a+b)\). Expanding confirms it: the cross terms \(+10ab\) and \(-10ab\) cancel, leaving \(100a^2-b^2\). B is not the correct conjugate pair and expands to \(100a^2+99ab-b^2\). C is a square of one binomial and expands to \(100a^2-20ab+b^2\), with extra terms and the wrong sign on \(b^2\). D treats \(100a\), rather than its square root \(10a\), as the factor base; it gives \(10000a^2-b^2\). Memory cue: take square roots first, then use one minus and one plus factor.