Factorisation

The correct answer is 3. A common factor must divide every term of the expression. The terms are 6x and 9. The greatest numerical factor common to 6 and 9 is 3, and there is no variable x in the second term, so x cannot be a common factor. Taking out 3 gives 6x + 9 = 3(2x + 3). This confirms that 3 is a factor of both terms and that the expression is correctly factorised. Option B is wrong because 6 does not divide 9. Option C is wrong because 9 does not divide 6x as a numerical factor in general. Option D is wrong because 9 has no factor x. To find a greatest common factor, first find the numerical GCF of the coefficients and then include only variables that occur in every term with the smallest common power.

The correct factorisation is 4(a + 2). Both terms, 4a and 8, have 4 as a common factor. Divide each term by 4: 4a ÷ 4 = a and 8 ÷ 4 = 2. Therefore, 4a + 8 = 4(a + 2). Multiplying back gives 4·a + 4·2 = 4a + 8, which verifies the answer. Option A would expand to 4a + 32, so it is too large in the constant term. Option C would expand to 8a + 32, changing both terms. Option D would give 4a + 8a = 12a, because the a is incorrectly placed outside the constant 8; it is not equivalent to the original expression. A safe checking habit is to multiply the proposed factorisation back. In factorisation, take out the greatest common factor and divide every term by it, not just one term.

Answer: C, x. The expression has two terms: xy and xz. To find a common factor, look for a number or variable that appears as a factor in every term. The variable x appears in xy and also in xz. Therefore, x can be taken outside the brackets: xy + xz = x(y + z). The remaining factors are y and z. Option A, y, is present only in xy, not in xz. Option B, z, is present only in xz, not in xy. Option D, yz, is not a factor of either term because xy has no z and xz has no y. A useful check is to multiply x(y + z) again; it returns xy + xz. Memory cue: a common factor must be present in every term, not merely in one term.

Answer: D, 5(m - 3). Both terms, 5m and -15, have 5 as a common factor. Divide each term by 5: 5m ÷ 5 = m and -15 ÷ 5 = -3. Therefore, taking 5 outside gives 5(m - 3). We can verify it by distributing 5: 5 × m + 5 × (-3) = 5m - 15, which is the original expression. Option A, 5(m + 3), expands to 5m + 15, so its sign is wrong. Option B, 15(m - 5), expands to 15m - 75, so neither coefficient is correct. Option C, m(5 - 15), expands to 5m - 15m, which is not the given expression. Memory cue: when a positive common factor is taken out, retain the negative sign inside the bracket.

Answer: A, 6m² + 9mn. A common factor must divide every term, so both its numerical part, 3, and its variable part, m, must occur in every term. In option A, 6m² = 3m × 2m and 9mn = 3m × 3n. Thus the expression can be written as 3m(2m + 3n). Option B fails because 6n has no m factor, even though the first term contains 3m. Option C fails because 3n has no m factor. Option D fails because 9n² has no m factor. Notice that a numerical coefficient alone is not enough; the variable must also be present with at least the required power. Memory cue: test a proposed common factor term by term by dividing each term by it; every quotient must be an algebraic expression without a leftover fraction or variable.

Answer: D, (x - 4)(x + 4). Since 16 = 4², the expression can be written as x² - 4². It is therefore a difference of two squares. Using a² - b² = (a - b)(a + b), with a = x and b = 4, we obtain x² - 16 = (x - 4)(x + 4). Option A incorrectly uses 16 instead of its square root, 4, as the second number. Option B expands to x² - 8x + 16, which has a middle term. Option C expands to x² + 8x + 16, also with a middle term and a different sign. Multiplying option D confirms the result: x² + 4x - 4x - 16 = x² - 16. Memory cue: take the square root of the constant before forming the two factors.

Answer: A, (a + b)². The identity for the square of a sum is (a + b)² = a² + 2ab + b². The given expression has the first square a², the last square b², and the middle term 2ab, exactly matching this identity. Therefore, a² + 2ab + b² = (a + b)². Option B would expand to a² - 2ab + b², so its middle sign is wrong. Option C is a difference of squares and expands to a² - b², so it has no 2ab term. Option D expands to a² + 2ab, leaving out b². To check the correct answer, multiply (a + b)(a + b): a² + ab + ab + b² = a² + 2ab + b². Memory cue: for a plus-square trinomial, the middle term is twice the product of the two square roots.

Answer: A, (p - q)(p + q). The expression p² - q² is a difference of two squares. Apply the identity a² - b² = (a - b)(a + b), taking a = p and b = q. This gives p² - q² = (p - q)(p + q). Option B expands to p² - 2pq + q², so it contains a middle term and has a positive q² term. Option C expands to p² + 2pq + q², which also has the wrong middle sign and a positive final term. Option D expands to p² - pq, so it does not reproduce the original expression. Multiplying option A shows the middle terms cancel: p² + pq - pq - q² = p² - q². Memory cue: a difference of squares always becomes difference times sum, not the square of a difference.

Answer: A, \(3(x+y)\). Factorisation means rewriting an expression as a product. Look at the two terms, \(3x\) and \(3y\). Both contain the common factor 3. Divide each term by 3: \(3x\div3=x\) and \(3y\div3=y\). Therefore, taking 3 outside the bracket gives \(3x+3y=3(x+y)\). Option A is correct because expanding it gives \(3x+3y\), exactly the original expression. Option B is wrong because \(3(x-y)=3x-3y\), which changes the plus sign to a minus sign. Option C is wrong because expanding it gives \(3x+xy\), not the given expression. Option D gives \(3y+xy\), so it is also not equivalent. Memory cue: when the same positive number multiplies every term, divide every term by that number and place the number outside the bracket. Always expand once to check.

Answer: A, \(a-2\). First identify the greatest common factor of the two terms \(7a^2\) and \(-14a\). Both contain 7a, so take \(7a\) outside. Divide the first term by it: \(7a^2\div7a=a\). Divide the second term by it: \(-14a\div7a=-2\). Thus, \(7a^2-14a=7a(a-2)\), and the part left inside the bracket is \(a-2\). Option A states exactly what remains. Option B is the complete factorised expression, not merely the bracketed remainder. Option C is wrong because the second term would need to be \(-14\), but division by \(7a\) gives \(-2\). Option D does not result from dividing both terms by the same common factor. Memory cue: “outside factor × inside remainder” must reproduce every original term; multiply \(7a(a-2)\) to check.

Answer: B, \((x+2)(x+3)\). For a monic quadratic \(x^2+bx+c\), look for two numbers whose sum is the coefficient of x and whose product is the constant term. Here we need numbers with sum 5 and product 6. The numbers 2 and 3 satisfy both conditions: \(2+3=5\) and \(2\times3=6\). Therefore, \(x^2+5x+6=(x+2)(x+3)\). Option A gives middle coefficient 7 and constant 6, so it expands to \(x^2+7x+6\). Option B expands to \(x^2+3x+2x+6=x^2+5x+6\), so it is correct. Option C gives \(x^2-5x+6\), with the wrong sign in the middle term. Option D gives \(x^2+11x+30\), so both its middle and constant terms are wrong. Memory cue: for \((x+p)(x+q)\), add p and q for the middle coefficient and multiply them for the constant term.

Answer: C, \((x-5)(x+5)\). Rewrite 25 as \(5^2\), so the expression becomes \(x^2-5^2\). This is a difference of squares. Applying \(a^2-b^2=(a-b)(a+b)\) with a=x and b=5 gives \((x-5)(x+5)\). Expanding confirms it: \(x^2+5x-5x-25=x^2-25\). Option A is wrong because its product is \(x^2-24x-25\), not the given expression. Option B gives \(x^2+10x+25\), because it is a square of a sum. Option D gives \(x^2-10x+25\), because it is a square of a difference. Both B and D contain a middle term, whereas the original expression has none. Memory cue: for a difference of squares, use the square root of the constant and write one sum factor and one difference factor.

Answer: D, \((3x-1)(3x+1)\). Express each term as a square: \(9x^2=(3x)^2\) and \(1=1^2\). Therefore \(9x^2-1=(3x)^2-1^2\). Use the difference-of-squares identity \(a^2-b^2=(a-b)(a+b)\), with a=3x and b=1. This gives \((3x-1)(3x+1)\). Expanding D gives \(9x^2+3x-3x-1=9x^2-1\). Option A expands to \(9x^2-6x+1\), so it introduces an unwanted middle term. Option B expands to \(9x^2+8x-1\), so it is not equivalent. Option C expands to \(9x^2+6x+1\), also with an unwanted middle term and the wrong constant sign. Memory cue: when two square terms have a minus between them, use one “minus” factor and one “plus” factor.

Answer: A, \(2(x+3)\). Taking 2 common means finding the expression left after dividing every term by 2. For the first term, \(2x\div2=x\). For the second term, \(6\div2=3\). Put the common factor outside and the results inside the bracket: \(2x+6=2(x+3)\). Option A is correct because expanding gives \(2x+6\). Option B would expand to \(2x+12\), so the constant term is too large. Option C does not represent taking 2 common from both terms; it also expands to \(2x+6x=8x\), not the original expression. Option D expands to \(6x+12\), so it is incorrect. Memory cue: divide every term by the factor taken outside; do not leave an unchanged term inside the bracket. Expansion is the fastest check.

Answer: B, 5. The question asks only for the greatest common numerical factor, so focus on the numbers 15 and 20; the variables p and q are different and are not part of the numerical factor. List the positive factors: factors of 15 include 1, 3, 5, 15; factors of 20 include 1, 2, 4, 5, 10, 20. The greatest number appearing in both lists is 5. Thus, 5 is the numerical HCF and \(15p+20q=5(3p+4q)\). Option A, 10, does not divide 15. Option B, 5, divides both 15 and 20 and is the greatest such number. Option C, 15, does not divide 20. Option D, 20, does not divide 15. Memory cue: for a numerical common factor, ignore unlike variables and find the HCF of the coefficients. Then verify by dividing both coefficients exactly.

The correct answer is A: 5(3p + 4q). Factorisation means writing an expression as a product by taking out a common factor. The numerical coefficients 15 and 20 have greatest common factor 5. There is no variable common to both terms, because one term contains p and the other contains q. Divide each term by 5: 15p ÷ 5 = 3p and 20q ÷ 5 = 4q. Therefore, 15p + 20q = 5(3p + 4q). Option A is correct because expanding it gives 15p + 20q. Option B gives 30p + 20q, so its first term is wrong. Option C gives 75p + 20q, so the factor has not been taken out correctly. Option D gives 15p + 60q, so its second term is wrong. A useful check is to multiply the outside factor by every term inside the bracket. Memory cue: divide every term by the common factor, not just one term.

The correct answer is A: x² + 6x + 9. A perfect-square trinomial has the form a² + 2ab + b² = (a + b)², or a² − 2ab + b² = (a − b)². Here x² is the square of x and 9 is the square of 3. The middle term must be twice their product: 2 × x × 3 = 6x. Thus x² + 6x + 9 = (x + 3)² = (x + 3)(x + 3). Option B fails because 8 is not 3², and its middle term does not match a square using the constant 8. Option C has the correct first and last square terms, but the middle term should be 6x, not 5x. Option D has a negative middle term, but its constant 8 still prevents it from being a square trinomial. Memory cue: square the first and last roots, then check whether the middle term is twice their product.

The correct answer is A: x² + 8x + 16. The first term is x² = (x)² and the last term is 16 = 4². For a square trinomial, the middle term must be 2 × x × 4 = 8x. Therefore x² + 8x + 16 = (x + 4)² = (x + 4)(x + 4). Option B has the same middle term but 12 is not a perfect square equal to 4², so it cannot be the required square. Option C has the correct last term, but its middle term is 6x instead of 8x. Option D also has 16 as its last term, but 4x is not twice x times 4. Expanding (x + 4)(x + 4) gives x² + 4x + 4x + 16 = x² + 8x + 16. Memory cue: for x² + bx + c to be (x + r)², c must be r² and b must be 2r.

The correct answer is D: (2x + 3)². Compare the expression with a² + 2ab + b². We have 4x² = (2x)² and 9 = 3². The middle term predicted by the identity is 2 × (2x) × 3 = 12x, exactly the given middle term. Hence 4x² + 12x + 9 = (2x + 3)², or (2x + 3)(2x + 3). Option A would expand to 4x² + 36x + 81, so both its middle and constant terms are wrong. Option B would expand to 16x² + 24x + 9, so its first coefficient is wrong. Option C would expand to 4x² − 12x + 9, giving the opposite sign for the middle term. Option D expands to 4x² + 6x + 6x + 9 = 4x² + 12x + 9, so it is correct. Memory cue: take the square root of the first and last terms, then use the sign of the middle term.

The correct answer is A: 3p² + 12p. A common factor must divide every term of an expression. In the first term, 3p² ÷ 3p = p. In the second term, 12p ÷ 3p = 4. Therefore both terms contain 3p, and the expression can be written as 3p(p + 4). Option B is not correct because 12 has no factor p; 12 ÷ 3p is not a polynomial term. Option C is not correct because p² has coefficient 1, so it is not divisible by 3p. Option D is not correct because 9 has no factor p, even though 6p² is divisible by 3p. Option A is confirmed by expanding 3p(p + 4) = 3p² + 12p. Memory cue: a factor is common only when it appears in every single term, not merely in one term or most terms.

The correct answer is A: a(b + c + d). Look at each term: ab, ac and ad. The letter a occurs in every term, so it is the common factor. Divide each term by a: ab ÷ a = b, ac ÷ a = c and ad ÷ a = d. Taking a outside gives ab + ac + ad = a(b + c + d). Option A is verified by distribution: a·b + a·c + a·d = ab + ac + ad. Option B would expand to ab + bc + bd, introducing terms that were not present. Option C would expand to ac + bc + cd, and option D would expand to ad + bd + cd; neither matches the original expression. A common mistake is to choose a letter that appears in only one term. Memory cue: circle the symbols repeated in every term, then place the remaining part inside the bracket.

The correct answer is A: y² − 49. Rewrite 49 as 7², giving y² − 7². Both terms are perfect squares and they are separated by subtraction, so the difference-of-squares identity applies: y² − 7² = (y − 7)(y + 7). Option B is y² + 7², a sum of squares, not a difference of squares. Option C is actually a perfect square trinomial: y² + 14y + 49 = (y + 7)², so it should use the square-of-a-sum identity. Option D is y² − 14y + 49 = (y − 7)², so it is also a perfect square trinomial, not the difference-of-squares form. The absence of a middle term in A is another useful clue. Memory cue: two squares with a minus sign split into conjugate factors, one with minus and one with plus.

The correct answer is A: 3 + 4 = 7 and 3 × 4 = 12. Multiply the binomials step by step: (x + 3)(x + 4) = x·x + x·4 + 3·x + 3·4 = x² + 4x + 3x + 12 = x² + 7x + 12. For a monic quadratic x² + bx + c, we look for two numbers whose sum is b and whose product is c. Here the numbers 3 and 4 have sum 7 and product 12. Option B reverses these two conditions, and 3 × 4 is not 7. Option C incorrectly uses a difference; 3 − 4 is −1, not 7. Option D also gives false relationships: 3 + 4 is 7, not 12, and 3 − 4 is −1. Memory cue: in (x + m)(x + n), the middle coefficient is m + n and the constant term is mn.

The correct answer is B: (x − 7)(x + 7). Rewrite the expression as x² − 49 = x² − 7². It is a difference of squares, so use a² − b² = (a − b)(a + b): x² − 7² = (x − 7)(x + 7). Riya’s expression (x − 7)² is a square of a difference. Expanding it gives x² − 14x + 49, which is not x² − 49 because it contains a middle term and has a positive constant. Option C expands to x² + 14x + 49, so both the middle term and constant sign are unsuitable. Option D expands to x² − 48x − 49, which is also different. Option B expands to x² + 7x − 7x − 49 = x² − 49, confirming it. Memory cue: a difference of squares produces two different conjugate binomials, not two identical binomials.

The direct answer is B: x²−9. The difference-of-squares identity applies to an expression of the form A²−B², meaning there must be two square terms separated by a minus sign. Here, x² is the square of x and 9 is the square of 3, so x²−9=x²−3². Therefore, using A²−B²=(A−B)(A+B), we get x²−9=(x−3)(x+3). Option A, x²+9, has a plus sign, so it is a sum of squares rather than a difference of squares. Option C, x²+6x+9, is the perfect-square trinomial (x+3)². Option D, x²−6x, has a common factor x, but it is not written as two complete squares in the required form. A useful memory cue is: two square terms with a minus sign become one minus factor and one plus factor.