Factorisation

Answer: A, x−7 metres. For a rectangle, area equals length multiplied by width, so the unknown side is (x²−49)/(x+7). Factor the numerator using the difference of squares: x²−49=x²−7²=(x+7)(x−7). Dividing by the known side x+7 leaves x−7, provided x+7 is nonzero. The condition x>7 also ensures x−7 is positive, as a physical side length must be. B is wrong because using x+7 for both sides would give area (x+7)²=x²+14x+49, not x²−49. C has the wrong algebraic degree and does not result from the division. D does not satisfy the factorisation and would not reproduce the given area. A useful memory cue is a²−b²=(a+b)(a−b): the two rectangle sides are the sum and difference factors.

Answer: B, 3(a+2). Look for the greatest common factor of all terms. Both 3a and 6 are divisible by 3, so take 3 outside the bracket: 3a+6=3(a+2). Expanding verifies the result: 3(a+2)=3a+6. A is wrong because it removes the factor 3 from the first term but leaves the constant unchanged, so a+6 is not equal to 3a+6. C is wrong because 6(a+3)=6a+18, which is much larger and has different coefficients. D is wrong because 3(a−2)=3a−6; the sign of the constant changes. The reliable method is to divide every original term by the common factor and place the resulting terms inside parentheses. Memory cue: factoring is the reverse of distributing; multiplying the outside factor back must reproduce every original term exactly.

The direct answer is A: p² − q². A difference of two squares has the form A² − B², meaning that two squared quantities are separated by a minus sign. Here p² is the square of p and q² is the square of q, and subtraction joins them. It can also be factorised as (p − q)(p + q), using the identity A² − B² = (A − B)(A + B). Option B is p² + q², so it is a sum, not a difference. Option C is p² + 2pq + q², which equals (p + q)², a perfect square trinomial. Option D is p² − 2pq + q², which equals (p − q)², also a perfect square trinomial rather than a difference of two squares. Memory cue: look specifically for square minus square, with no middle term.

The direct answer is A: x² + 8x + 16 = (x + 4)² = (x + 4)(x + 4). When two identical binomials are multiplied, the result has the form (x + a)² = x² + 2ax + a². The first term is x², the constant term must be a², and the middle coefficient must be 2a. In A, the constant 16 is 4² and 2 × 4 = 8, so every part matches. Option B has the required first two terms but 12 is not 4², so it cannot be a repeated binomial square. Option C is a difference of squares, (x − 4)(x + 4), whose factors are not identical. Option D has no middle term and is not a square of a binomial. A useful check is to expand the proposed repeated factor before accepting it.

The direct answer is A: (x − 3)². Use the identity (x − r)² = x² − 2rx + r². The constant term 9 tells us that r can be 3, because 3² = 9. Since the middle coefficient is −6, we need −2r = −6, giving r = 3. Therefore (x − 3)² = x² − 6x + 9, exactly the given expression. Option B gives (x + 3)² = x² + 6x + 9, so its middle term has the wrong sign. Option C gives x² − 12x + 36, so both the middle and constant terms differ. Option D gives x² + 18x + 81, which also does not match. A quick memory cue is: halve the magnitude of the middle coefficient and keep its sign inside the bracket; −6 becomes −3.

The direct answer is B: 2x(x + 4). To take out a common factor, identify what is present in every term. The terms are 2x² and 8x. Both contain 2 and x, so their greatest common factor is 2x. Divide each term by 2x: 2x² ÷ 2x = x and 8x ÷ 2x = 4. Putting the factor outside gives 2x(x + 4). Check by multiplying: 2x·x + 2x·4 = 2x² + 8x. Option A expands to 2x + 16, because x was not removed correctly. Option C does not show division by the common factor and includes an incorrect extra x inside. Option D expands to 8x² + 16x, which is too large. Memory cue: take the greatest common numerical factor and the lowest common power of every common variable.

The direct answer is C: (a + 3)(a + 4). For a monic quadratic a² + ba + c, find two numbers whose product is c and whose sum is b. Here the product must be 12 and the sum must be 7. The numbers 3 and 4 satisfy both conditions: 3 × 4 = 12 and 3 + 4 = 7. Thus a² + 7a + 12 = a² + 3a + 4a + 12 = (a + 3)(a + 4). Option A has product 12 but sum 13, giving a² + 13a + 12. Option B has product 12 but sum 8, giving the wrong middle coefficient. Option D has product 12 but sum −7, producing a² − 7a + 12. Memory cue: for a² + ba + c, the pair must satisfy both sum = b and product = c.

The direct answer is D: (m − 3)(m + 3). First recognise that 9 = 3², so m² − 9 becomes m² − 3². This is a difference of two squares. Apply A² − B² = (A − B)(A + B), with A = m and B = 3: m² − 3² = (m − 3)(m + 3). Expanding option D gives m² + 3m − 3m − 9 = m² − 9, so it is verified. Option A gives m² − 9m, not the original expression. Option B gives m² − 8m − 9 and introduces an unwanted middle term. Option C gives (m − 3)² = m² − 6m + 9, which is a square of a difference, not a difference of squares. Memory cue: a² − b² always produces conjugate factors with opposite signs.

The direct answer is B: (x + 4)(x + 5). For x² + 9x + 20, find two numbers whose product is 20 and whose sum is 9. The numbers 4 and 5 satisfy both conditions: 4 × 5 = 20 and 4 + 5 = 9. Therefore (x + 4)(x + 5) expands to x² + 5x + 4x + 20 = x² + 9x + 20. Option A has product 20 but sum 12, so it gives x² + 12x + 20. Option C has product 20 but sum −9, so it gives a negative middle term. Option D has product 20 but sum 21, producing the wrong middle coefficient. The safest check is always to multiply the selected factors and compare all three terms.

The direct answer is A: 5(p + 3q). Look for a factor present in every term. The numerical coefficients are 5 and 15, and their greatest common factor is 5. The variables p and q are different, so no variable is common to both terms. Divide each term by 5: 5p ÷ 5 = p and 15q ÷ 5 = 3q. Hence 5p + 15q = 5(p + 3q). Verify by distributing: 5·p + 5·3q = 5p + 15q. Option B expands to 5p + 75q, so the second term is incorrect. Option C incorrectly treats p as common even though the second term contains q, not p. Option D uses 15, but 15 cannot be taken from 5p because 5 is not divisible by 15. Memory cue: a common factor must divide every term exactly.

The direct answer is A: 7m(m + 2). Start with the expanded expression represented by the choices: 7m² + 14m. Both terms share 7m, so factor it out: 7m² ÷ 7m = m and 14m ÷ 7m = 2. This gives 7m(m + 2). The bracket m + 2 has no further common numerical factor or common variable, so the factorisation is complete. Option B is still expanded, so its common factor has not been taken out. Option C is only partly factorised because 7m + 14 still has a common factor 7; it can become 7m(m + 2). Option D is also incomplete because m² + 2m still has common factor m; it becomes 7m(m + 2). Memory cue: after factoring, inspect the bracket and continue if every term inside still shares a factor.

The direct answer is C: \((x-7)(x+7)\). The expression has the form of a difference of two squares. First write the constant 49 as \(7^2\): \(x^2-49=x^2-7^2\). Now use the identity \(a^2-b^2=(a-b)(a+b)\), with \(a=x\) and \(b=7\). Therefore, \(x^2-7^2=(x-7)(x+7)\). Option A is incorrect because its product is \(x^2-48x-49\), not the given expression. Option B gives \(x^2-14x+49\), so it has an unwanted middle term and the wrong constant sign. Option C is correct because multiplying gives \(x^2+7x-7x-49=x^2-49\). Option D gives \(x^2+14x+49\), which also has a middle term. Memory cue: a difference of squares always becomes two conjugate factors, one with plus and one with minus.

The direct answer is D: \((y+4)^2\). Begin with the identity \((a+b)^2=a^2+2ab+b^2\). In the given trinomial, the first term is \(y^2\), the last term is \(16=4^2\), and twice the product of y and 4 is \(2\cdot y\cdot4=8y\). Thus \(y^2+8y+16=y^2+2(y)(4)+4^2=(y+4)^2\). Option A gives \(y^2+16y+64\), so its middle and constant terms do not match. Option B gives \(y^2-8y+16\), with the wrong sign in the middle term. Option C gives \(y^2+10y+16\), because the cross terms are 8y and 2y. Option D reproduces all three terms exactly. A useful check is to take the square root of the constant term and then verify the middle term as twice that number times y.

The direct answer is A: \((a-b)^2\). This trinomial matches the identity \((u-v)^2=u^2-2uv+v^2\). Substitute \(u=a\) and \(v=b\): \((a-b)^2=a^2-2ab+b^2\). Hence the given expression factorises as \((a-b)^2\). Option B is wrong because \((a+b)^2=a^2+2ab+b^2\), whose middle term has a plus sign. Option C represents the difference of squares, \(a^2-b^2\), and does not contain the term \(-2ab\). Option D expands only to \(a^2-2ab\), so it misses the final term \(b^2\). Option A is verified by multiplying: \((a-b)(a-b)=a^2-ab-ab+b^2=a^2-2ab+b^2\). Common confusion: the sign inside the square controls the sign of the middle term; a minus inside produces a negative middle term.

The direct answer is B: \(5x(2x-3)\). To factor by taking out a common factor, find what every term shares. The numbers 10 and 15 have greatest common factor 5, and both terms contain x. Therefore the greatest common factor is \(5x\). Divide each term by \(5x\): \(10x^2\div5x=2x\), and \(-15x\div5x=-3\). Thus \(10x^2-15x=5x(2x-3)\). Option A has a plus sign and expands to \(10x^2+15x\), not the given expression. Option C expands to \(10x^2-30x\), so its second coefficient is wrong. Option D expands to \(30x^2-15x\), so its first coefficient is wrong. Option B can be checked by distribution: \(5x\cdot2x=10x^2\) and \(5x\cdot(-3)=-15x\). Memory cue: take out the numerical common factor and the smallest power of each variable present in every term.

The direct answer is A: a common binomial factor in both groups. In grouping, terms are arranged into two or more groups so that each group can be factorised. After taking out a factor from each group, the important test is whether the remaining bracket is identical, including signs. For example, \(x(a+b)+y(a+b)\) contains the common binomial \((a+b)\), so it becomes \((x+y)(a+b)\). Option B is not sufficient because equal exponents do not guarantee that the groups share a factor. Option C is unnecessary; negative terms can also be grouped and factorised. Option D is false because grouping can be used with expressions having different numbers of terms, not only exactly three. The key is to make the same binomial appear in both groups, possibly by taking a negative sign out of one group. Memory cue: group, factor each group, then look for the same bracket.

The direct answer is A: \(x^2-y^2=(x-y)(x+y)\). Rewrite 25 as \(5^2\), so \(a^2-25=a^2-5^2\). This is a difference of two squares, and the identity gives \(a^2-5^2=(a-5)(a+5)\). Option B is the identity for the square of a sum and is used for a three-term expression with a positive middle term. Option C is the identity for the square of a difference and also requires a three-term expression containing \(-2xy\). Neither applies directly to two squares separated by subtraction. Option D concerns a sum of cubes, but the expression contains squares, not cubes, and it is a difference rather than a sum. Option A is therefore the suitable identity, even though its letters x and y are general symbols; they can represent a and 5. Memory cue: two perfect squares with a minus sign between them signal difference of squares.

The direct answer is B: \(7x(y+2)\). Look for the greatest common factor in both terms. In \(7xy\) and \(14x\), the common numerical factor is 7 and the common variable factor is x, so the greatest common factor is \(7x\). Divide each term by \(7x\): \(7xy\div7x=y\), and \(14x
div7x=2\). Therefore \(7xy+14x=7x(y+2)\). Option A does not contain the correct common factor placement and expands to \(7y+98x\). Option C expands to \(7xy+14xy\), which incorrectly gives the second term an extra y. Option D expands to \(14xy+98x\), so both coefficients are wrong. Option B is confirmed by multiplying: \(7x
y+7x
times2=7xy+14x\). A reliable habit is to divide every original term by the factor taken outside, then multiply back to verify. Common confusion: the factor 14 is not common to the first term because \(7xy\) is not divisible by 14 as an integer coefficient.

The direct answer is A: \((x+1)(x+2)\). For a monic quadratic of the form \(x^2+bx+c\), find two numbers whose sum is b and whose product is c. Here the required sum is 3 and the required product is 2. The numbers 1 and 2 satisfy both conditions, so \(x^2+3x+2=(x+1)(x+2)\). Verify by multiplication: \(x\cdot x=x^2\), the cross terms are \(2x+x=3x\), and the constant product is \(1
times2=2\). Option B gives \(x^2+5x+6\), so both the middle and final terms are wrong. Option C gives \(x^2-3x+2\), with the wrong sign in the middle term. Option D gives \(x^2+4x+3\), so its sum and product do not match. The safest method is to multiply each proposed pair back out. Memory cue: for \(x^2+bx+c\), the two constants must add to b and multiply to c.

The direct answer is B: \(4pq(3p-2q)\). Find the greatest common factor by considering numbers and variables separately. The greatest common numerical factor of 12 and 8 is 4. Both terms contain p and q; the smallest powers common to both are p and q, so the variable part is pq. Thus the greatest common factor is \(4pq\). Divide: \(12p^2q\div4pq=3p\), and \(-8pq^2
div4pq=-2q\). Hence \(12p^2q-8pq^2=4pq(3p-2q)\). Option A is algebraically equivalent because it expands to the same expression, but it does not remove the greatest common factor; 2pq is not the greatest possible common factor. Option C does not reproduce the terms and has an incorrect coefficient. Option D is also algebraically equivalent to the original expression, but it leaves q inside and therefore does not take out the full common factor. Option B is the expected fully factorised form. Always multiply back to check both terms and signs.

The direct answer is C: \((z-2)^2\). Recognise the expression as a perfect-square trinomial. The first term is \(z^2\), the last term is \(4=2^2\), and the middle term is \(-4z=-2
times z imes2\). Therefore it matches \((u-v)^2=u^2-2uv+v^2\), with u=z and v=2: \(z^2-4z+4=(z-2)^2\). Option A gives \(z^2+4z+4\), so its middle sign is wrong. Option B is only a partially rearranged expression; expanding it gives \(z^2-4z+4\), but it is not the requested compact factorised product of the displayed options. Option D expands to \(z^2-3z-4\), which is different. Option C can be checked directly: \((z-2)(z-2)=z^2-2z-2z+4=z^2-4z+4\). Memory cue: when the first and last terms are squares, check whether the middle term is plus or minus twice their square roots multiplied together.

The correct answer is A: (5p − 2)² = 25p² − 20p + 4. Start with the identity (a − b)² = a² − 2ab + b². Here a = 5p and b = 2. Therefore, a² = 25p², −2ab = −2 × 5p × 2 = −20p, and b² = 4. Combining these terms gives 25p² − 20p + 4, exactly the expression in the question. Option A is correct because all three terms match. Option B omits the middle term, so it is not the square of a binomial. Option C has an incorrect middle coefficient; the product must be doubled. Option D has the wrong sign because squaring a difference produces a negative middle term. Memory cue: for (a − b)², remember “square, minus double product, square.”

The correct answer is A: (a + 2)(a + 3). To factor a² + 5a + 6, find two numbers whose product is the constant term 6 and whose sum is the coefficient of a, 5. The numbers 2 and 3 satisfy both conditions: 2 × 3 = 6 and 2 + 3 = 5. Thus a² + 5a + 6 = (a + 2)(a + 3). Verify by multiplying: a(a + 3) + 2(a + 3) = a² + 3a + 2a + 6 = a² + 5a + 6. Option B gives a middle coefficient of 7 because 1 + 6 = 7. Option C gives a negative middle coefficient, −5a, and option D gives a much larger product and middle coefficient. Memory cue: for a² + ba + c, look for two numbers that multiply to c and add to b.

The correct answer is A: (4n − 5)(4n + 5). Recognise 16n² − 25 as a difference of two squares. Since 16n² = (4n)² and 25 = 5², use the identity a² − b² = (a − b)(a + b). Substitution gives (4n)² − 5² = (4n − 5)(4n + 5). Expanding this answer gives 16n² + 20n − 20n − 25 = 16n² − 25, so it is verified. Option B is a square of a binomial and would contain a middle term −40n. Option C incorrectly uses 16n instead of its square root 4n, producing 256n² − 25. Option D has n rather than 4n and produces n² − 25. Memory cue: in a difference of squares, take the square root of each term before making the two brackets.

The correct answer is C: 4(2x + 3). Factorisation means taking a common factor outside the bracket. The terms 8x and 12 have common factors 1, 2, and 4; the greatest common factor is 4. Divide each term by 4: 8x ÷ 4 = 2x and 12 ÷ 4 = 3. Therefore 8x + 12 = 4(2x + 3). Expanding confirms it: 4 × 2x + 4 × 3 = 8x + 12. Option A is mathematically equal to the original expression, but it is not simplest because 2 is not the greatest common factor; the bracket can still be factored. Option B expands to 8x + 96, not the given expression. Option D expands to 96x + 12. Memory cue: for the simplest factorised form, remove the greatest common factor from every term.