Factorisation

The direct answer is A: \((x-5)(x-8)\). For a rectangle model, multiply the proposed side lengths and compare the middle and constant terms. \((x-5)(x-8)=x^2-8x-5x+40=x^2-13x+40\), exactly matching the expression. Equivalently, the two numbers must have product 40 and sum -13; -5 and -8 satisfy both conditions. Option B gives \(x^2+13x+40\), so the middle sign is wrong. Option C gives \(x^2-14x+40\), so its middle coefficient is -14 rather than -13. Option D gives \(x^2+14x+40\), wrong in both sign and coefficient. A common confusion is to focus only on the constant product; both the product and the signed sum must be checked. Memory cue: for \(x^2+bx+c\), find two signed numbers with sum b and product c.

The direct answer is A, (2x−5)(x+3). Expand it by multiplying every rectangle cell: (2x)(x)=2x², (2x)(3)=6x, (−5)(x)=−5x, and (−5)(3)=−15. Combining the two middle contributions gives 6x−5x=x, so the total is 2x²+x−15, exactly the target polynomial. In the rectangle model, the signs of the side labels control the signs of the four regions. Option B gives (2x+5)(x−3)=2x²−6x+5x−15=2x²−x−15, so its middle term is negative. Option C gives (x−5)(2x+3)=2x²+3x−10x−15=2x²−7x−15. Option D gives 2x²+13x−15, not the required expression, because its factors produce the wrong middle coefficient. The constant −15 suggests factors −5 and +3, while the leading term 2x² suggests 2x and x. Memory cue: in a rectangle factorisation model, multiply all four cells, including their signs, before combining like terms.

Direct answer: A, (a+b)(x+y). Group the first two terms and the last two terms: ax+ay+bx+by=a(x+y)+b(x+y). Both groups contain the identical binomial (x+y). Taking it common gives (a+b)(x+y). Expanding confirms the result: a(x+y)+b(x+y)=ax+ay+bx+by. Option B expands to ax+ab+xy+by, introducing ab and xy, which are not the given terms. Option C expands to ab+ay+bx+xy, also a different collection of terms. Option D contains only products with ab and does not reproduce the four original terms; it also omits the required dependence on a and b separately. A common mistake is to choose brackets because they look similar without expanding them. The safest method is to group, extract a common binomial, and then multiply back to check every term.

Direct answer: B, (p+q)(x-y). Group the terms according to p and q: px-py+qx-qy=p(x-y)+q(x-y). The same bracket, (x-y), is common in both parts. Taking it out gives (p+q)(x-y). Expansion confirms: p(x-y)+q(x-y)=px-py+qx-qy. Option A expands to px+py-qx-qy, so the signs of py and qx do not match. Option C expands to px-pq+xy-qy, which contains the unrelated products pq and xy. Option D expands to pq-py-qx+xy and is also not the given expression. Pay special attention to the minus sign: both terms containing y are negative, so the common bracket must be (x-y), not (x+y). Memory cue: after extracting the common bracket, the remaining outside coefficients are p and q, which therefore combine as p+q.

Direct answer: C, \(3(a-2b)(a+2b)\). First take the common factor 3 from both terms: \(3a^2-12b^2=3(a^2-4b^2)\). Now recognise a difference of squares: \(a^2-4b^2=a^2-(2b)^2\). Using \(A^2-B^2=(A-B)(A+B)\), this becomes \((a-2b)(a+2b)\). Therefore the complete factorisation is \(3(a-2b)(a+2b)\). Option A stops after taking out 3 and is incomplete. Option B uses 4b, which would produce \(a^2-16b^2\) and hence \(3a^2-48b^2\). Option D expands to \(3a^2-9ab-12b^2\), introducing an unwanted ab term. Complete factorisation means continue until no common factor or applicable identity remains. A useful check is to multiply the two binomials first: their middle terms cancel and their last terms give \(-4b^2\).

Direct answer: A, a difference of two perfect squares, \(A^2-B^2\). The identity is \(A^2-B^2=(A-B)(A+B)\). It requires two terms, each of which is a perfect square, separated by a minus sign. For example, \(x^2-9=x^2-3^2=(x-3)(x+3)\). Option B has a plus sign; the expression \(A^2+B^2\) is not factored by the difference-of-squares identity. Option C has three terms and does not have the required two-square structure. Option D is a difference of cubes, not squares; it belongs to a different identity, \(A^3-B^3=(A-B)(A^2+AB+B^2)\). A frequent confusion is to focus only on the minus sign and forget that both terms must also be squares. Memory cue: “square minus square gives difference times sum.”

Answer: B. To factorise \(2x^2+11x+15\), look for two factors whose first terms multiply to \(2x^2\), whose constants multiply to 15, and whose cross-products add to \(11x\). In B, \((2x+5)(x+3)=2x^2+6x+5x+15=2x^2+11x+15\). Therefore B is exactly correct. Option A gives \(2x^2+31x+15\), because the cross-products are 30x and x. Option C gives \(2x^2+13x+15\), since 10x+3x=13x. Option D gives \(x^2+11x+30\), so even its leading and constant terms do not match. The safest method is to expand each proposed pair; factorisation must reproduce every original term.

Answer: A. The given expression has the pattern \(u^2-2uv+v^2=(u-v)^2\), with \(u=7a\) and \(v=5b\). Thus it is \((7a-5b)^2\). Expanding confirms this: \((7a-5b)^2=49a^2-70ab+25b^2\). The student’s product \((7a-5b)(7a+5b)\) uses the difference-of-squares identity and expands to \(49a^2-25b^2\); its middle terms cancel, so it cannot represent the given trinomial. Option B would produce a positive middle term, \(+70ab\). Option C is false because A provides a factorisation. Option D expands to \(49a^2-49ab+25b^2\), not the original expression. Always verify the middle term.

Answer: A. First find the greatest common factor of both terms: \(12x^2-27y^2=3(4x^2-9y^2)\). The expression inside the brackets is a difference of squares: \(4x^2=(2x)^2\) and \(9y^2=(3y)^2\). Therefore, \(4x^2-9y^2=(2x-3y)(2x+3y)\), giving the complete factorisation \(3(2x-3y)(2x+3y)\). Option D stops after taking out 3, so it is not complete. Option B expands to a different expression. Option C uses an incorrect common factor and incorrect binomials; its expansion does not match. The key habit is to take out a common factor first, then check whether the remaining expression is another special form.

Answer: C. Begin with the difference of squares: \(x^4-81=(x^2)^2-9^2=(x^2-9)(x^2+9)\). The factor \(x^2-9\) is itself another difference of squares: \(x^2-3^2=(x-3)(x+3)\). Therefore the complete factorisation is \((x-3)(x+3)(x^2+9)\). Option A is only an intermediate factorisation, not complete, because \(x^2-9\) can still be factored. Option B incorrectly uses \(x^2-81=(x-9)(x+9)\), which is not the factor present. Option D gives \((x^2-9)^2\), which expands to \(x^4-18x^2+81\), not the original expression. Memory cue: for a difference of squares, factor it repeatedly until no factor has the same pattern.

Answer: A. First recognise the first three terms as a perfect square: \(x^2-2xy+y^2=(x-y)^2\). The whole expression is therefore \((x-y)^2-25=(x-y)^2-5^2\). Applying \(U^2-V^2=(U-V)(U+V)\), with \(U=x-y\) and \(V=5\), gives \((x-y-5)(x-y+5)\). Option B uses \((x+y)^2\), whose middle term would be +2xy, not −2xy. Option C is only a rewritten expression and also has the wrong sign before 25; the original has subtraction. Option D does not produce the required quadratic expression when multiplied. The important order is: identify the trinomial square first, then apply the difference-of-squares identity.

The correct answer is A: \((3x+4)(2x-3)\). To check a factorisation, multiply the two binomials carefully. In A, the product is \(3x\cdot2x+3x(-3)+4(2x)+4(-3)=6x^2-9x+8x-12=6x^2-x-12\). Thus every term matches. Option B expands to \(6x^2-24x+3x-12=6x^2-21x-12\), so its middle coefficient is wrong. Option C gives \(6x^2-8x+6x-12=6x^2-2x-12\), also incorrect. Option D gives \(6x^2+9x-8x-12=6x^2+x-12\), whose middle term has the wrong sign. A useful method is to multiply the leading coefficients to obtain \(6x^2\), multiply the constants to obtain \(-12\), and then verify that the cross terms combine to \(-x\). Do not judge only by the numbers in the factors; the signs and cross terms must both be checked.

The correct answer is A. Without regrouping, the first three terms of A are already a perfect square: \(p^2-10p+25=(p-5)^2\). The last term is \(49q^2=(7q)^2\), and the sign between these two square expressions is minus. Thus A becomes \((p-5)^2-(7q)^2\), which factors directly as \((p-5-7q)(p-5+7q)\). Option B has a plus sign between the square expressions, so it is a sum of squares, not the required difference. Option C has \(-49q\), which is not the square \((7q)^2\), and its middle sign also changes the perfect square. Option D has \(p^2-25ba\) but the remaining terms do not form one square without regrouping, so it does not fit the requested direct pattern. First identify a complete square and then check the separating sign. Memory cue: complete square minus complete square gives two conjugate factors.

The correct answer is A. Use \((a-b)^3=a^3-3a^2b+3ab^2-b^3\). The first term \(8x^3\) is \((2x)^3\), and the last term \(-27y^3
a\) is \(-(3y)^3\). Put \(a=2x\) and \(b=3y\): \(-3a^2b=-3(4x^2)(3y)=-36x^2y\), while \(3ab^2=3(2x)(9y^2)=54xy^2\). Hence option A is \((2x-3y)^3\). Option B has the wrong sign for the second term, so it does not match either simple cube pattern. Option C has the wrong sign on the final term. Option D has incorrect coefficients in both middle terms. A useful check is that the coefficients 1, 3, 3, 1 appear with signs determined by the binomial. Memory cue: for \((a-b)^3\), the signs are minus, plus, minus after the first term.

The correct answer is A: x³ − 8y³. Recognise 8y³ as (2y)³. Thus x³ − 8y³ = x³ − (2y)³. The difference-of-cubes identity is U³ − V³ = (U − V)(U² + UV + V²). Using U = x and V = 2y gives (x − 2y)(x² + 2xy + 4y²), so x − 2y is a factor. Option B is a sum of cubes and contains x + 2y, not x − 2y. Option C is not a difference of cubes. Option D is the second factor produced after factoring A, but it does not itself contain x − 2y as a factor over the usual polynomial factors. Memory cue: for a difference of cubes, the first factor keeps the same minus sign.

The correct answer is D: x² + x − 1. For a quadratic ax² + bx + c, the discriminant is Δ = b² − 4ac. If the expression factors into linear factors with integer coefficients, its discriminant must be a perfect square. For D, Δ = 1² − 4(1)(−1) = 5, and 5 is not a perfect square; hence integer-coefficient linear factorisation is impossible. Option A factors as (x + 2)(x + 3). Option B is (x − 3)(x + 3). Option C is (x + 1)². Thus A, B, and C all factorise over the integers. Memory cue: a non-square discriminant rules out integer linear factors, although other forms of roots may still exist.

The correct answer is A: x² + 5x + 5. For ax² + bx + c, calculate Δ = b² − 4ac. In A, Δ = 5² − 4(1)(5) = 25 − 20 = 5. Since 5 is not a perfect square, the roots are not rational and the trinomial cannot be written as two linear binomials with integer coefficients. Option B factors as (x + 3)(x + 4). Option C factors as (x − 4)(x + 3). Option D factors as (2x + 1)(x + 3), because the product is 2x² + 7x + 3. Therefore only A fails the required type of factorisation. Memory cue: for integer linear factors, a perfect-square discriminant is a quick necessary check.

The direct answer is C: \(m^2+16\) cannot be split into two non-constant polynomials with integer coefficients. The difference-of-squares identity works only for subtraction: \(A^2-B^2=(A-B)(A+B)\). Option A is \(x^2-7^2=(x-7)(x+7)\). Option B is \((5a)^2-(2b)^2=(5a-2b)(5a+2b)\). Option D is \((3p)^2-q^2=(3p-q)(3p+q)\). In C, the expression is a sum of squares, \(m^2+4^2\), not a difference. Over the integers, there is no corresponding identity that produces two non-constant factors. A common confusion is to write \((m-4)(m+4)\), but that product equals \(m^2-16\), with a minus sign, not \(m^2+16\). Always check the sign before applying the difference-of-squares formula.

The direct answer is A: \(6x^2+11x+3=(3x+1)(2x+3)\). To use middle-term splitting, multiply the leading and constant coefficients: \(6\times3=18\). We need two integers whose product is 18 and whose sum is 11; they are 9 and 2. Split the middle term: \(6x^2+9x+2x+3\). Grouping gives \(3x(2x+3)+1(2x+3)=(3x+1)(2x+3)\). Checking the other choices by possible factor pairs of 18 shows no pair produces the required middle coefficient 10, 9, or 8 together with the necessary grouping into integer binomials. The important procedure is: calculate \(ac\), find a pair with sum \(b\), split the middle term, and factor by grouping. Always expand the final answer to verify it.

The direct answer is A. Expanding \((x+m)(x+n)\) gives \(x^2+(m+n)x+mn\). Comparing coefficients with \(x^2+px+q\), we get \(m+n=p\) and \(mn=q\). Since \(q<0\), the product \(mn\) is negative, so one of \(m,n\) is positive and the other is negative. Since \(p>0\), their sum is positive. When two numbers have opposite signs, their sum is positive only when the positive number has the greater magnitude. Therefore A must be true. B and D are impossible because numbers with the same sign have a positive product or, if both negative, a negative sum. C would produce a negative sum, not the given positive \(p\). Memory cue: negative product means opposite signs; the sign of the sum tells which magnitude is larger.

The direct answer is D: \((4a-5)(3a+2)\). Multiply the first terms: \(4a\cdot3a=12a^2\). Multiply the last terms: \((-5)(2)=-10\). Now calculate the cross-products: \(4a\cdot2=8a\) and \((-5)(3a)=-15a\). Their sum is \(8a-15a=-7a\), so the complete expansion is \(12a^2-7a-10\). Option A gives a middle term \(-14a\), so it is incorrect. Option B gives \(12a^2-26a+10\), with both the middle and constant terms wrong. Option C gives \(12a^2-26a-10\), so only its constant term matches. D matches all three terms. A reliable method is to check the leading product, constant product, and then the signed cross-products; signs must be retained throughout.

Correct answer: B, (3y + 4)(y + 2). Multiply the leading coefficient 3 by the constant 8 to get 24. We need two numbers whose product is 24 and whose sum is 14: they are 12 and 2. Split the middle term: 3y² + 14y + 8 = 3y² + 12y + 2y + 8. Grouping gives 3y(y + 4) + 2(y + 4) = (3y + 2)(y + 4), which appears to be option A—so this reveals that the supplied key B is incorrect. In fact, option A expands to 3y² + 14y + 8, while B expands to 3y² + 10y + 8. Option C expands to 3y² + 14y + 16, and D expands to 3y² + 11y + 8. Therefore the unambiguous correct answer is A, not B. Memory cue: always expand the selected factors; the middle coefficient is the quickest check.

The correct answer is B: (x − 3)(x + 3)(x² + 9). Begin by seeing x⁴ as (x²)² and 81 as 9². Using the difference-of-squares identity p² − q² = (p − q)(p + q), we get x⁴ − 81 = (x² − 9)(x² + 9). The factor x² − 9 can be factored again because it is another difference of squares: x² − 9 = x² − 3² = (x − 3)(x + 3). The factor x² + 9 is not factorised further over the real numbers. Therefore the complete factorisation is (x − 3)(x + 3)(x² + 9). Option A is only an intermediate step, not complete. Option C incorrectly treats x⁴ − 81 as x² − 81. Option D gives a square of x² − 9, which would produce different terms. Memory cue: after applying an identity, always check whether any factor can be broken further.

The correct answer is A: (3x − 2y)(4x − 5y). Multiply the first terms: 3x·4x = 12x², matching the leading term. Multiply the last terms: (−2y)(−5y) = 10y², matching the constant term. Now check the cross-products: 3x(−5y) = −15xy and (−2y)(4x) = −8xy. Their sum is −23xy, exactly the required middle term. Thus the product is 12x² − 23xy + 10y². Option B gives cross-products −6xy and −20xy, whose sum is −26xy. Option C gives a last product of 10y² but its cross-products total −22xy, not −23xy. Option D gives the correct first and last products but the cross-products total −22xy. The safest method is to verify all three resulting terms, especially the middle term where sign errors are common.

The correct answer is A: (x − y − z)(x − y + z). First inspect the first three terms: x² − 2xy + y² = (x − y)², using p² − 2pq + q² = (p − q)². The whole expression becomes (x − y)² − z². Now apply the difference-of-squares identity r² − s² = (r − s)(r + s), with r = x − y and s = z. This gives [(x − y) − z][(x − y) + z] = (x − y − z)(x − y + z). Option B uses (x + y)² and would produce +2xy, not −2xy. Option C is not fully factorised and also has z instead of z². Option D leaves the expression partly expanded rather than factorised into the required product. A useful check is to multiply the two factors: the z terms cancel and the original four-term expression returns.