Factorisation

The correct answer is A: x^2 + 10x + 25 = (x + 5)^2. In the identity a^2 + 2ab + b^2, take a = x and b = 5. Then a^2 = x^2, 2ab = 2(x)(5) = 10x, and b^2 = 25. Thus all parts match exactly. Option B has the correct first two terms but its constant is 20 instead of 25, so it is not a perfect square of this form. Option C has the correct square terms but a negative middle term; it equals (x − 5)^2 and uses the difference identity. Option D has 5x rather than the required 10x, so its middle term is too small. The safest check is to identify the square roots of the first and last terms and then verify that twice their product gives the middle term, including its sign.

The correct answer is A: (7x + y)^2. Use the identity (p + q)^2 = p^2 + 2pq + q^2. The first term 49x^2 is (7x)^2, and the last term y^2 is y^2. Now check the middle term: 2(7x)(y) = 14xy, which is positive and exactly matches the expression. Therefore 49x^2 + 14xy + y^2 = (7x + y)^2. Option B would expand to 49x^2 − 14xy + y^2, so its middle sign is wrong. Option C would have first term 2401x^2, not 49x^2, because (49x)^2 is much larger. Option D is the difference of squares and expands to (7x)^2 − y^2 = 49x^2 − y^2; it has no middle term. A helpful memory cue is: take the square roots of the first and last terms, then use the sign of the middle term between them.

The correct answer is B: (9a − b)^2. Apply (p − q)^2 = p^2 − 2pq + q^2. Let p = 9a and q = b. Then p^2 = 81a^2, −2pq = −2(9a)(b) = −18ab, and q^2 = b^2. These are exactly the three terms in the given expression, so the factorisation is (9a − b)^2. Option A gives a positive middle term, 18ab, and therefore does not match. Option C has an incorrect first term: (81a)^2 is 6561a^2, not 81a^2. Option D expands using the difference-of-squares identity to 81a^2 − b^2, so it lacks the middle term. Common confusion: a negative middle term does not mean a difference of squares automatically; if there are three matching terms, check for the square of a difference first.

The correct answer is A: (2x + 1)(x + 2). Verify it by multiplying each term: (2x + 1)(x + 2) = 2x·x + 2x·2 + 1·x + 1·2 = 2x^2 + 4x + x + 2 = 2x^2 + 5x + 2. Thus the leading term, middle term, and constant all match. Option B expands to 2x^2 + 2x + 2x + 2 = 2x^2 + 4x + 2, so the middle coefficient is wrong. Option C expands to x^2 + 5x + 4, so both the leading and constant terms differ. Option D gives 2x^2 − 5x + 2, with the wrong sign for the middle term. A reliable method is to use the outer and inner products: in the correct pair, 4x + x must combine to 5x, while the first and last products must be 2x^2 and 2.

The correct answer is A: x^2 + 10x + 25 = (x + 5)^2. Compare the expression with a^2 + 2ab + b^2. Choose a = x and b = 5. Then a^2 = x^2, 2ab = 2(x)(5) = 10x, and b^2 = 25. Every part matches, so the identity applies directly. Option B has 24 instead of 25, so the last term is not the square of 5. Option C has 5x instead of 10x; the middle term is not twice the product of the two square roots. Option D has a negative middle term and a constant 24, so it matches neither the plus-square nor the minus-square pattern. To avoid confusion, do not check only the first and last terms: always verify the middle term as twice the product of their square roots, with the correct sign.

The correct answer is A: (5x + 1)(x + 3). Multiply the proposed factors carefully: (5x + 1)(x + 3) = 5x·x + 5x·3 + 1·x + 1·3 = 5x^2 + 15x + x + 3 = 5x^2 + 16x + 3. The leading term and constant term are correct, and the two cross-products combine to give the required 16x. Option B expands to 5x^2 + 5x + 3x + 3 = 5x^2 + 8x + 3, so its middle term is too small. Option C expands to x^2 + 8x + 15, which has the wrong leading and constant terms. Option D expands to 5x^2 − 15x − x + 3 = 5x^2 − 16x + 3, so the middle sign is wrong. Memory cue: for two binomials, check the first product, the last product, and then add the two cross-products to test the middle term.

The direct answer is A: \((2x+3)(3x+2)\). To factorise, first check the product of the leading and constant coefficients: \(6\times6=36\). We need two middle products whose sum is \(13x\); they are \(4x\) and \(9x\). Thus \(6x^2+13x+6=6x^2+4x+9x+6\), and grouping gives \(2x(3x+2)+3(3x+2)=(2x+3)(3x+2)\). Option A expands to \(6x^2+4x+9x+6=6x^2+13x+6\). B expands to \(6x^2+15x+6\), so its middle coefficient is wrong. C gives \(6x^2+12x+6\), also wrong. D gives positive leading and constant terms but a negative middle term, so it cannot match. Memory cue: always multiply the proposed factors back to check all three terms.

The direct answer is A: \((2a+5)(a-3)\). Multiply the leading coefficient and constant: \(2\times(-15)=-30\). We need two numbers with product \(-30\) and sum \(-1\): they are \(5\) and \(-6\). Split the middle term: \(2a^2-a-15=2a^2+5a-6a-15\). Grouping gives \(a(2a+5)-3(2a+5)=(2a+5)(a-3)\). Checking option A directly gives \(2a^2-6a+5a-15=2a^2-a-15\). B gives a positive constant, not the required negative one. C gives \(2a^2+7a-15\), so its middle term is wrong. D gives \(2a^2-31a+15\). A negative constant usually means the two constant signs in the factors are opposite, but the middle coefficient must still be checked.

The direct answer is A: \(x^2-81\). The difference-of-squares identity is \(P^2-Q^2=(P-Q)(P+Q)\). Here, \(x^2=x^2\) and \(81=9^2\), so \(x^2-81=x^2-9^2=(x-9)(x+9)\). Expanding confirms the result: \((x-9)(x+9)=x^2+9x-9x-81=x^2-81\). Option B is a sum of squares, not a difference, and it does not factorise by this integer identity. Option C is instead the perfect square \((x+9)^2\), because its middle term is \(18x\). Option D is not a difference of two perfect squares in the required form; its middle term prevents that identity. Memory cue: look for two square terms separated by a minus sign, then use one minus and one plus factor.

The direct answer is B: \((2x-3)^2\), or equivalently \((2x-3)(2x-3)\). Recognize the pattern \((P-Q)^2=P^2-2PQ+Q^2\). The first term \(4x^2\) is \((2x)^2\), and the last term \(9\) is \(3^2\). The middle term should be \(-2(2x)(3)=-12x\), exactly as given. Therefore \(4x^2-12x+9=(2x-3)^2\). A would produce \(+12x\), not \(-12x\). C expands to \(4x^2-13x+9\), so its middle coefficient is incorrect. D is the difference \((2x)^2-3^2=4x^2-9\), which has no middle term. The sign of the middle term is the key: a negative middle term requires subtraction inside the square.

The direct answer is A: \((x+7)^2\). Use the identity \((P+Q)^2=P^2+2PQ+Q^2\). In this expression, the first term is \(x^2\), so take \(P=x\). The constant term is \(49=7^2\), so take \(Q=7\). The middle term predicted by the identity is \(2(x)(7)=14x\), which matches. Hence \(x^2+14x+49=(x+7)^2\). Option B expands to \(x^2-14x+49\), so its middle sign is wrong. C expands to \(x^2+63x+686\), not the given expression. D expands to \(x^2+14x+45\), where the constant term is wrong. A useful check is to take the square roots of the first and last terms and verify twice their product as the middle term.

The direct answer is A: \(b^2=4c\). A monic perfect-square trinomial has the form \((x+r)^2=x^2+2rx+r^2\). Comparing coefficients with \(x^2+bx+c\), we get \(b=2r\) and \(c=r^2\). Squaring the first relation gives \(b^2=4r^2=4c\). Therefore the necessary condition is \(b^2=4c\), and when this holds, \(x^2+bx+c=(x+b/2)^2\). Option B incorrectly makes the middle coefficient directly equal to four times the constant. Option C misses the factor 4; for example, with \(r=1\), we have \(b=2,c=1\), so \(b^2=4\), not \(1\). Option D is unrelated to the perfect-square pattern. Memory cue: halve the middle coefficient and square it; that must equal the constant term.

The direct answer is B: \(m^2-36n^2\). Rewrite the second term as a square: \(36n^2=(6n)^2\). Therefore the expression is \(m^2-(6n)^2\), exactly the difference-of-squares pattern. Applying \(A^2-B^2=(A-B)(A+B)\), with \(A=m\) and \(B=6n\), gives \((m-6n)(m+6n)\). Expanding cancels the cross terms and returns \(m^2-36n^2\). A is a sum of squares, so the stated identity does not apply. C is the perfect square \((m+6n)^2\), because its middle term is \(12mn\). D has \(36n\), not \(36n^2\), so the second term is not a perfect square of the required form. Memory cue: both terms must be squares and the sign between them must be minus.

Answer: A, x² + 10x + 25 = (x + 5)². The basic identity is a² + 2ab + b² = (a + b)². Here the first term x² is the square of x, the last term 25 is the square of 5, and the middle term must be 2 × x × 5 = 10x. Therefore the expression has exactly the required pattern and is a perfect-square trinomial. A is correct because it becomes (x + 5)². B is wrong because x² + 10x + 20 has 20 instead of the required 25. C is wrong because the middle term should be 10x, not 5x. D is wrong because although its middle term is −10x, the constant term would need to be 25 for (x − 5)²; with 20 it is not a square. Memory cue: square the end terms and check whether the middle term is twice their square roots multiplied together.

Answer: A, because a² + 2ab + b² = (a + b)². To understand this from the beginning, multiply (a + b) by itself: a² + ab + ab + b² = a² + 2ab + b². Thus a perfect-square trinomial has three parts: the square of the first term, twice the product of the two terms, and the square of the second term. A contains all three parts in the correct signs and coefficients. B is wrong because its middle term is only ab, not 2ab. C is wrong because the signs do not match the identity for (a+b)²; it also has a negative final term. D is wrong because it has only two terms and lacks the required middle term. Do not confuse (a+b)² with a²+b²: squaring a sum creates the two cross-products ab, which combine to 2ab.

Answer: A. We can check a proposed factorisation by expanding it. For A, (2x + 5)(x + 2) = 2x·x + 2x·2 + 5·x + 5·2 = 2x² + 4x + 5x + 10 = 2x² + 9x + 10. Both the leading term and constant term are correct, and the middle terms add to 9x. B gives 2x² + 20x + x + 10 = 2x² + 21x + 10, so its middle coefficient is wrong. C gives x² + 9x + 20, so its leading coefficient is wrong. D gives 2x² − 4x − 5x + 10 = 2x² − 9x + 10, which has the opposite sign for the middle term. The safest method is to multiply every term in the first bracket by every term in the second bracket and collect like terms.

Answer: A, (2y + 1)². Compare the expression with p² + 2pq + q². The first term 4y² is (2y)², the last term 1 is 1², and twice their product is 2 × 2y × 1 = 4y, exactly the middle term. Hence 4y² + 4y + 1 = (2y + 1)². B is wrong because (2y − 1)² = 4y² − 4y + 1; its middle term is negative. C expands to 4y² + 4y + y + 1 = 4y² + 5y + 1, so it does not match. D is a difference of squares: (2y+1)(2y−1) = 4y² − 1, with no 4y term. The sign of the middle term is an important clue: a positive middle term requires the plus form here.

Answer: A. The difference-of-squares identity needs exactly two perfect-square terms separated by a minus sign. In A, 25m² = (5m)² and 36n² = (6n)², so 25m² − 36n² = (5m)² − (6n)² = (5m − 6n)(5m + 6n). B has a plus sign between the squares, so the stated identity does not apply. C is actually (5m − 6n)² because its middle term is −2·5m·6n = −60mn; it is a square of a binomial, not the requested two-square difference form. D is (5m + 6n)² for the same reason, with a positive middle term. A useful warning is that the identity a²−b² produces two factors with opposite signs, while a binomial square produces a middle term.

Answer: A, since x² + 10x + 25 = (x + 5)². Use the pattern u² + 2uv + v² = (u+v)². Here u=x and v=5: u²=x², 2uv=10x, and v²=25. All three parts agree, so A is a perfect-square trinomial. B has the correct first two terms but its constant 20 is not 5², so it cannot equal (x+5)². C has the correct end terms but its middle term is 5x rather than 2·x·5=10x. D has a negative middle term, which suggests (x−5)², but that square expands to x²−10x+25, not a constant 20. Checking both the coefficient and the sign prevents quick visual errors. The word trinomial also means that three terms must be present.

Answer: A, because x² + 10x + 25 = (x+5)(x+5) = (x+5)². When two identical binomials are multiplied, (u+v)² gives u²+2uv+v². Taking u=x and v=5 produces x²+10x+25. B is not correct: changing 25 to 20 destroys the square pattern. C is not correct because the cross-term should be 10x, not 5x. D is not correct for this question because it has only two terms and lacks the cross-term; x²+25 is not generally equal to (x+5)². Notice that identical binomials may have either a plus or minus sign, but the corresponding middle term and constant must agree. A quick check is to take the square roots of the first and last terms and then calculate twice their product. This prevents confusing a sum of squares with a square of a sum.

Answer: A, (x−3)(x−7). For a monic quadratic x²+bx+c, find two numbers whose product is c and whose sum is b. Here the product must be 21 and the sum must be −10. The numbers −3 and −7 satisfy both conditions: (−3)(−7)=21 and −3+(−7)=−10. Therefore x²−10x+21=(x−3)(x−7). Expanding A gives x²−7x−3x+21=x²−10x+21. B gives a positive middle term, x²+10x+21, so its signs are wrong. C gives x²−22x+21, so the sum is incorrect. D gives x²−10x+24, so its middle term is right but its constant term is wrong. Memory cue: with a positive constant and a negative middle coefficient, both integer factors are negative.

The direct answer is B: \((x+4)(x+5)\). For a monic quadratic, choose two numbers whose product is the constant term 20 and whose sum is the coefficient 9. The numbers 4 and 5 work because \(4\times5=20\) and \(4+5=9\). Thus \(x^2+9x+20=(x+4)(x+5)\). Expanding B gives \(x^2+5x+4x+20=x^2+9x+20\). A has product 20 but sum 12, so it gives \(x^2+12x+20\). C has product 20 and sum -9, giving the wrong middle sign. D has product 20 but sum 21, producing \(x^2+21x+20\). The common mistake is checking only the constant term; always check both product and sum. Since both required values are positive, both factors use plus signs.

The direct answer is A: \((a-4)(a+4)\). Recognise a difference of squares: \(p^2-q^2=(p-q)(p+q)\). Since \(16=4^2\), the expression becomes \(a^2-4^2\). Applying the identity gives \((a-4)(a+4)\). Expanding confirms this: \(a^2+4a-4a-16=a^2-16\). B expands to \(a^2-6a-16\), so it introduces an unwanted middle term. C expands to \(a^2+8a+16\), which is a sum-square expression rather than the given difference. D expands to \(a^2+6a-16\), again with an unwanted middle term. The important point is that the two factors must have opposite signs so that the middle terms cancel. Memory cue: difference of squares equals difference of roots times sum of roots.

The direct answer is A: \(x^2-49\). A difference of squares has the form \(a^2-b^2\), and its identity is \(a^2-b^2=(a-b)(a+b)\). Here, \(x^2=x^2\) and \(49=7^2\), so \(x^2-49=x^2-7^2=(x-7)(x+7)\). Option A therefore becomes a product of two binomials. Option B is a sum of squares, \(x^2+7^2\), so the difference-of-squares identity does not apply. Option C is \(x^2-14x+49=(x-7)^2\), a perfect-square trinomial, not simply a difference of two squares. Option D is \((x+7)^2\), because its middle term is positive. Memory cue: look for two perfect squares separated by a minus sign; the factors then have opposite signs.

The direct answer is D: \((m+5)(m+6)\). For a monic quadratic \(m^2+bm+c\), find two numbers whose sum is \(b\) and whose product is \(c\). We need numbers with sum \(11\) and product \(30\); these are 5 and 6. Therefore, \(m^2+11m+30=(m+5)(m+6)\). Expanding confirms it: \(m^2+6m+5m+30=m^2+11m+30\). Option A uses 3 and 8: their product is 24, not 30. Option B uses 4 and 7: their product is 28, not 30. Option C has product 30, but the sum of the constants is \(-11\), producing a negative middle term. Option D has both the required sum and product. Memory cue: for \((m+r)(m+s)\), check both \(r+s\) and \(rs\), not just one.