Factorisation

The direct answer is B: x^2-9x+20=(x-4)(x-5). To factor a monic quadratic, find two numbers whose product is the constant term 20 and whose sum is the coefficient of x, -9. The numbers -4 and -5 satisfy both conditions: (-4)(-5)=20 and -4+(-5)=-9. Multiplying the factors confirms the result: (x-4)(x-5)=x^2-5x-4x+20=x^2-9x+20. A uses -2 and -10; their product is 20, but their sum is -12, so the middle term would be -12x. C gives a positive middle term, producing x^2+9x+20. D has product 20 but sum -21, so it also fails. In a rectangle or algebra-tile model, the two factors appear as the side expressions. Memory cue: for a positive constant and a negative middle coefficient, both factor numbers must be negative.

The correct answer is B: the sides are \((x+5)\) and \((x+7)\). In the rectangle model, multiplying the side lengths gives the total area. Expand step by step: \((x+5)(x+7)=x^2+7x+5x+35=x^2+12x+35\). The two cross-rectangles would have areas \(7x\) and \(5x\), while the corner rectangle has area \(5\times7=35\). Thus the listed pieces are exactly \(x^2\), \(5x\), \(7x\), and \(35\), with no overlap or missing part. A has constants 12 and 35, giving the wrong cross terms and constant product. C would produce negative middle terms, not the positive \(5x\) and \(7x\) pieces. D has constant product 35 but its cross terms total \(36x\), so it also fails. Memory cue: for \(x^2+px+q\), find two numbers whose sum is \(p\) and product is \(q\).

Direct answer: Option A is correct. To factor x² + 18x + 80 using a rectangle model, find two numbers whose product is 80 and whose sum is 18. The numbers 8 and 10 satisfy both conditions. The four area cells are x·x = x², x·10 = 10x, 8·x = 8x, and 8·10 = 80. Their sum is x² + 10x + 8x + 80 = x² + 18x + 80, so the sides are (x + 8) and (x + 10). Option B has product 80 but sum 24, producing 24x. Option C has product 80 but sum 42. Option D has product 80, but sum 21. Thus each wrong option matches the constant term while failing the middle coefficient. Memory cue: in a monic quadratic, check both the product and the sum, not just the product.

The correct answer is B: x² − 16. This is the difference of two perfect squares: x² − 16 = x² − 4². Using a² − b² = (a − b)(a + b), we obtain x² − 16 = (x − 4)(x + 4), a product of two binomials. Option A is a sum of squares, x² + 3², and the difference-of-squares identity does not apply. Option C can be written as (x + 4)² + 1, so it is not a product of two real binomials in the intended school-level factorisation. Option D becomes (x + 2)² + 1, so it also has no real binomial factorisation of this type. Notice that B has two square terms separated by a minus sign and no middle term. Memory cue: difference of squares needs two perfect squares with subtraction; a plus sign is not interchangeable with a minus sign.

Correct answer: A. Find the greatest common factor of the two terms. The factors of 6x include 3, and 9 is also divisible by 3, so take 3 outside: \(6x+9=3(6x\div3+9\div3)=3(2x+3)\). Expanding A gives \(6x+9\), confirming it. B expands to \(6x+54\), so the second term is too large. C expands to \(9x+54\), changing both coefficients. D expands to \(6x+27\), so its constant term is incorrect. A useful habit is to divide every term by the factor taken outside; do not divide only one term. You can always verify a factorisation by distributing the outside factor across the bracket. Memory cue: the numbers inside the bracket are what remain after dividing each original term by the common factor.

Correct answer: A. Look for a factor common to both terms. Since \(x^2=x\cdot x\) and \(5x=5 hickspace ext{×} hickspace x\), both terms contain \(x\). Taking it outside leaves one x from the first term and 5 from the second: \(x^2+5x=x(x+5)\). Multiplying back gives \(x hickspace ext{×} hickspace x+x hickspace ext{×} hickspace5=x^2+5x\). B expands to \(5x+5\), so it loses the required \(x^2\) term. C expands to \(x^3+5x\), adding an extra power of x. D expands to \(x^3+5x^2\), also changing both powers. The common-factor rule is: take out the greatest factor shared by every term, then place the quotients inside brackets. Memory cue: after factorisation, distribute the outside factor to check that every original term returns.

Answer: C, c = (b/2)². Begin with the basic perfect-square identity: (x + k)² = x² + 2kx + k². Comparing this with x² + bx + c, the coefficient of x must satisfy b = 2k. Hence k = b/2. The constant term must then be k², so c = (b/2)². This is the required relationship. Option A incorrectly squares b itself; the coefficient b is twice the number inside the binomial, so its square is not generally the constant term. Option B gives only half of b, not its square, and also has the wrong type of relationship. Option D rearranges the numbers incorrectly and is not equivalent to the perfect-square condition. For example, x² + 6x + 9 = (x + 3)² because 9 = (6/2)². Memory cue: halve the middle coefficient, then square it; the result must equal the constant term.

Answer: A, x² + 5x + 6. For a monic quadratic x² + bx + c, look for two integers whose product is c and whose sum is b. For this expression, the required product is 6 and the required sum is 5. The integers 2 and 3 satisfy both conditions: 2 × 3 = 6 and 2 + 3 = 5. Therefore x² + 5x + 6 = (x + 2)(x + 3). Expanding confirms it: x² + 3x + 2x + 6 = x² + 5x + 6. Option B would require integers with product 7 and sum 4; the integer factor pairs of 7 are 1 and 7, whose sum is 8. Option C would require product 2 and sum 2; the pair 1 and 2 sums to 3. Option D also fails because the factor pair 1 and 1 has sum 2, not 1. Memory cue: for x² + bx + c, find “multiply to c, add to b.”

Answer: A, x² + 10x + 25. Use the identity x² + 2kx + k² = (x + k)². In option A, 10x = 2 × x × 5 and 25 = 5², so x² + 10x + 25 = (x + 5)² = (x + 5)(x + 5). Option B has the correct middle term for k = 5, but its constant term is 20 rather than 25, so it is not a perfect square. Option C has constant term 25, suggesting k = 5, but then the middle term should be 10x, not 5x. Option D has a negative middle term, but with constant 20 it still fails the condition; x² − 10x + 25 would be (x − 5)². The sign of the middle term tells whether the repeated binomial uses plus or minus, while its magnitude must match twice the square-root number. Memory cue: the last term is the square of half the middle coefficient.

Answer: B, (a − 3)(a − 4). For a² − 7a + 12, find two integers whose product is 12 and whose sum is −7. The numbers −3 and −4 work because (−3)(−4) = 12 and −3 + (−4) = −7. Thus a² − 7a + 12 = (a − 3)(a − 4). Expanding gives a² − 4a − 3a + 12 = a² − 7a + 12. Option A has positive numbers and expands to a² + 7a + 12, so its middle sign is wrong. Option C has the correct signs but gives a² − 8a + 12, so the middle coefficient is wrong. Option D expands to a² + 8a + 12, wrong in both the sign and the middle coefficient. Since the constant term is positive and the middle term is negative, both integer factors must be negative. Memory cue: positive product plus negative sum means both numbers are negative.

Answer: B, 4x² − 25. Rewrite each term as a square: 4x² = (2x)² and 25 = 5². Therefore 4x² − 25 = (2x)² − 5². Applying r² − s² = (r − s)(r + s), we get (2x − 5)(2x + 5). Option A is a sum of squares, x² + 3², so the difference-of-squares identity does not apply. Option C is a perfect-square trinomial because x² + 6x + 9 = (x + 3)²; it has three terms rather than two separated squares. Option D has a common factor, since 8x² + 12x = 4x(2x + 3), but it is not in the difference-of-two-squares pattern asked for. The essential clues are exactly two square terms and a minus sign between them. Memory cue: convert coefficients into squares before choosing the identity; 4x² is (2x)², not 4 times x² left unexamined.

The direct answer is B: \((3a-5b)(3a+5b)\). First recognise that \(9a^2=(3a)^2\) and \(25b^2=(5b)^2\). Thus the expression has the form \(x^2-y^2\), where \(x=3a\) and \(y=5b\). The difference-of-squares identity is \(x^2-y^2=(x-y)(x+y)\), so \(9a^2-25b^2=(3a-5b)(3a+5b)\). Checking by multiplication gives \(9a^2+15ab-15ab-25b^2=9a^2-25b^2\). A is wrong because \((3a-5b)^2\) produces a middle term \(-30ab\) and a positive \(25b^2\). C does not expand to the given expression. D produces a positive middle term and positive \(25b^2\). Memory cue: a difference of two squares becomes the product of the difference and the sum; it is not the square of one binomial.

The direct answer is A: \((2x+3)^2\). Begin by comparing the expression with the identity \((p+q)^2=p^2+2pq+q^2\). Here, \(4x^2=(2x)^2\) and \(9=3^2\). The middle term required for a perfect square is \(2(2x)(3)=12x\), which is exactly present. Therefore \(4x^2+12x+9=(2x+3)^2\). A is correct. B would have leading term \(16x^2\), not \(4x^2\). C gives a middle term \(-12x\), so its sign is wrong. D gives leading term \(x^2\), not \(4x^2\). Expanding the correct answer confirms it: \((2x+3)(2x+3)=4x^2+12x+9\). Memory cue: for a perfect-square trinomial, the middle term must be twice the product of the square roots of the first and last terms.

The direct answer is C: \((5r-3)^2\). The first term is \(25r^2=(5r)^2\), and the last term is \(9=3^2\). For the identity \((m-n)^2=m^2-2mn+n^2\), the middle term should be \(-2(5r)(3)=-30r\), exactly as given. Hence \(25r^2-30r+9=(5r-3)^2\). A is incorrect because \((5r+3)^2\) gives \(+30r\), not \(-30r\). B has the wrong leading term and does not produce the required polynomial. D expands to \(25r^2-50r+9\), so its middle coefficient is wrong. Always verify a factorisation by expanding it, especially when signs are involved. The sign of the middle term tells us whether the binomial uses a plus or minus sign.

The direct answer is A: \((a+b)(x+y)\). Use grouping: the first two terms have a common factor \(a\), and the last two have a common factor \(b\). Thus \(ax+ay+bx+by=a(x+y)+b(x+y)\). Now \((x+y)\) is common to both grouped terms, so take it outside: \((a+b)(x+y)\). Expanding this result gives \(ax+ay+bx+by\), confirming it. B expands to \(ab+ay+bx+xy\), which contains different terms. C expands to \(ab+ax+by+xy\), also not the original expression. D would give only \(abx+aby\), and it cannot produce the four required terms. A useful method is to group terms so that the same binomial appears twice; then factor out that common binomial.

The direct answer is A: \((p+q)(x+y)\). Group the terms containing \(x\) and the terms containing \(y\): \(px+qx+py+qy=x(p+q)+y(p+q)\). The same binomial \((p+q)\) is common in both terms. Taking it outside gives \((p+q)(x+y)\). Expanding confirms the result: \((p+q)(x+y)=px+py+qx+qy\), which is the same set of terms as the original expression, only in a different order. B expands to \(pq+py+qx+xy\), so it introduces terms that were not given. C expands to \(pq+px+qy+xy\), also incorrect. D gives only two terms, \(pqx+pqy\), and omits the required separate coefficients. Grouping is successful when the same factor appears after factoring each group.

The direct answer is A: \(x^2+10x+25=(x+5)^2\). A perfect-square trinomial has the form \(a^2+2ab+b^2=(a+b)^2\) or \(a^2-2ab+b^2=(a-b)^2\). Here the first term is \(x^2\, so the first square root is \(x\), and the last term \(25\) has square root \(5\). Twice their product is \(2\cdot x\cdot5=10x\), exactly the middle term. B has the correct middle term but should have constant term \(25\), not \(20\). C has the correct first and last terms, but its middle term should be \(10x\), not \(5x\). D has a negative constant and cannot match this square form. The quick check is: halve the middle coefficient, square it, and compare with the constant term.

The direct answer is A: \(x^2-16=(x-4)(x+4)\). Since \(16=4^2\), the expression is \(x^2-4^2\), a difference of squares. Applying \(m^2-n^2=(m-n)(m+n)\) gives \((x-4)(x+4)\). The student confused a difference of squares with the square of a difference. A square such as \((x-4)^2\) expands to \(x^2-8x+16\), which has a middle term and a positive constant; it is not \(x^2-16\). B repeats the original error. C expands to \(x^2-15x-16\), not the given expression. D is false because A is an integer factorisation. Memory cue: \(m^2-n^2\) has two different factors, one with minus and one with plus; \((m-n)^2\) has a middle term \(-2mn\).

The direct answer is A: \((u+7)(u-6)\). For a monic quadratic \(u^2+bu+c\), find two numbers whose product is the constant term \(c\) and whose sum is the middle coefficient \(b\). Here the numbers must multiply to \(-42\) and add to \(+1\). The pair is \(+7\) and \(-6\), because \(7(-6)=-42\) and \(7+(-6)=1\). Therefore \(u^2+u-42=(u+7)(u-6)\). Expanding gives \(u^2-6u+7u-42=u^2+u-42\). B has product \(-42\) but sum \(-1\), producing the wrong middle sign. C has the correct product but sum \(-11\). D has product \(-42\) but sum \(41\). Always check both product and sum, not just the constant term.

The direct answer is A: \(x^2-12x+36=(x-6)^2=(x-6)(x-6)\). For two identical linear factors, the expression must be a perfect square. Comparing with \((x-k)^2=x^2-2kx+k^2\), the middle coefficient \(-12\, gives \(2k=12\), so \(k=6\). The constant term must then be \(6^2=36\), which matches A. B has the same first two terms but its constant term is 24, not 36, so it is not a perfect square. C has a positive middle term and a negative constant, so it cannot equal the square of a real linear binomial in this form. D is a difference of squares, \((x-6)(x+6)\), whose factors are different, not identical. A quick check is to take half the magnitude of the middle coefficient and square it.

The direct answer is A: \(6xy(2x+3y)\). Find the greatest common factor of the numerical coefficients 12 and 18, which is 6. Both terms contain at least one \(x\) and one \(y\), so the greatest common monomial factor is \(6xy\). Divide each term by it: \(12x^2y\div6xy=2x\) and \(18xy^2\div6xy=3y\). Hence the factorised form is \(6xy(2x+3y)\). Multiplication verifies it: \(6xy\cdot2x=12x^2y\) and \(6xy\cdot3y=18xy^2\). B and C are algebraically equivalent to the original expression after expansion, but they do not take the greatest common factor in the intended form. D expands to \(12x^2y+12xy^2\), so its second coefficient is wrong. The common factor must divide every term exactly.

The correct answer is A: 5ab(3a − 2b). To factor by the common-factor method, first compare the terms 15a^2b and −10ab^2. The greatest common numerical factor of 15 and 10 is 5. Both terms contain a, and both contain b, so the greatest common factor is 5ab. Divide each term by 5ab: 15a^2b ÷ 5ab = 3a, and −10ab^2 ÷ 5ab = −2b. Therefore, the expression becomes 5ab(3a − 2b). Option A is correct because multiplying back gives 15a^2b − 10ab^2. Option B is not the complete common-factor form because it leaves an extra b inside the bracket and does not represent the original expression correctly when expanded. Option C gives 10a^2b − 10ab^2, so its first term is wrong. Option D gives 15a^2b − 30ab^2, so its second coefficient is wrong. Memory cue: take the smallest power of every variable common to all terms, then divide each term carefully.

The correct answer is A: 10a^2b + 15ab^2. Find the HCF in two parts. First, the numerical coefficients 10 and 15 have HCF 5. Next, both terms contain a and b. The smallest power of a present in both terms is a^1, and the smallest power of b is b^1. Thus the algebraic HCF is 5ab. Dividing the expression by 5ab gives 2a + 3b, confirming that 10a^2b + 15ab^2 = 5ab(2a + 3b). Option B does not have b in its first term, so ab cannot be common. Option C does not have a in its second term, so ab cannot be common. Option D has numerical HCF 6, together with ab, so its HCF is 6ab rather than 5ab. The important rule is to use the HCF of the coefficients and the lowest exponent of each variable common to every term.

The correct answer is B: x^2 − 6xy + 9y^2 = (x − 3y)^2. The relevant identity is (p − q)^2 = p^2 − 2pq + q^2. Here p = x and q = 3y. Then p^2 = x^2, −2pq = −2(x)(3y) = −6xy, and q^2 = 9y^2. All three terms therefore match option B exactly. Option A has the same square terms but a positive middle term, so it represents (x + 3y)^2, not a square of a difference. Option C has a negative final term, so it does not have the perfect-square pattern. Option D is a difference of squares, (x − 3y)(x + 3y), rather than the square of one difference. A useful warning is that the middle sign determines whether the binomial is a sum or a difference: a negative middle term means the two terms are subtracted.

The correct answer is B: (x − y)^2. Start with the identity (p − q)^2 = p^2 − 2pq + q^2. Compare it with x^2 − 2xy + y^2. Taking p = x and q = y gives p^2 = x^2, −2pq = −2xy, and q^2 = y^2, so every term matches. Therefore x^2 − 2xy + y^2 = (x − y)^2. Option A is (x + y)^2, whose expansion is x^2 + 2xy + y^2; its middle sign is wrong. Option C is (x − y)(x + y), which uses the difference-of-squares identity and expands to x^2 − y^2, so it has no −2xy term. Option D expands to xy − y^2 and does not even contain the required x^2 term. Memory cue: when the first and last terms are squares and the middle term is negative twice their roots multiplied, use the square of a difference.