Factorisation

The direct answer is A. Factorisation begins by finding the greatest common factor of all terms. The numerical coefficients \(15\) and \(20\) have greatest common factor \(5\). Both terms also contain \(x\), and the common algebraic factor is \(x\); \(y\) is not common because the first term has no \(y\). Therefore the greatest common factor is \(5x\). Divide each term by \(5x\): \(15x^2÷5x=3x\) and \(−20xy÷5x=−4y\). Hence \(15x^2−20xy=5x(3x−4y)\). A is the complete factorisation. B takes out only \(5\) and leaves a further common factor \(x\) inside, so it is not complete. C takes out \(x\) but leaves a numerical common factor \(5\). D expands to \(20x^2−20xy\), not the original expression. The safest check is to multiply \(5x\) by each term in the bracket. Memory cue: a complete common-factor answer has no non-unit common factor left inside the bracket.

The direct answer is A. Because the coefficient of \(x^2\) is \(1\), look for two integers whose product is the constant term \(−24\) and whose sum is the middle coefficient \(5\). The pair \(8\) and \(−3\) works: \(8×(−3)=−24\) and \(8+(−3)=5\). Therefore \(x^2+5x−24=(x+8)(x−3)\). Expanding confirms it: \(x^2−3x+8x−24=x^2+5x−24\). B uses \(−8\) and \(3\), whose sum is \(−5\), so it gives \(x^2−5x−24\). C has product \(−24\) but sum \(2\), producing the wrong middle term. D has product \(−24\) but sum \(10\), also incorrect. When the constant is negative, the two numbers have opposite signs; the positive number must have the larger absolute value here because the middle coefficient is positive. Memory cue: for \(x^2+bx+c\), find \(p\) and \(q\) with \(pq=c\) and \(p+q=b\), then write \((x+p)(x+q)\).

The direct answer is A. Inspect both terms: $6a^2b$ and $9ab^2$. The greatest common numerical factor of 6 and 9 is 3. Both terms contain a, and both contain b, so the common algebraic factor is $ab$. The greatest common factor is therefore $3ab$. Divide each term by it: $6a^2b\div3ab=2a$ and $9ab^2\div3ab=3b$. Hence $6a^2b+9ab^2=3ab(2a+3b)$. Option B is mathematically equivalent after further simplification, but it is not the complete factorisation because a common factor $b$ remains inside. C is also equivalent but leaves the common numerical factor 3 inside. D has a minus sign and expands to $6a^2b-9ab^2$, not the given sum. A complete factorisation extracts the greatest common factor. Memory cue: compare coefficients, then take the smallest exponent of every variable common to all terms.

The direct answer is A. A difference of two perfect squares has the form $P^2−Q^2$: both terms must be perfect squares and they must be separated by a minus sign. Here $9a^2=(3a)^2$ and $16b^2=(4b)^2$, so $9a^2−16b^2=(3a)^2−(4b)^2$. It can further factorise as $(3a−4b)(3a+4b)$. B contains two squares but uses addition, so it is a sum of squares, not a difference of squares. C has $16b$ rather than $16b^2$; without the square on b, the second term is not generally a perfect square monomial. D is a perfect-square trinomial, since $a^2+8a+16=(a+4)^2$, but it is not written as the difference of two squares. Memory cue: check two things together—each term is a square and the sign between them is minus.

The direct answer is A. Compare the expression with the perfect-square identity $p^2+2pq+q^2=(p+q)^2$. Here $y^2$ is $p^2$, 36 is $q^2$, so $p=y$ and $q=6$. The middle term is $2pq=2×y×6=12y$, exactly as required. Therefore $y^2+12y+36=(y+6)^2$. Expanding A gives $y^2+6y+6y+36=y^2+12y+36$. B expands to $y^2−12y+36$, so its middle sign is wrong. C expands to $y^2+15y+36$, not 12y. D expands to $y^2+13y+36$, also incorrect. The key is to use the square root of the constant term and then check twice the product with the first term. Memory cue: a positive middle term means the repeated binomial has a plus sign; a negative middle term would require a minus sign.

The direct answer is B: $(m-5)^2$. Begin with the identity $a^2-2ab+b^2=(a-b)^2$. In the given expression, $m^2$ is the square of m, and 25 is the square of 5. Now check the middle term: $-2×m×5=-10m$, exactly as given. Therefore $m^2-10m+25=(m-5)^2$. We can verify by expanding: $(m-5)(m-5)=m^2-5m-5m+25=m^2-10m+25$. A gives $(m+5)^2=m^2+10m+25$, so its middle sign is wrong. C expands to $m^2+15m-250$, not the original expression. D gives $m^2-50m+625$, so both its middle and constant terms are wrong. Memory cue: for a perfect square, take the square roots of the first and last terms, then use the middle sign.

The direct answer is A: \((2x-3)(2x+3)\). Recognise \(4x^2-9\) as a difference of two squares. The first term is \(4x^2=(2x)^2\) and the second is \(9=3^2\). Apply \(P^2-Q^2=(P-Q)(P+Q)\), with \(P=2x\) and \(Q=3\). Thus \(4x^2-9=(2x-3)(2x+3)\). Expanding confirms it: \(4x^2+6x-6x-9=4x^2-9\), because the middle terms cancel. B incorrectly uses 4x and 9 as the square roots and expands to \(16x^2-81\). C is a square, and its expansion is \(4x^2-12x+9\), which contains a middle term. D expands to \(4x^2-9x-9\), also not the given expression. Memory cue: for two squares with a minus sign, write the difference and the sum of their actual square roots.

The direct answer is A: \(3x(x+4)\). Look at each term, \(3x^2\) and \(12x\), and find their greatest common factor. Both contain 3 and x, so the greatest common factor is \(3x\). Divide each term by \(3x\): \(3x^2÷3x=x\) and \(12x÷3x=4\). Therefore \(3x^2+12x=3x(x+4)\). Expanding gives \(3x^2+12x\), confirming the result. B is equal to \(3x^2+12x\), but it leaves x common inside the bracket, so it is not the simplest factorised form. C is also an incomplete extraction because 3 is still common inside \(3x+12\). D expands to \(12x^2+36x\), which is different. The useful habit is to take out the greatest common factor, not merely any common factor.

The direct answer is C: \(x^2-25\). Rewrite 25 as \(5^2\); then \(x^2-25=x^2-5^2\). This has the exact pattern \(P^2-Q^2\), so use \(P^2-Q^2=(P-Q)(P+Q)\). The factorisation is \((x-5)(x+5)\). Expanding gives \(x^2+5x-5x-25=x^2-25\). A has a plus sign between the squares, so it is not the difference-of-squares pattern. B is the perfect square \((x+3)^2\), because its middle term is \(6x\). D is a quadratic trinomial, but it does not consist of just two square terms separated by a minus sign and is not a perfect-square identity of the required type. A common confusion is to notice the number 25 but forget to check the sign. The minus sign between the squares is essential.

The direct answer is A: \((4a-b)(4a+b)\). Recognise the expression as a difference of two squares. Since \(16a^2=(4a)^2\) and \(b^2=b^2\), the two square roots are \(4a\) and \(b\). Apply \(P^2-Q^2=(P-Q)(P+Q)\): \(16a^2-b^2=(4a-b)(4a+b)\). Expanding gives \(16a^2+4ab-4ab-b^2=16a^2-b^2\), so the result is verified. B treats \(16a\) as the square root of \(16a^2\), but the correct square root is \(4a\); its expansion would be \(256a^2-b^2\). C is a square of a difference and would produce the extra middle term \(-8ab\). D expands to \(16a^2+6ab-b^2\), so it is not equal to the original expression. Always find the actual square roots before applying the identity.

The direct answer is A: \(x^2+10x+25\). A perfect-square trinomial follows \(x^2+2xy+y^2=(x+y)^2\). Since \(25=5^2\) and \(2×x×5=10x\), the expression becomes \(x^2+2×x×5+5^2=(x+5)^2\). Expanding confirms it: \(x^2+5x+5x+25=x^2+10x+25\). B has the correct first two terms but its constant is \(24\) rather than \(25\), so it cannot be \((x+5)^2\). C has the correct first and last terms for \((x+5)^2\), but its middle coefficient is \(9\) instead of \(10\). D has neither the required constant term \(25\) nor the matching middle term. A useful test is to take the square roots of the first and last terms and check whether twice their product equals the middle term.

The direct answer is A: 7x(x + 2y). Factorisation begins by finding the greatest common factor of all terms. Both 7x² and 14xy contain 7x: 7x² ÷ 7x = x and 14xy ÷ 7x = 2y. Taking 7x outside gives 7x(x + 2y). Expanding verifies the result: 7x·x + 7x·2y = 7x² + 14xy. Option B incorrectly changes the second term to 14x² because it uses 2x instead of 2y. Option C gives 7x² + 14x, which loses the variable y. Option D expands to 14x² + 14xy, so its first coefficient is twice the required value. Memory cue: divide every term by the common factor and write exactly what remains inside the bracket; never change a variable while taking out a factor.

Direct answer: A, (a−b)(a+b). The basic identity for the difference of two squares is a²−b²=(a−b)(a+b). We can verify it by multiplication: (a−b)(a+b)=a²+ab−ab−b²=a²−b². The middle terms cancel because one factor has a plus sign and the other has a minus sign. Option A therefore reproduces the exact expression. Option B represents a square of a difference and expands to a²−2ab+b², which has a middle term and a positive final term. Option C expands to a²+2ab+b², so both the middle sign and the final sign differ. Option D contains an extra factor and does not generally equal a²−b²; it is not the standard identity. Memory cue: difference means minus between the squares, but the two factors use opposite signs.

Direct answer: A, x²+10x+25. Compare it with the identity (u+v)²=u²+2uv+v². Here u=x and v=5: u²=x², 2uv=2×x×5=10x, and v²=25. Therefore x²+10x+25=(x+5)²=(x+5)(x+5), so the two binomial factors are identical. Option B has 20 instead of the required 25, so it is not a perfect square. Option C is a difference of squares, x²−25=(x−5)(x+5), but its factors are not identical and it is not a trinomial. Option D has the correct first and last square terms, but its middle term would need to be 10x, not 5x. Expanding A confirms the result. Memory cue: for x²+bx+c to be (x+k)², c must be k² and b must be 2k.

The correct answer is A: take the common factor outside. Suppose an expression is 6x² + 9x. Both terms contain 3x, so remove that common factor: 6x² + 9x = 3x(2x + 3). This changes a sum into a product and usually makes the remaining factor easier to factorise further. Option B is not a general algebraic operation for factorisation; exponents are combined only under specific multiplication rules. Option C changes the value and is unrelated to taking a factor. Option D changes the equation or expression and is not a factorisation step. The common factor may be a number, a variable, or a whole bracket. Memory cue: before using identities or splitting terms, check the HCF of every term and take it out once.

The correct answer is A: 49p² − 25q². The identity U² − V² = (U − V)(U + V) applies when both terms are perfect squares and they are separated by subtraction. Here 49p² = (7p)² and 25q² = (5q)². Therefore 49p² − 25q² = (7p)² − (5q)² = (7p − 5q)(7p + 5q). Option B has addition, so it is not a difference of squares. Option C has 25q rather than 25q², so the second term is not generally a square of a monomial. Option D has coefficients and variables arranged as 7p² and 5q², which are not generally perfect squares in the required form. Memory cue: check both the exponent 2 and the minus sign before applying this identity.

The direct answer is B: \(25a^2-49b^2=(5a)^2-(7b)^2\). The difference-of-squares identity is \(A^2-B^2=(A-B)(A+B)\). Therefore B factorises as \((5a-7b)(5a+7b)\). Option A is \((3p-4)^2\), a perfect-square trinomial, so the difference-of-squares identity is not the most appropriate first method. Option C is a quadratic trinomial; it can be factorised by finding two numbers with product 10 and sum 7, not by the difference-of-squares identity. Option D is a sum of cubes, because \(8m^3=(2m)^3\) and \(27n^3=(3n)^3\); it belongs to a different identity. A useful warning is that the signs matter: the identity needs subtraction between two complete squares, not addition. Look for two square terms separated by a minus sign before applying the formula.

The direct answer is B: \(5x^2-45=5(x-3)(x+3)\). First take out the common factor 5: \(5(x^2-9)\). Now recognize \(x^2-9=x^2-3^2\) as a difference of two squares. Using \(A^2-B^2=(A-B)(A+B)\), it becomes \((x-3)(x+3)\). Therefore the complete factorisation is \(5(x-3)(x+3)\). Option A incorrectly treats 9 as the square root of 9; \((x-9)(x+9)=x^2-81\), not \(x^2-9\). Option C changes the common factor and does not expand to the original expression. Option D takes out 5 but stops too early, because \(x^2-45\) is not the result after correctly extracting 5; the remaining constant is 9. The common error is forgetting that the square root of 9 is 3, not 9. Always factor out the common number first, then inspect the remaining expression.