Factorisation

Here (64a^3=(4a)^3) and (125b^3=(5b)^3). Exam tip: in the difference formula, the second bracket has positive (20ab).

A perfect-square trinomial has the form \(x^2+2ax+a^2\). Here, \(10x=2\times5\times x\) and \(25=5^2\), so \(x^2+10x+25=(x+5)^2\). In option B, the middle term is \(10x\), but the constant term should be \(25\), not \(15\). Exam tip: square half of the middle-term coefficient and compare it with the constant term.

The direct answer is A: the factorised form is \(7x(x-4y)\). Factorisation means writing an expression as a product of simpler factors. In \(7x^2-28xy\), the first term is \(7x\cdot x\), and the second term is \(7x\cdot(-4y)\). Therefore, taking the common factor \(7x\) outside gives \(7x(x-4y)\). Multiplying back confirms it: \(7x\cdot x=7x^2\) and \(7x\cdot(-4y)=-28xy\). Option A is correct because both the number 7 and the variable x have been taken out. Option B, \(7(x^2-4y)\), is wrong because multiplying it gives \(7x^2-28y\), which lacks x in the second term. Option C, \(x(7x-28)\), gives \(7x^2-28x\), not \(-28xy\); the y has disappeared. Option D, \(7xy(x-4)\), gives \(7x^2y-28xy\), adding an unwanted y to the first term. A useful exam cue is to find the greatest common factor of every term and then check by multiplication.

The direct answer is option A: \((x+2y−3z)(x+2y+3z)\). First recognise \(x^2+4xy+4y^2\) as \((x+2y)^2\), using the square identity. Also, \(9z^2=(3z)^2\). The expression therefore becomes \((x+2y)^2−(3z)^2\). Apply the difference-of-squares identity \(U^2−V^2=(U−V)(U+V)\), with \(U=x+2y\) and \(V=3z\). This gives \((x+2y−3z)(x+2y+3z)\), option A. Option B uses \(x−2y\), whose square would produce \(-4xy\), so it has the wrong middle sign. Option C is not a complete factorisation of the difference of two squares and also writes \(-3z\) instead of subtracting \((3z)^2\). Option D does not expand to the required four-term expression and omits the correct squared structure. The order of the two factors may be reversed without changing the product, but their signs must remain as shown. Memory cue: make the perfect square first, then use difference of squares.

Write it as ((a-b)(a+b)-(a-b)), then take ((a-b)) common. Exam tip: also convert the last two terms into a group.

Here \(9a^2=(3a)^2\) and \(16b^2=(4b)^2\). The middle term is \(-2(3a)(4b)=-24ab\), so it matches \((3a-4b)^2\). In exams, always check the sign of the middle term.

((6x-5)(3x+2)) gives (18x^2+12x-15x-10). Exam tip: confirm that the middle term becomes (-21x).

In \(a^2+14a+49\), \(49=7^2\) and the middle term is \(14a=2\times a\times7\). Hence it equals \((a+7)^2\). In B, the constant term is not \(7^2\). Exam tip: check the \(2ab\) term.

It is ((2a)^3+3(2a)^2b+3(2a)b^2+b^3). Exam tip: identify cube coefficients (1), (3), (3), (1).

\(49a^2-16b^2=(7a)^2-(4b)^2\). Using \(A^2-B^2=(A-B)(A+B)\), it becomes \((7a-4b)(7a+4b)\). Option C is a perfect square, not a difference. Exam tip: first identify whether both terms are squares.

The first three terms form the square of a binomial: 4x²+4xy+y²=(2x+y)². Also, 16=4². Therefore the expression becomes (2x+y)²−4². Applying the difference-of-squares identity u²−v²=(u−v)(u+v), with u=2x+y and v=4, gives (2x+y−4)(2x+y+4). Hence option A is the required factorised form.

Expanding the factors confirms the result: [(2x+y)−4][(2x+y)+4]=(2x+y)²−16=4x²+4xy+y²−16. Option D stops at an intermediate form and is not fully factorised. Option B has the wrong sign in the xy term because it uses 2x−y, while option C does not reproduce the original expression.

For \(x^2+xy+y^2\), integer values \(a,b\) would need \(a+b=1\) and \(ab=1\), but no integer pair satisfies both. Option A uses difference of squares. Exam tip: check the required sum and product of coefficients.

\(4p^2-25q^2=(2p)^2-(5q)^2\), so it is a difference of squares and factorises as \((2p-5q)(2p+5q)\). Options C and D are perfect-square trinomials. Exam tip: identify square terms first, then check for a minus sign between them.

((5m+2n)(m-3n)) gives (5m^2-15mn+2mn-6n^2). Exam tip: use opposite signs for a negative constant term.

Here (x^2+6x+9=(x+3)^2), then difference of squares applies. Exam tip: first convert the trinomial into a perfect square.

A perfect-square trinomial has the form \(a^2-2ab+b^2=(a-b)^2\). Here, \(4x^2=(2x)^2\), \(9y^2=(3y)^2\), and the middle term is \(-2(2x)(3y)=-12xy\). Therefore, \(4x^2-12xy+9y^2=(2x-3y)^2\). In option B, the last term would need to be \(9y^2\); in option C, the middle term would need to be \(-12xy\). Exam tip: take the square roots of the first and last terms, then verify the middle term using \(\pm2ab\).

Grouping gives (a(a+b)+c(a+b)), so the factor is ((a+b)(a+c)). Exam tip: learn grouping in four-term expressions.

In \(x^2+10x+25\), \(25=5^2\) and the middle term is \(2\times x\times5=10x\). Hence it equals \((x+5)^2\). In \(x^2+10x+24\), the constant term is not 25. Exam tip: verify the middle term for a perfect square.

The direct answer is A: \((x-8)^2\). A perfect-square trinomial follows \(a^2-2ab+b^2=(a-b)^2\). Here \(x^2=x^2\), the constant term is \(64=8^2\), and the middle term is \(-16x=-2\cdot x\cdot8\). Hence \(x^2-16x+64=(x-8)^2\). Expanding confirms this: \((x-8)^2=x^2-16x+64\). Option A is correct. Option B, \((x+8)^2\), expands to \(x^2+16x+64\), so its middle sign is wrong. Option C, \((x-4)^2\), expands to \(x^2-8x+16\), so both the middle coefficient and constant are wrong. Option D, \((x-16)(x+4)\), expands to \(x^2-12x-64\), not the given expression. The reliable method is to take the square roots of the first and last terms, then check whether twice their product has the correct sign and coefficient. Remember that a negative middle term requires a minus sign inside the square.

The expression is a quadratic trinomial. To factor it, we need two binomials whose first terms multiply to give the leading term, whose last terms multiply to give the constant term, and whose cross-products add to the middle term. This is a useful way to check every proposed factorisation rather than relying only on appearance.

For choice A, expand \((2x+3)(5x+8)\). The products are \(10x^2\), \(16x\), \(15x\), and \(24\). The middle terms add to \(31x\), so the result is exactly \(10x^2+31x+24\). Therefore A is correct. The other choices do not produce the required middle coefficient or constant term.

((7x-2)(2x-3)) gives (14x^2-21x-4x+6). Exam tip: check that the middle term becomes (-23x), not (-25x).

Grouping all terms gives a common factor like (x^2(x+y)+y^2(x+y)+xy(x+y)). Exam tip: take ((x+y)) common and add the remaining terms.

This is a trinomial square with terms (a), (2b), and (c). Exam tip: confirm using pair terms (4ab), (2ac), and (4bc).

First (4x^2-12xy+9y^2=(2x-3y)^2), then use difference of squares with (25=5^2). Exam tip: form the perfect square and factor further.

Grouping gives (x(x+3y-4)+2y(x+3y-4)). Exam tip: first make groups that produce a common bracket.