Factorisation

Here (8+9=17) and (8\cdot9=72). In exams choose factors of the constant whose sum is the middle coefficient.

Here (-9+(-10)=-19) and ((-9)(-10)=90). In exams use two negative signs when the constant is positive and the middle term is negative.

\(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so it is a perfect-square binomial. In option B, the constant should be 25. Exam tip: match the middle term with \(2ab\).

((3x+1)(x-5)=3x^2-14x-5). In exams look for factors with opposite signs when the constant is negative.

(4a^2=(2a)^2), (9b^2=(3b)^2), and the middle term is (-12ab). In exams the sign of the middle term is decisive in a perfect square.

\(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\). Hence, it is the product of two identical binomials, \((x+5)(x+5)\). In option B, the middle term \(10x\) requires \(5\), so the constant term should be \(25\), not \(20\). Exam tip: use the identity \(a^2+2ab+b^2=(a+b)^2\).

In \((3m+2n)(4m-5n)\), the cross-products are \(-15mn\) and \(+8mn\), giving \(-7mn\). Option B gives \(+7mn\), so its middle-term sign is wrong. Exam tip: always add the cross-products to verify factorisation.

The direct answer is D: \((a-5)(b+2)\). Group the first two and last two terms: \(ab+2a-5b-10=a(b+2)-5(b+2)\). Both groups contain the common factor \(b+2\), so taking it out gives \((a-5)(b+2)\). Option D therefore reproduces all four original terms. Option A, \((a-5)(b-2)\), expands to \(ab-2a-5b+10\), so the signs of the a-term and constant are wrong. Option B expands to \(ab-2a+5b-10\), with the signs of the a-term and b-term wrong. Option C gives \((a-5)(b+10)=ab+10a-5b-50\), so neither the second term nor the constant matches. D works because the repeated binomial \(b+2\) appears after grouping. Memory cue: when grouping, look for identical brackets, including their signs.

Here, \(4p^2=(2p)^2\) and \(9q^2=(3q)^2\). The middle term is \(-2(2p)(3q)=-12pq\), so A is \((2p-3q)^2\). In exams, always check the sign of the middle term.

The direct answer is A: \(27a^3+64b^3=(3a+4b)(9a^2-12ab+16b^2)\). Use the sum-of-cubes identity \(p^3+q^3=(p+q)(p^2-pq+q^2)\). Put \(p=3a\) and \(q=4b\). Then \(p^3=27a^3\), \(q^3=64b^3\), \(p+q=3a+4b\), and the second factor becomes \(9a^2-12ab+16b^2\). Option A is correct. Option B uses the difference-of-cubes first factor \(3a-4b\), so it does not represent a sum. Option C has an incorrect structure and does not follow the cube identity. Option D has the wrong sign in the middle term: for a sum of cubes, the second factor has a negative middle term, not a positive one. Multiplying A confirms the original expression because the cross terms cancel appropriately. Memory cue: sum of cubes has plus, then plus-minus-plus; difference has minus, then all plus.

The identity \(p^2+2pq+q^2=(p+q)^2\) applies, so A is correct. In \((p-q)^2\), the middle term is \(-2pq\). Exam tip: check the middle-term sign first.

It is ((x-2)^3) because the expansion gives (x^3-6x^2+12x-8). In exams match the signs of a perfect cube in order.

First take out (2y), then (x^2+3x+2=(x+1)(x+2)). In exams check the trinomial after taking the common factor.

The direct answer is D: \(4x^2-20x+25=(2x-5)^2\). Use \(a^2-2ab+b^2=(a-b)^2\). The first term is \(4x^2=(2x)^2\), and the last term is \(25=5^2\). The middle term is \(-2(2x)(5)=-20x\), exactly as required. Therefore the expression is \( (2x-5)(2x-5)\). Option A, \( (2x+5)^2\), would produce a positive middle term \(+20x\), so its sign is wrong. Option B, \( (4x-5)^2\), would begin with \(16x^2\), not \(4x^2\). Option C gives \( (2x-10)(2x-5)\), whose product has a different middle term and constant term. Option D is correct because both square roots and the negative middle sign match. Memory cue: a negative middle term means the two terms in the square are subtracted.

The two square terms come from (3p) and (4q), and the middle term is negative. In exams match all three terms of the perfect square.

First (x^2-2xy+y^2=(x-y)^2), then apply difference of squares. In exams group first and apply the identity.

First (a^2+2ab+b^2=(a+b)^2), then apply difference of squares. In exams treat the perfect-square group as one term.

Complete factorisation means continuing until no factor can be split further over the usual integer factors. First, \(x^4-16\) is a difference of squares: \( (x^2)^2-4^2=(x^2-4)(x^2+4)\). The factor \(x^2-4\) is also a difference of squares, so it becomes \( (x-2)(x+2)\). Hence the complete factorisation is \( (x-2)(x+2)(x^2+4)\), option B.

The expression \(x^2+4\) cannot be factored further using integer linear factors, while stopping at \( (x^2-4)(x^2+4)\) is only partial factorisation. Option C uses 4 and 16 incorrectly as a difference of squares, and D does not expand to the original polynomial. Thus the supplied answer is correct.

It is ((4x^2)^2-(9y^2)^2), and (4x^2-9y^2) factors again. In exams check difference of squares repeatedly.

\(49p^2=(7p)^2\) and \(36q^2=(6q)^2\), with a minus sign between them. Thus A becomes \((7p-6q)(7p+6q)\). In exams, check for two perfect squares separated by subtraction.

A perfect-square trinomial has the form \(a^2+2ab+b^2=(a+b)^2\). Here, \(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so option A is correct. In option B, the constant term would need to be \(25\), while option C is a difference of squares. Exam tip: take the square roots of the first and last terms and check whether the middle term is twice their product.

Expanding \((3a-b)(a+2b)\) gives \(3a^2+6ab-ab-2b^2=3a^2+5ab-2b^2\). Therefore, option A is correct. Expanding option B gives a middle term of \(-5ab\), whereas the given expression has \(+5ab\). Exam tip: while checking factors, add the two cross-products and verify the sign of the middle term.

The first and last terms are \((3x)^2\) and \((4y)^2\). Using \((a-b)^2=a^2-2ab+b^2\), the middle term is \(-2(3x)(4y)=-24xy\), so A is correct. In exams, always check the sign of the middle term.

Since \(4x^2=(2x)^2\) and \(25=5^2\), it is a difference of squares: \((2x-5)(2x+5)\). Option B is a perfect-square trinomial. Exam tip: check whether both terms are perfect squares first.

((3x-2)(5x-4)=15x^2-22x+8). In exams both signs can be negative when the constant is positive and the middle term is negative.