Factorisation

Correct answer: A. ((x+8)(x+9))

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

Correct answer: B. ((x-9)(x-10))

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.

Correct answer: A. \(x^2+10x+25\)

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

Correct answer: D. ((3x+1)(x-5))

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

Correct answer: A. ((2a-3b)^2)

(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.

Correct answer: A. \(x^2+10x+25\)

\(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\). 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 \(a^2+2ab+b^2=(a+b)^2\).

Correct answer: A. \((3m+2n)(4m-5n)\)

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.

Correct answer: D. ((a-5)(b+2))

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.

Correct answer: A. \(4p^2-12pq+9q^2\)

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.

Correct answer: A. ((3a+4b)(9a^2-12ab+16b^2))

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.

Correct answer: A. \((p+q)^2\)

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.

Correct answer: C. ((x-2)^3)

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.

Correct answer: B. (2y(x+1)(x+2))

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

Correct answer: D. ((2x-5)^2)

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.

Correct answer: A. ((3p-4q)^2)

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.

Correct answer: A. ((x-y-z)(x-y+z))

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

Correct answer: A. ((a+b+c)(a+b-c))

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

Correct answer: B. ((x-2)(x+2)(x^2+4))

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.

Correct answer: B. ((2x-3y)(2x+3y)(4x^2+9y^2))

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

Correct answer: A. \(49p^2-36q^2\)

\(49p^2=(7p)^2\) 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.

Correct answer: A. \(x^2+10x+25\)

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.

Correct answer: A. \((3a-b)(a+2b)\)

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.

Correct answer: A. क्योंकि \((3x-4y)^2=9x^2-24xy+16y^2\)

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.

Correct answer: C. \(4x^2-25\)

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.

Correct answer: A. ((3x-2)(5x-4))

((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.