What is the correct factorised form of (a^3-b^3)?
Correct answer: B
For difference of cubes, the first factor is (a-b) and the second is (a^2+ab+b^2). Exam tip: keep sum and difference formulas separate.
View question detailsMathematics
Factorisation
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Hard · Level 70 · difference of cubes,identity,factorisation
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- ((a+b)(a^2-ab+b^2))
- ((a-b)(a^2+ab+b^2))
- ((a-b)(a^2-ab+b^2))
- ((a+b)(a^2+ab+b^2))
Hard · Level 70 · sum of cubes,two variables,factorisation
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- ((2m+3n)(4m^2-6mn+9n^2))
- ((2m-3n)(4m^2+6mn+9n^2))
- ((8m+27n)(m^2-mn+n^2))
- ((2m+3n)(4m^2+6mn+9n^2))
Hard · Level 70 · difference of cubes,factorisation,identity
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- ((5x-y)(25x^2+5xy+y^2))
- ((5x+y)(25x^2-5xy+y^2))
- ((125x-y)(x^2+xy+y^2))
- ((5x-y)(25x^2-5xy+y^2))
Hard · Level 70 · combined identities,difference of squares,factorisation
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- ((x-y-z)(x-y+z))
- ((x+y-z)(x+y+z))
- ((x-y)^2-z)
- ((x-z)(x+z)-2xy+y^2)
Hard · Level 70 · mixed identity,perfect square,difference of squares
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- ((a+b-c)(a+b+c))
- ((a-b-c)(a-b+c))
- ((a+b)^2-c)
- ((a+c)(a-c)+2ab+b^2)
Hard · Level 70 · grouping,perfect square,difference of squares
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- ((x+1-y)(x+1+y))
- ((x-y)(x+y)+1)
- ((x+y+1)^2)
- ((x-1-y)(x-1+y))
Hard · Level 70 · perfect square,difference of squares,factorisation
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- ((2x+3-y)(2x+3+y))
- ((2x-3-y)(2x-3+y))
- ((4x+9-y)(4x+9+y))
- ((2x+3)^2-y)
Hard · Level 70 · factorisation,algebraic identities,grouping terms,common binomial,class 9 mathematics
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- \((a+2)(2b+3)\)
- \((a+3)(2b+2)\)
- \((a+2)(2b+6)\)
- \((2a+3)(b+2)\)
Hard · Level 70 · factorisation by grouping,algebraic identities,common factor,polynomials,class 9 mathematics
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- \((3x+5)(y-2)\)
- \((3x-5)(y+2)\)
- \((3x+2)(y-5)\)
- \((3x-10)(y+1)\)
Hard · Level 70 · factorisation, algebraic identities, perfect square trinomial, class 9 mathematics, quadratic expressions
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- \(4x^2-12x+9\)
- \(4x^2-12x-9\)
- \(4x^2-6x+9\)
- \(4x^2+12x-9\)
Hard · Level 70 · factorisation, algebraic identities, sum of cubes, class 9 mathematics, polynomial classification
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- \(125x^3+64y^3\)
- \(125x^3-64y^3\)
- \(25x^2+16y^2\)
- \(125x^3+16y^3\)
Hard · Level 70 · algebraic identities, factorisation, perfect square trinomial, quadratic expressions, class 9 mathematics
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- 9p² − 24pq + 16q²
- 9p² − 24pq + 14q²
- 9p² + 20pq + 16q²
- 8p² − 24pq + 16q²
Hard · Level 70 · factorisation, algebraic identities, difference of squares, class 9 mathematics, polynomials
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- \(x^2-25y^2\)
- \(x^2+25y^2\)
- \(x^2+10xy+25y^2\)
- \(x^2-10xy+25y^2\)
Hard · Level 70 · factorisation, algebraic identities, difference of squares, perfect squares, class 9 mathematics
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- \(25p^2-16q^2\)
- \(25p^2+16q^2\)
- \(25p^2-40pq+16q^2\)
- \(25p^2+40pq+16q^2\)
Hard · Level 70 · factorisation, algebraic identities, difference of cubes, class 9 mathematics, polynomial identities
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- \((a-b)(a^2+ab+b^2)\)
- \((a-b)(a^2-ab+b^2)\)
- \((a+b)(a^2-ab+b^2)\)
- \((a+b)(a^2+ab+b^2)\)
Hard · Level 70 · common factor,cubes,factorisation
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- (27(8x^3-y^3))
- (9(24x^3-3y^3))
- (3(72x^3-9y^3))
- (27(6x^3-y^3))
Hard · Level 70 · factorisation, algebraic identities, difference of cubes, class 9 mathematics, polynomial factorisation
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- \(u^2+uv+v^2\)
- \(u^2-uv+v^2\)
- \(u^2+2uv+v^2\)
- \(u^2-2uv+v^2\)
Hard · Level 70 · algebraic identities, factorisation, sum of cubes, class 9 mathematics, polynomial identities
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- \(p^3+q^3\)
- \(p^3-q^3\)
- \(p^2-q^2\)
- \(p^2+2pq+q^2\)
Hard · Level 70 · factorisation, algebraic identities, perfect square trinomial, binomial square, class 9 mathematics
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- \(9p^2-24pq+16q^2\)
- \(9p^2-12pq+16q^2\)
- \(9p^2-24pq-16q^2\)
- \(6p^2-24pq+16q^2\)
Hard · Level 70 · perfect square,difference of squares,factorisation
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- ((x-2-y)(x-2+y))
- ((x+2-y)(x+2+y))
- ((x-y)(x+y)-4x+4)
- ((x-4-y)(x-4+y))
Factorise (8m^3+27n^3).
Correct answer: A
This is ((2m)^3+(3n)^3). Exam tip: in (a^3+b^3), the middle term of the second bracket is negative.
View question detailsWhat is the factorised form of (125x^3-y^3)?
Correct answer: A
The expression is a difference of two cubes. Since \(125x^3=(5x)^3\), it can be written as \((5x)^3-y^3\). The identity for a difference of cubes is \(a^3-b^3=(a-b)(a^2+ab+b^2)\). The signs in the second factor are especially important: all its terms are positive.
Put \(a=5x\) and \(b=y\). Then the factorisation becomes \((5x-y)(25x^2+5xy+y^2)\). Multiplying confirms the middle terms cancel and the result is \(125x^3-y^3\). Therefore option A is correct. Option D has the wrong sign in its middle term, and the other options do not represent the original difference of cubes.
Which is the correct factorised form of (x^2-2xy+y^2-z^2)?
Correct answer: A
First (x^2-2xy+y^2=(x-y)^2), then use difference of squares. Exam tip: form the perfect square first.
View question detailsWhat will be the factors of (a^2+2ab+b^2-c^2)?
Correct answer: A
First (a^2+2ab+b^2=(a+b)^2), then take the difference with (c^2). Exam tip: solve mixed identities in two steps.
View question detailsIn which form can (x^2-y^2+2x+1) be factorised?
Correct answer: A
Since (x^2+2x+1=(x+1)^2), the expression becomes ((x+1)^2-y^2). Exam tip: group terms correctly.
View question detailsWhat is the correct factorised form of (4x^2+12x+9-y^2)?
Correct answer: A
(4x^2+12x+9=(2x+3)^2). Exam tip: identify the perfect square first and then use difference of squares.
View question detailsChoose the correct factorised form of (2ab+3a+4b+6).
Correct answer: A
Group the terms: \(2ab+3a+4b+6=a(2b+3)+2(2b+3)\). The common binomial is \((2b+3)\), so the factorised form is \((a+2)(2b+3)\). Option D expands to \(2ab+4a+3b+6\), so it is not correct. Exam tip: expand your answer once to verify it against the original expression.
View question detailsFactorise (3xy-6x+5y-10) by grouping.
Correct answer: A
Group the terms as \(3xy-6x+5y-10=3x(y-2)+5(y-2)\). The common factor is \((y-2)\), so the factorisation is \((y-2)(3x+5)=(3x+5)(y-2)\). Expanding option B gives \(3xy+6x-5y-10\), which is not the given expression. Exam tip: verify a grouped factorisation by expanding the final factors once.
View question detailsWhich of the following trinomials is a perfect-square trinomial and will factorise into two identical binomials?
Correct answer: A
\(4x^2-12x+9=(2x)^2-2(2x)(3)+3^2\), so it is \((2x-3)^2\) and has identical binomial factors. In B and D, the negative constant prevents this form. Exam tip: check the middle term using twice the product of square roots.
View question detailsWhich of the following expressions can be factorised directly using the identity for the sum of cubes, \(a^3+b^3=(a+b)(a^2-ab+b^2)\)?
Correct answer: A
Since \(125x^3=(5x)^3\) and \(64y^3=(4y)^3\), option A is a sum of two cubes. Option B is a difference of cubes, so it needs another identity. Exam tip: first check whether both terms are perfect cubes.
View question detailsWhich of the following expressions is a perfect-square trinomial and can be written as a product of two identical binomials?
Correct answer: A
In option A, 9p² = (3p)² and 16q² = (4q)², while the middle term −24pq equals −2(3p)(4q). Hence it is (3p − 4q)². In exams, compare the middle term with ±2ab.
View question detailsWhich of the following expressions can be factorised using only the identity for the difference of squares, \(a^2-b^2=(a-b)(a+b)\)?
Correct answer: A
\(x^2-25y^2=x^2-(5y)^2\), so it factorises as \((x-5y)(x+5y)\). Options C and D are perfect-square trinomials. In exams, check for a minus sign between two square terms.
View question detailsWhich of the following expressions can be identified and factorised as a difference of two perfect squares, \(A^2-B^2\)?
Correct answer: A
\(25p^2-16q^2=(5p)^2-(4q)^2\), so it factorises as \((5p-4q)(5p+4q)\). Options C and D are perfect-square trinomials, while B is a sum of squares. Exam tip: first identify square roots of both terms.
View question detailsWhich of the following is the correct factorised form of the difference of two cubes, \(a^3-b^3\)?
Correct answer: A
\(a^3-b^3=(a-b)(a^2+ab+b^2)\). On expansion, the middle terms cancel, leaving \(a^3-b^3\). Option B has the wrong sign in the middle term. Exam tip: for a difference, the first factor is always \(a-b\).
View question detailsAfter taking out the common factor, how can (216x^3-27y^3) be written first?
Correct answer: A
Both terms have common factor (27), and (216x^3=27\cdot8x^3). Exam tip: taking out the common factor first makes it easier.
View question detailsWhen factorising the expression \(u^3-v^3\), which quadratic factor occurs along with the linear factor \(u-v\)?
Correct answer: A
The difference-of-cubes identity is \(u^3-v^3=(u-v)(u^2+uv+v^2)\), so the required quadratic factor is \(u^2+uv+v^2\). The factor \(u^2-uv+v^2\) is used for a sum of cubes. Exam tip: for a difference, the middle term of the quadratic factor is positive.
View question detailsWhich of the following expressions produces the quadratic factor \(p^2-pq+q^2\) when factorised?
Correct answer: A
Use the identity \(p^3+q^3=(p+q)(p^2-pq+q^2)\), so A is correct. For \(p^3-q^3\), the quadratic factor has \(+pq\), not \(-pq\). Exam tip: check the middle sign first.
View question detailsWhich of the following polynomials can be factorised as the square of a binomial?
Correct answer: A
\(9p^2=(3p)^2\) and \(16q^2=(4q)^2\), while the middle term is \(-2(3p)(4q)=-24pq\). Hence it is \((3p-4q)^2\). In exams, always match the sign of the middle term.
View question detailsWhat is obtained by factorising (x^2-4x-y^2+4)?
Correct answer: A
Since (x^2-4x+4=(x-2)^2), the expression becomes ((x-2)^2-y^2). Exam tip: use (+4) to form a perfect square.
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