What is the factorised form of (a^2+4b^2+9c^2+4ab+12bc+6ac)?
Correct answer: C
All middle terms form (2ab)-type pairs of (a), (2b), and (3c). Exam tip: check a three-term square pairwise.
View question detailsMathematics
Factorisation
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Hard · Level 71 · perfect square trinomial,three variables,factorisation
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- ( (a+2b-3c)^2 )
- ( (a-2b+3c)^2 )
- ( (a+2b+3c)^2 )
- ( (a+4b+9c)^2 )
Hard · Level 71 · perfect square,three terms,signs
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- ( (m+n+p)^2 )
- ( (m-n-p)^2 )
- ( (m-n+p)^2 )
- ( (m+n-p)^2 )
Hard · Level 71 · quadratic,sign choice,factorisation
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- ( (x+9)(x-5) )
- ( (x-9)(x+5) )
- ( (x+15)(x-3) )
- ( (x-1)(x+45) )
Hard · Level 71 · quadratic,ac method,factorisation
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- ( (3x-10)(x-1) )
- ( (3x-2)(x-5) )
- ( (3x+2)(x+5) )
- ( (x-10)(3x-1) )
Hard · Level 71 · factorisation, grouping terms, algebraic identities, common binomial, class 9 mathematics
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- \(ax+ay+bx+by\)
- \(x^2-16\)
- \(p^2+10p+25\)
- \(12m^2+18m\)
Hard · Level 71 · factorisation,algebraic identities,quadratic expressions,binomial products,grade 9 mathematics
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- \((7a+2b)(a-b)\)
- \((7a-2b)(a+b)\)
- \((7a-b)(a+2b)\)
- \((7a+b)(a-2b)\)
Hard · Level 71 · cubic identity,three variables,factorisation
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- ( x^2+y^2+z^2-xy-yz-zx )
- ( x^2+y^2+z^2+xy+yz+zx )
- ( x^2-y^2+z^2 )
- ( x+y+z )
Hard · Level 71 · factorisation, algebraic identities, perfect square trinomial, binomial square, class 9 mathematics
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- \(a^2+6ab+9b^2\)
- \(a^2+6ab+6b^2\)
- \(a^2+3ab+9b^2\)
- \(a^2-6ab-9b^2\)
Hard · Level 71 · cubic identity,three variables,advanced factorisation
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- ( (4x+5y+6z)(16x^2+25y^2+36z^2-20xy-30yz-24xz) )
- ( (4x-5y+6z)(16x^2+25y^2+36z^2) )
- ( (4x+5y-6z)^3 )
- ( (8x+25y+36z)(x^2+y^2+z^2) )
Hard · Level 71 · algebraic identities,factorisation,difference of squares,perfect squares,class 9 mathematics
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- 25x^2+20xy+4y^2
- a^2+9b^2
- 49p^2-q^2
- 16m^2-40mn+25n^2
Hard · Level 71 · common factor,sum of cubes,factorisation
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- ( 2(x+2)(x^2-2x+4) )
- ( (2x+16)(x^2-2x+4) )
- ( 2(x-2)(x^2+2x+4) )
- ( (x+2)(2x^2-4x+8) )
Hard · Level 71 · factorisation, algebraic identities, difference of squares, class 9 mathematics, polynomial expressions
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- \(9x^2-16y^2\)
- \(9x^2+16y^2\)
- \(9x^2-24xy+16y^2\)
- \(9x^2+24xy+16y^2\)
Hard · Level 71 · biquadratic,complete factorisation,difference of squares
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- ( (x^2-1)(x^2-9) )
- ( (x^2+1)(x^2+9) )
- ( (x-1)(x+1)(x-3)(x+3) )
- ( (x^2-3)^2 )
Hard · Level 71 · perfect square,difference of squares,factorisation
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- ( (3x-5y-7z)(3x-5y+7z) )
- ( (9x+25y-49z)(x+y+z) )
- ( (3x+5y+7z)^2 )
- ( (3x+5y-7z)(3x+5y+7z) )
Hard · Level 71 · cyclic expression,factorisation,advanced algebra
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- ( -(a-b)(b-c)(c-a) )
- ( (a+b+c)(ab+bc+ca) )
- ( (a-b)(b-c)(c-a) )
- ( abc(a-b-c) )
Hard · Level 71 · factorisation, algebraic identities, quadratic trinomials, error analysis, class 9 mathematics
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- \((3x+2)(2x-1)\)
- \((3x-2)(2x+1)\)
- \((6x-2)(x+1)\)
- \((3x-1)(2x+2)\)
Hard · Level 71 · algebraic identities, factorisation, difference of squares, class 9 mathematics, polynomial expressions
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- \(49p^2-64q^2\)
- \(49p^2+64q^2\)
- \(49p^2-64pq\)
- \(49p-64q\)
Hard · Level 71 · factorisation, algebraic identities, sum of cubes, class 9 mathematics, error analysis
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- The factorisation is correct; it is based on the identity \(a^3+b^3=(a+b)(a^2-ab+b^2)\).
- The middle sign should be positive, so the correct factorisation is \((2p+q)(4p^2+2pq+q^2)\).
- This factorisation is correct only when \(q\) is negative.
- The given expression is a difference of cubes, so the first factor should be \(2p-q\).
Hard · Level 71 · perfect square,difference of squares,factorisation
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- ( (5x-2y-6)(5x-2y+6) )
- ( (5x+2y-6)(5x+2y+6) )
- ( (25x-4y-36)(x+y+1) )
- ( (5x-2y+6)^2 )
Hard · Level 71 · factorisation, algebraic identities, grouping method, polynomials, class 9 mathematics
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- 3x^2+6x+5x+10
- 3x^2+6x+5x+9
- 3x^2+5x+6x+8
- 3x^2+7x+5x+10
Which is the correct factorisation of (m^2+n^2+p^2-2mn-2np+2mp)?
Correct answer: C
This is the perfect square of (m), (-n), and (p). Exam tip: decide bracket signs from the signs of mixed terms.
View question detailsChoose the factorised form of (x^2+4x-45).
Correct answer: A
The numbers (9) and (-5) have sum (4) and product (-45). Exam tip: with opposite signs, the larger magnitude gives the middle term sign.
View question detailsWhich is the correct factorisation of (3x^2-17x+10)?
Correct answer: B
Expanding gives ( (3x-2)(x-5)=3x^2-17x+10 ). Exam tip: choose the pair with product (30) and sum (-17).
View question detailsWhich of the following expressions can be factorised most directly by suitable grouping of terms?
Correct answer: A
In \(ax+ay+bx+by\), group the terms as \(a(x+y)+b(x+y)\), then take \((x+y)\) common to get \((a+b)(x+y)\). \(x^2-16\) instead uses the difference-of-squares identity. Exam tip: look for a repeated bracket after grouping.
View question detailsChoose the correct factorisation of (7a^2-5ab-2b^2).
Correct answer: A
Expanding \((7a+2b)(a-b)\) gives \(7a^2-7ab+2ab-2b^2=7a^2-5ab-2b^2\). Hence, option A is correct. Option B produces a middle term of \(+5ab\), not \(-5ab\). Exam tip: expand the factors and check the sign of the middle term carefully.
View question detailsIf (x+y+z) is a factor of (x^3+y^3+z^3-3xyz), what is the other factor?
Correct answer: A
This is the identity (x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)). Exam tip: remember the three-variable cubic identity.
View question detailsWhich of the following expressions can be factorised as the product of two identical binomials?
Correct answer: A
In \(a^2+6ab+9b^2\), the first and last terms are \(a^2\) and \((3b)^2\). The middle term is \(2\times a\times3b=6ab\), so it equals \((a+3b)^2\). Exam tip: always verify the middle coefficient.
View question detailsHow can (64x^3+125y^3+216z^3-180xyz) be written?
Correct answer: A
It is ( (4x)^3+(5y)^3+(6z)^3-3(4x)(5y)(6z) ). Exam tip: apply the identity (a^3+b^3+c^3-3abc).
View question detailsWhich of the following expressions can be identified as a difference of two perfect squares?
Correct answer: C
In 49p^2-q^2, 49p^2=(7p)^2 and q^2 is also a perfect square, with a minus sign between them. Hence it is a difference of squares. Options A and D are perfect-square trinomials. Exam tip: check for two square terms separated by ‘−’.
View question detailsChoose the complete factorised form of (2x^3+16).
Correct answer: A
First take out common factor (2) and apply sum of cubes to (x^3+8). Exam tip: do not forget the common factor.
View question detailsWhich expression can be factorised using the identity for the difference of two squares?
Correct answer: A
\(9x^2-16y^2=(3x)^2-(4y)^2\), so it fits \(a^2-b^2=(a-b)(a+b)\): \((3x-4y)(3x+4y)\). Option B has a sum, not a difference. Exam tip: check for two perfect squares separated by a minus sign.
View question detailsWhat is the complete factorised form of (x^4-10x^2+9)?
Correct answer: C
First we get ((x^2-1)(x^2-9)), and both are differences of squares. Exam tip: factor till the final complete form.
View question detailsChoose the factorised form of (9x^2+30xy+25y^2-49z^2).
Correct answer: D
First write (9x^2+30xy+25y^2) as ((3x+5y)^2). Exam tip: use (49z^2=(7z)^2) for difference of squares.
View question detailsWhat is the factorised form of (ab(a-b)+bc(b-c)+ca(c-a))?
Correct answer: A
Expanding and grouping the terms gives (-(a-b)(b-c)(c-a)). Exam tip: handle signs carefully in cyclic expressions.
View question detailsA student claims that \(6x^2-x-2=(3x+2)(2x-1)\). Which is the correct factorisation of \(6x^2-x-2\) that corrects the student's error?
Correct answer: B
\((3x-2)(2x+1)=6x^2+3x-4x-2=6x^2-x-2\), so B is correct. The student's factors give a middle term of \(+x\), not \(-x\). Exam tip: expand factors to verify the middle term.
View question detailsWhich of the following expressions is an example of the difference of squares identity?
Correct answer: A
Since \(49p^2=(7p)^2\) and \(64q^2=(8q)^2\), option A has the form \(a^2-b^2\), the difference of squares. Option B is a sum of squares, not this identity. Exam tip: check for two perfect squares separated by a minus sign.
View question detailsA student has factorised \(8p^3+q^3\) as \((2p+q)(4p^2-2pq+q^2)\). Which statement about the solution is correct?
Correct answer: A
Here, \(8p^3=(2p)^3\) and \(q^3=q^3\). Applying the sum-of-cubes identity gives the stated factorisation. Exam tip: in \(a^3+b^3\), the middle term of the second factor has a negative sign.
View question detailsWhich is the correct factorisation of (25x^2-20xy+4y^2-36)?
Correct answer: A
First write (25x^2-20xy+4y^2) as ((5x-2y)^2). Exam tip: use (36=6^2) for difference of squares.
View question detailsIn which of the following polynomials does grouping the first two and the last two terms produce the same binomial as a common factor?
Correct answer: A
In option A, grouping gives 3x(x+2)+5(x+2). The common binomial is (x+2), so it factorises as (3x+5)(x+2). Exam tip: check that both groups contain the same binomial.
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