Which is the complete factorised form of (a^4-16b^4)?
Correct answer: B
First (a^4-16b^4=(a^2-4b^2)(a^2+4b^2)), then (a^2-4b^2) factors further. Exam tip: stop only after complete factorisation.
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Factorisation
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Hard · Level 70 · higher powers,complete factorisation,difference of squares
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- ((a^2-4b^2)(a^2+4b^2))
- ((a-2b)(a+2b)(a^2+4b^2))
- ((a-4b)(a+4b))
- ((a-2b)^2(a+2b)^2)
Hard · Level 70 · factorisation, quadratic expressions, algebraic identities, coefficient comparison, class 9 mathematics
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- \(m+n=p\) and \(mn=q\)
- \(m+n=q\) and \(mn=p\)
- \(m-n=p\) and \(mn=q\)
- \(m+n=p+q\) and \(mn=1\)
Hard · Level 70 · quadratic factorisation,signs,checking
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- ((5x+3)(2x-1))
- ((5x-3)(2x+1))
- ((10x-3)(x+1))
- ((2x+3)(5x-1))
Hard · Level 70 · hard quadratic,factorisation,middle term
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- ((3x+2)(5x-4))
- ((3x-2)(5x+4))
- ((15x+4)(x-2))
- ((5x+2)(3x-4))
Hard · Level 70 · factorisation, quadratic expressions, algebraic identities, integer coefficients, linear factors
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- \(x^2+5x+6\)
- \(x^2-9\)
- \(x^2+2x+2\)
- \(2x^2+7x+3\)
Hard · Level 70 · factorisation, algebraic identities, common factor, difference of squares, class 9 mathematics
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- \(x^3-4x\)
- \(x^3+4x\)
- \(x^2-4\)
- \(x^3-4\)
Hard · Level 70 · algebraic identities, factorisation, difference of cubes, class 9 mathematics, cubic expressions
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- \(p^2+pq+q^2\)
- \(p^2-pq+q^2\)
- \(p^2-q^2\)
- \(p^2+2pq+q^2\)
Hard · Level 69 · factorisation,quadratic polynomials,algebraic identities,error analysis,class 9 mathematics
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- \((x-5)(x-4)\)
- \((x-10)(x+2)\)
- \((x-20)(x+1)\)
- \((x+5)(x+4)\)
Hard · Level 70 · binomial cube,factorisation,algebraic identities
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- ((3a+2b)^3)
- ((3a-2b)^3)
- ((27a+8b)^3)
- ((a+2b)^3)
Hard · Level 70 · mixed identities,difference of squares,factorisation
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- ((x^2+y^2-4)(x^2+y^2+4))
- ((x^2-y^2-4)(x^2-y^2+4))
- ((x+y-4)(x+y+4))
- ((x^2+y^2)^2-4)
Hard · Level 71 · difference of squares,complete factorisation,algebraic identities
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- ( (x-2)(x+2)(x^2+4) )
- ( (x^2-4)(x^2+4) )
- ( (x-4)(x+4)(x^2+1) )
- ( (x-2)^2(x+2)^2 )
Hard · Level 71 · difference of fourth powers,factorisation,identity
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- ( (a^2-b^2)(a^2+b^2) )
- ( (a-b)^2(a+b)^2 )
- ( (a-b)(a+b)(a^2+b^2) )
- ( (a-b)(a^3+b^3) )
Hard · Level 71 · mixed identity,perfect square,difference of squares
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- ( (x-y-3z)(x-y+3z) )
- ( (x+y-3z)(x+y+3z) )
- ( (x+y-9z)(x+y+z) )
- ( (x+y+3z)^2 )
Hard · Level 71 · perfect square,difference of squares,factorisation
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- ( (2p+3q-5r)(2p+3q+5r) )
- ( (2p-3q)^2-25r^2 )
- ( (4p-9q-25r)(p+q+r) )
- ( (2p-3q-5r)(2p-3q+5r) )
Hard · Level 71 · difference of cubes,common factor,factorisation
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- ( (x-2y)(x+2y)^2 )
- ( (x+2y)(x-2y)^2 )
- ( (x-2y)(x^2+2xy+4y^2) )
- ( (x+2y)(x^2-2xy+4y^2) )
Hard · Level 71 · quadratic in two variables,middle term,factorisation
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- ( (2x+3y)(x+y) )
- ( (2x+y)(x+3y) )
- ( (x+2y)(2x+3y) )
- ( (2x-y)(x-3y) )
Hard · Level 71 · quadratic,signs,factorisation
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- ( (3x-4)(2x+3) )
- ( (6x-4)(x+3) )
- ( (3x+4)(2x-3) )
- ( (2x-4)(3x+3) )
Hard · Level 71 · quadratic in variables,ac method,factorisation
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- ( (3a+b)(4a-b) )
- ( (12a+b)(a-b) )
- ( (3a-b)(4a+b) )
- ( (4a- b)(3a+b) ) / ( (4a-b)(3a+b) )
Hard · Level 71 · biquadratic,factorisation,substitution
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- ( (x^2+1)(x^2+6) )
- ( (x^2+2)(x^2+3) )
- ( (x+2)(x+3) )
- ( (x^2-2)(x^2-3) )
Hard · Level 71 · biquadratic,middle term,factorisation
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- ( (y^2-4)(y^2-9) )
- ( (y^2+4)(y^2+9) )
- ( (y-4)(y-9) )
- ( (y^2-6)(y^2-6) )
If the quadratic expression \(x^2+px+q\) is factorised in the form \((x+m)(x+n)\), which relation between \(m\) and \(n\) is correct?
Correct answer: A
Expanding \((x+m)(x+n)\) gives \(x^2+(m+n)x+mn\). Comparing corresponding terms gives \(m+n=p\) and \(mn=q\). Option B wrongly interchanges the sum and product. Exam tip: match the coefficient of \(x\) with the sum first.
View question detailsWhat is the correct factorised form of (10x^2+x-3)?
Correct answer: A
((5x+3)(2x-1)) gives (10x^2-5x+6x-3). Exam tip: identify the small difference for the middle term (x).
View question detailsWhat are the factors of (15x^2-2x-8)?
Correct answer: D
((5x+2)(3x-4)) gives (15x^2-20x+6x-8). Exam tip: choose the pair that gives (-2x), not (-14x).
View question detailsWhich of the following quadratic expressions cannot be expressed as a product of two linear factors with integer coefficients?
Correct answer: C
\(x^2+2x+2=(x+1)^2+1\), so it has no linear factors with integer coefficients. In contrast, option A is \((x+2)(x+3)\). Exam tip: test factor pairs of the constant term systematically.
View question detailsFor which of the following expressions does taking out a common factor first leave a difference of squares?
Correct answer: A
In \(x^3-4x\), take out the common factor \(x\): \(x(x^2-4)\). Now \(x^2-4=x^2-2^2\), a difference of squares. In \(x^3+4x\), the remaining expression is a sum. Exam tip: check the HCF first.
View question detailsWhich quadratic factor is obtained when the expression \(p^3-q^3\) is factorised?
Correct answer: A
The identity for a difference of cubes is \(p^3-q^3=(p-q)(p^2+pq+q^2)\), so A is the quadratic factor. On expansion, the middle terms cancel. Exam tip: in the quadratic factor for a difference of cubes, the middle term is positive.
View question detailsA student claims that the factorisation of \(x^2-9x+20\) is \((x-10)(x+2)\). Identify the error and choose the correct factorisation.
Correct answer: A
We need two factors of \(20\) whose sum is \(-9\): \(-5\) and \(-4\). Thus, \((x-5)(x-4)=x^2-9x+20\). The student's factors give constant term \(-20\), not \(+20\). Exam tip: verify both the middle and constant terms.
View question detailsIn which form will (27a^3+54a^2b+36ab^2+8b^3) be written?
Correct answer: A
It is of the form ((3a)^3+3(3a)^2(2b)+3(3a)(2b)^2+(2b)^3). Exam tip: match the (3AB) terms carefully in cube identities.
View question detailsWhat is the correct factorised form of (x^4+2x^2y^2+y^4-16)?
Correct answer: A
First (x^4+2x^2y^2+y^4=(x^2+y^2)^2), and then (16=4^2). Exam tip: solve mixed identities in two steps.
View question detailsWhat is the complete factorised form of (x^4-16)?
Correct answer: A
First write (x^4-16) as ((x^2)^2-4^2) and then factor (x^2-4). Exam tip: do not stop before the complete factorisation.
View question detailsChoose the complete factorisation of (a^4-b^4).
Correct answer: C
In (a^4-b^4), first use difference of squares and then factor (a^2-b^2). Exam tip: (a^2+b^2) is usually not factored further here.
View question detailsWhich is the correct factorisation of (x^2+2xy+y^2-9z^2)?
Correct answer: B
First convert (x^2+2xy+y^2) into ((x+y)^2). Exam tip: after forming a perfect square, apply difference of squares.
View question detailsWhat is the factorised form of (4p^2-12pq+9q^2-25r^2)?
Correct answer: D
First write (4p^2-12pq+9q^2) as ((2p-3q)^2). Exam tip: identify ((a-b)^2) from the negative middle term.
View question detailsChoose the factorised form of (x^3-8y^3+6xy(x-2y)).
Correct answer: A
Write (x^3-8y^3) as ((x-2y)(x^2+2xy+4y^2)) and take the common factor. Exam tip: identify the common binomial first.
View question detailsWhich is the correct factorisation of (2x^2+7xy+3y^2)?
Correct answer: B
Split the middle term (7xy) as (6xy+xy). Exam tip: choose the pair with product (6) and sum (7).
View question detailsWhat will be the factorised form of (6x^2-x-12)?
Correct answer: C
Expanding gives ( (3x+4)(2x-3)=6x^2-x-12 ). Exam tip: verify by expansion when signs are tricky.
View question detailsChoose the correct factorisation of (12a^2-7ab-b^2).
Correct answer: D
For (-7ab), write (12a^2-8ab+ab-b^2). Exam tip: when options look similar, confirm by expansion.
View question detailsConvert (x^4+5x^2+6) into factors.
Correct answer: B
Treat (x^2) as a new term and factor like a quadratic. Exam tip: for (x^4) type questions, think (u=x^2).
View question detailsWhat is the correct factorised form of (y^4-13y^2+36)?
Correct answer: A
Taking (y^2) as one quantity, the product must be (36) and the sum (-13). Exam tip: identify the pair (-4) and (-9).
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