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Chapter 11

Factorisation — Exercise 11.1

Class - 8 ML Aggarwal Understanding ICSE Mathematics

Exercise 11.1

Question 1(i)

Factorise the following polynomial:

8xy3+12x2y28xy^3 + 12x^2y^2

Answer

Factorising,

⇒ 8xy3+12x2y28xy^3 + 12x^2y^2

HCF of 8xy38xy^3 and 12x2y212x^2y^2:

HCF of numerical coefficients = HCF of 8 and 12 = 4

HCF of literal coefficients = HCF of xy3xy^3 and x2y2x^2y^2

= product of each common literal raised to the lowest power

= xy2xy^2

∴ HCF of 8xy38xy^3 and 12x2y212x^2y^2 = 4 × xy2xy^2 = 4xy24xy^2

Dividing each term by 4xy24xy^2 and writing 4xy24xy^2 outside the bracket,

⇒ 4xy2(2y+3x)4xy^2(2y + 3x)

Hence, 8xy3+12x2y2=4xy2(2y+3x)\bm{8xy^3 + 12x^2y^2 = 4xy^2(2y + 3x)}

Question 1(ii)

Factorise the following polynomial:

15ax3−9ax215ax^3 - 9ax^2

Answer

Factorising,

⇒ 15ax3−9ax215ax^3 - 9ax^2

HCF of 15ax315ax^3 and 9ax29ax^2:

HCF of numerical coefficients = HCF of 15 and 9 = 3

HCF of literal coefficients = HCF of ax3ax^3 and ax2ax^2

= product of each common literal raised to the lowest power

= ax2ax^2

∴ HCF of 15ax315ax^3 and 9ax29ax^2 = 3 × ax2ax^2 = 3ax23ax^2

Dividing each term by 3ax23ax^2 and writing 3ax23ax^2 outside the bracket,

⇒ 3ax2(5x−3)3ax^2(5x - 3)

Hence, 15ax3−9ax2=3ax2(5x−3)\bm{15ax^3 - 9ax^2 = 3ax^2(5x - 3)}

Question 2(i)

Factorise the following polynomial:

21py2−56py21py^2 - 56py

Answer

Factorising,

⇒ 21py2−56py21py^2 - 56py

HCF of 21py221py^2 and 56py56py:

HCF of numerical coefficients = HCF of 21 and 56 = 7

HCF of literal coefficients = HCF of py2py^2 and pypy

= product of each common literal raised to the lowest power

= pypy

∴ HCF of 21py221py^2 and 56py56py = 7 × pypy = 7py7py

Dividing each term by 7py7py and writing 7py7py outside the bracket,

⇒ 7py(3y−8)7py(3y - 8)

Hence, 21py2−56py=7py(3y−8)\bm{21py^2 - 56py = 7py(3y - 8)}

Question 2(ii)

Factorise the following polynomial:

4x3−6x24x^3 - 6x^2

Answer

Factorising,

⇒ 4x3−6x24x^3 - 6x^2

HCF of 4x34x^3 and 6x26x^2:

HCF of numerical coefficients = HCF of 4 and 6 = 2

HCF of literal coefficients = HCF of x3x^3 and x2x^2

= product of each common literal raised to the lowest power

= x2x^2

∴ HCF of 4x34x^3 and 6x26x^2 = 2 × x2x^2 = 2x22x^2

Dividing each term by 2x22x^2 and writing 2x22x^2 outside the bracket,

⇒ 2x2(2x−3)2x^2(2x - 3)

Hence, 4x3−6x2=2x2(2x−3)\bm{4x^3 - 6x^2 = 2x^2(2x - 3)}

Question 3(i)

Factorise the following polynomial:

25abc2−15a2b2c25abc^2 - 15a^2b^2c

Answer

Factorising,

⇒ 25abc2−15a2b2c25abc^2 - 15a^2b^2c

HCF of 25abc225abc^2 and 15a2b2c15a^2b^2c:

HCF of numerical coefficients = HCF of 25 and 15 = 5

HCF of literal coefficients = HCF of abc2abc^2 and a2b2ca^2b^2c

= product of each common literal raised to the lowest power

= abcabc

∴ HCF of 25abc225abc^2 and 15a2b2c15a^2b^2c = 5 × abcabc = 5abc5abc

Dividing each term by 5abc5abc and writing 5abc5abc outside the bracket,

⇒ 5abc(5c−3ab)5abc(5c - 3ab)

Hence, 25abc2−15a2b2c=5abc(5c−3ab)\bm{25abc^2 - 15a^2b^2c = 5abc(5c - 3ab)}

Question 3(ii)

Factorise the following polynomial:

x2yz+xy2z+xyz2x^2yz + xy^2z + xyz^2

Answer

Factorising,

⇒ x2yz+xy2z+xyz2x^2yz + xy^2z + xyz^2

HCF of x2yzx^2yz, xy2zxy^2z and xyz2xyz^2:

HCF of numerical coefficients = HCF of 1, 1 and 1 = 1

HCF of literal coefficients = HCF of x2yzx^2yz, xy2zxy^2z and xyz2xyz^2

= product of each common literal raised to the lowest power

= xyzxyz

∴ HCF of x2yzx^2yz, xy2zxy^2z and xyz2xyz^2 = 1 × xyzxyz = xyzxyz

Dividing each term by xyzxyz and writing xyzxyz outside the bracket,

⇒ xyz(x+y+z)xyz(x + y + z)

Hence, x2yz+xy2z+xyz2=xyz(x+y+z)\bm{x^2yz + xy^2z + xyz^2 = xyz(x + y + z)}

Question 4(i)

Factorise the following polynomial:

8x3−6x2+10x8x^3 - 6x^2 + 10x

Answer

Factorising,

⇒ 8x3−6x2+10x8x^3 - 6x^2 + 10x

HCF of 8x38x^3, 6x26x^2 and 10x10x:

HCF of numerical coefficients = HCF of 8, 6 and 10 = 2

HCF of literal coefficients = HCF of x3x^3, x2x^2 and xx

= product of each common literal raised to the lowest power

= xx

∴ HCF of 8x38x^3, 6x26x^2 and 10x10x = 2 × xx = 2x2x

Dividing each term by 2x2x and writing 2x2x outside the bracket,

⇒ 2x(4x2−3x+5)2x(4x^2 - 3x + 5)

Hence, 8x3−6x2+10x=2x(4x2−3x+5)\bm{8x^3 - 6x^2 + 10x = 2x(4x^2 - 3x + 5)}

Question 4(ii)

Factorise the following polynomial:

14mn+22m−62p14mn + 22m - 62p

Answer

Factorising,

⇒ 14mn+22m−62p14mn + 22m - 62p

HCF of 14mn14mn, 22m22m and 62p62p:

HCF of numerical coefficients = HCF of 14, 22 and 62 = 2

HCF of literal coefficients = HCF of mnmn, mm and pp

= product of each common literal raised to the lowest power

= 1, as no literal is common to all the three terms

∴ HCF of 14mn14mn, 22m22m and 62p62p = 2 × 1 = 2

Dividing each term by 2 and writing 2 outside the bracket,

⇒ 2(7mn+11m−31p)2(7mn + 11m - 31p)

Hence, 14mn+22m−62p=2(7mn+11m−31p)\bm{14mn + 22m - 62p = 2(7mn + 11m - 31p)}

Question 5(i)

Factorise the following polynomial:

18p2q2−24pq2+30p2q18p^2q^2 - 24pq^2 + 30p^2q

Answer

Factorising,

⇒ 18p2q2−24pq2+30p2q18p^2q^2 - 24pq^2 + 30p^2q

HCF of 18p2q218p^2q^2, 24pq224pq^2 and 30p2q30p^2q:

HCF of numerical coefficients = HCF of 18, 24 and 30 = 6

HCF of literal coefficients = HCF of p2q2p^2q^2, pq2pq^2 and p2qp^2q

= product of each common literal raised to the lowest power

= pqpq

∴ HCF of 18p2q218p^2q^2, 24pq224pq^2 and 30p2q30p^2q = 6 × pqpq = 6pq6pq

Dividing each term by 6pq6pq and writing 6pq6pq outside the bracket,

⇒ 6pq(3pq−4q+5p)6pq(3pq - 4q + 5p)

Hence, 18p2q2−24pq2+30p2q=6pq(3pq−4q+5p)\bm{18p^2q^2 - 24pq^2 + 30p^2q = 6pq(3pq - 4q + 5p)}

Question 5(ii)

Factorise the following polynomial:

27a3b3−18a2b3+75a3b227a^3b^3 - 18a^2b^3 + 75a^3b^2

Answer

Factorising,

⇒ 27a3b3−18a2b3+75a3b227a^3b^3 - 18a^2b^3 + 75a^3b^2

HCF of 27a3b327a^3b^3, 18a2b318a^2b^3 and 75a3b275a^3b^2:

HCF of numerical coefficients = HCF of 27, 18 and 75 = 3

HCF of literal coefficients = HCF of a3b3a^3b^3, a2b3a^2b^3 and a3b2a^3b^2

= product of each common literal raised to the lowest power

= a2b2a^2b^2

∴ HCF of 27a3b327a^3b^3, 18a2b318a^2b^3 and 75a3b275a^3b^2 = 3 × a2b2a^2b^2 = 3a2b23a^2b^2

Dividing each term by 3a2b23a^2b^2 and writing 3a2b23a^2b^2 outside the bracket,

⇒ 3a2b2(9ab−6b+25a)3a^2b^2(9ab - 6b + 25a)

Hence, 27a3b3−18a2b3+75a3b2=3a2b2(9ab−6b+25a)\bm{27a^3b^3 - 18a^2b^3 + 75a^3b^2 = 3a^2b^2(9ab - 6b + 25a)}

Question 6(i)

Factorise the following polynomial:

15a(2p−3q)−10b(2p−3q)15a(2p - 3q) - 10b(2p - 3q)

Answer

Factorising,

⇒ 15a(2p−3q)−10b(2p−3q)15a(2p - 3q) - 10b(2p - 3q)

HCF of 15a(2p−3q)15a(2p - 3q) and 10b(2p−3q)10b(2p - 3q):

HCF of numerical coefficients = HCF of 15 and 10 = 5

HCF of the remaining factors = HCF of a(2p−3q)a(2p - 3q) and b(2p−3q)b(2p - 3q)

= product of each common factor raised to the lowest power

= (2p−3q)(2p - 3q)

∴ HCF of 15a(2p−3q)15a(2p - 3q) and 10b(2p−3q)10b(2p - 3q) = 5 × (2p−3q)(2p - 3q) = 5(2p−3q)5(2p - 3q)

Dividing each term by 5(2p−3q)5(2p - 3q) and writing 5(2p−3q)5(2p - 3q) outside the bracket,

⇒ 5(2p−3q)(3a−2b)5(2p - 3q)(3a - 2b)

Hence, 15a(2p−3q)−10b(2p−3q)=5(2p−3q)(3a−2b)\bm{15a(2p - 3q) - 10b(2p - 3q) = 5(2p - 3q)(3a - 2b)}

Question 6(ii)

Factorise the following polynomial:

3a(x2+y2)+6b(x2+y2)3a(x^2 + y^2) + 6b(x^2 + y^2)

Answer

Factorising,

⇒ 3a(x2+y2)+6b(x2+y2)3a(x^2 + y^2) + 6b(x^2 + y^2)

HCF of 3a(x2+y2)3a(x^2 + y^2) and 6b(x2+y2)6b(x^2 + y^2):

HCF of numerical coefficients = HCF of 3 and 6 = 3

HCF of the remaining factors = HCF of a(x2+y2)a(x^2 + y^2) and b(x2+y2)b(x^2 + y^2)

= product of each common factor raised to the lowest power

= (x2+y2)(x^2 + y^2)

∴ HCF of 3a(x2+y2)3a(x^2 + y^2) and 6b(x2+y2)6b(x^2 + y^2) = 3 × (x2+y2)(x^2 + y^2) = 3(x2+y2)3(x^2 + y^2)

Dividing each term by 3(x2+y2)3(x^2 + y^2) and writing 3(x2+y2)3(x^2 + y^2) outside the bracket,

⇒ 3(x2+y2)(a+2b)3(x^2 + y^2)(a + 2b)

Hence, 3a(x2+y2)+6b(x2+y2)=3(x2+y2)(a+2b)\bm{3a(x^2 + y^2) + 6b(x^2 + y^2) = 3(x^2 + y^2)(a + 2b)}

Question 7(i)

Factorise the following polynomial:

6(x+2y)3+8(x+2y)26(x + 2y)^3 + 8(x + 2y)^2

Answer

Factorising,

⇒ 6(x+2y)3+8(x+2y)26(x + 2y)^3 + 8(x + 2y)^2

HCF of 6(x+2y)36(x + 2y)^3 and 8(x+2y)28(x + 2y)^2:

HCF of numerical coefficients = HCF of 6 and 8 = 2

HCF of the remaining factors = HCF of (x+2y)3(x + 2y)^3 and (x+2y)2(x + 2y)^2

= product of each common factor raised to the lowest power

= (x+2y)2(x + 2y)^2

∴ HCF of 6(x+2y)36(x + 2y)^3 and 8(x+2y)28(x + 2y)^2 = 2 × (x+2y)2(x + 2y)^2 = 2(x+2y)22(x + 2y)^2

Dividing each term by 2(x+2y)22(x + 2y)^2 and writing 2(x+2y)22(x + 2y)^2 outside the bracket,

⇒ 2(x+2y)2(3(x+2y)+4)2(x + 2y)^2(3(x + 2y) + 4)

⇒ 2(x+2y)2(3x+6y+4)2(x + 2y)^2(3x + 6y + 4)

Hence, 6(x+2y)3+8(x+2y)2=2(x+2y)2(3x+6y+4)\bm{6(x + 2y)^3 + 8(x + 2y)^2 = 2(x + 2y)^2(3x + 6y + 4)}

Question 7(ii)

Factorise the following polynomial:

14(a−3b)3−21p(a−3b)14(a - 3b)^3 - 21p(a - 3b)

Answer

Factorising,

⇒ 14(a−3b)3−21p(a−3b)14(a - 3b)^3 - 21p(a - 3b)

HCF of 14(a−3b)314(a - 3b)^3 and 21p(a−3b)21p(a - 3b):

HCF of numerical coefficients = HCF of 14 and 21 = 7

HCF of the remaining factors = HCF of (a−3b)3(a - 3b)^3 and p(a−3b)p(a - 3b)

= product of each common factor raised to the lowest power

= (a−3b)(a - 3b)

∴ HCF of 14(a−3b)314(a - 3b)^3 and 21p(a−3b)21p(a - 3b) = 7 × (a−3b)(a - 3b) = 7(a−3b)7(a - 3b)

Dividing each term by 7(a−3b)7(a - 3b) and writing 7(a−3b)7(a - 3b) outside the bracket,

⇒ 7(a−3b)[2(a−3b)2−3p]7(a - 3b)[2(a - 3b)^2 - 3p]

Hence, 14(a−3b)3−21p(a−3b)=7(a−3b)[2(a−3b)2−3p]\bm{14(a - 3b)^3 - 21p(a - 3b) = 7(a - 3b)[2(a - 3b)^2 - 3p]}

Question 8

Factorise the following polynomial:

10a(2p+q)3−15b(2p+q)2+35(2p+q)10a(2p + q)^3 - 15b(2p + q)^2 + 35(2p + q)

Answer

Factorising,

⇒ 10a(2p+q)3−15b(2p+q)2+35(2p+q)10a(2p + q)^3 - 15b(2p + q)^2 + 35(2p + q)

HCF of 10a(2p+q)310a(2p + q)^3, 15b(2p+q)215b(2p + q)^2 and 35(2p+q)35(2p + q):

HCF of numerical coefficients = HCF of 10, 15 and 35 = 5

HCF of the remaining factors = HCF of a(2p+q)3a(2p + q)^3, b(2p+q)2b(2p + q)^2 and (2p+q)(2p + q)

= product of each common factor raised to the lowest power

= (2p+q)(2p + q)

∴ HCF of 10a(2p+q)310a(2p + q)^3, 15b(2p+q)215b(2p + q)^2 and 35(2p+q)35(2p + q) = 5 × (2p+q)(2p + q) = 5(2p+q)5(2p + q)

Dividing each term by 5(2p+q)5(2p + q) and writing 5(2p+q)5(2p + q) outside the bracket,

⇒ 5(2p+q)[2a(2p+q)2−3b(2p+q)+7]5(2p + q)[2a(2p + q)^2 - 3b(2p + q) + 7]

Hence, 10a(2p+q)3−15b(2p+q)2+35(2p+q)=5(2p+q)[2a(2p+q)2−3b(2p+q)+7]\bm{10a(2p + q)^3 - 15b(2p + q)^2 + 35(2p + q) = 5(2p + q)[2a(2p + q)^2 - 3b(2p + q) + 7]}

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