Download Factors oF aLgebraic eXpressions

10
Factorisation
introduction
In previous classes, you have learnt that composite numbers can be expressed as the product of prime
numbers.
For example:
42 = 2 × 3 × 7
Here 2, 3 and 7 are prime factors of 42.
Similarly, algebraic expressions can also be expressed as the product of irreducible factors. (Irreducible
factors means a factor which cannot be expressed further as a product of factors).
Inthischapter,weshalllearndifferentmethodsoffindingthefactorsofanalgebraicexpressions.
Factors oF aLgebraic eXpressions
In class VII, you have learnt that terms of an algebraic expression are formed as the product of factors.
For example, in the algebraic expression 7xy + 8y, the term 7xy is formed by the product of 7, x and
y. We say that 7, x and y are factors of 7xy.
Similarly, the product of 3a2 and 5a + 4b = 3a2(5a + 4b) = 15a3 + 12a2b, we say that 3a2 and 5a + 4b
are factors of 15a3 + 12a2b.
Also the product of 2x + 7 and 2x – 7 = (2x + 7)(2x – 7) = 4x2 – 49; we say that 2x + 7 and 2x – 7
are factors of 4x2 – 49. We write it as
4x2 – 49 = (2x + 7)(2x – 7).
Thus, when an algebraic expression can be written as the product of two or more expressions, then
each of these expressions is called a factor of the given expression.
To find factors of a given expression means to obtain two or more expressions whose product is the
given expression.
The process of finding two or more expressions whose product is the given expression is called
factorisation.
Thus, factorisation is the reverse process of multiplication.
For example:
(i)
(ii)
(iii)
(iv)
Product
7xy(5xy – 3) = 35x2y2 – 21xy
(4a + 5b)(4a – 5b) = 16a2 – 25b2
(p + 3)(p – 7) = p2 – 4p – 21
(2x + 3)(3x – 5) = 6x2 – x – 15
Factors
35x2y2 – 21xy = 7xy(5xy – 3)
16a2 – 25b2 = (4a + 5b)(4a – 5b)
p2 – 4p – 21 = (p + 3)(p – 7)
6x2 – x – 15 = (2x + 3)(3x – 5)
In this chapter, we shall learn the factorisation of polynomials by the following methods:
• Taking out common factors
• Grouping
Learning Mathematics–VIII
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• Using algebraic identities
• Factors of trinomials of the type
(i)x2 + px + q where p, q are integers
(ii)ax2 + bx + c where a, b, c are integers.
Before taking up factorisation, we would like to introduce the concept of H.C.F.
H.C.F. of two or more polynomials (with integral coefficients) is the largest common factor of the given
polynomials.
H.C.F. of two or more monomials = (H.C.F. of their numerical coefficients) ×
(H.C.F. of their literal coefficients)
H.C.F. of literal coefficients = product of each common literal raised to the lowest power
Remark
Here it is understood that the numerical coefficients of the monomials (under consideration)
are integers and the powers of the literals involved in the monomials are positive
integers.
For example:
(i)H.C.F. of 6x2y2 and 8xy3 :
H.C.F. of numerical coefficients = H.C.F. of 6 and 8 = 2
H.C.F. of literal coefficients = H.C.F. of x2y2 and xy3
= product of each common literal raised to the lowest power = xy2
∴
H.C.F. of 6x2y2 and 8xy3 = 2 × xy2 = 2xy2.
(ii)H.C.F. of 15a3b2c3, 12a4bc4 and 18a5b3c2 :
H.C.F. of numerical coefficients = H.C.F. of 15, 12 and 18 = 3
H.C.F. of literal coefficients = H.C.F. of a3b2c3, a4bc4 and a5b3c2
= product of each common literal raised to the lowest power = a3bc2
∴H.C.F. of the given monomials = 3 × a3bc2 = 3a3bc2.
Factorising by taking out common factors
If the different terms of the given polynomial have common factors, then the given polynomial can
be factorised by the following procedure :
(i)Find the H.C.F. of all the terms of the given polynomial.
(ii)Divide each term of the given polynomial by H.C.F. Enclose the quotient within the brackets and
keep the common factor outside the bracket.
Example 1. Factorise the following polynomials:
(ii) 15ab2 – 21a2b
(i)24x3 – 32x2
(iii) 14x2y2 – 10x2y + 8xy2
Solution.
(i)H.C.F. of 24x3 and 32x2 is 8x2
24x3 – 32x2 = 8x2(3x – 4)
(ii)H.C.F. of 15ab2 and 21a2 b is 3ab
∴
15ab2 – 21a2 b = 3ab(5b – 7a)
(iii)H.C.F. of 14x2y2, 10x2y and 8xy2 is 2xy
∴
14x2y2 – 10x2y + 8xy2 = 2xy(7xy – 5x + 4y)
Divide each term by 8x2 and
keep 8x2 outside the bracket
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179
Example 2. Factorise the following:
(i)3x(y + 2z) + 5a(y + 2z)
(ii)10(p – 2q)3 + 6(p – 2q)2 – 20(p – 2q)
Solution.
(i)H.C.F. of the expressions
Divide each expression by
y + 2z and keep y + 2z
3x(y + 2z) and 5a(y + 2z) is y + 2z
outside the bracket
∴ 3x(y + 2z) + 5a(y + 2z) = (y + 2z)(3x + 5a)
(ii)H.C.F. of the expressions 10(p – 2q)3, 6(p – 2q)2 and 20(p – 2q) is 2(p – 2q)
∴ 10(p – 2q)3 + 6(p – 2q)2 – 20(p – 2q)
= 2(p – 2q)[5(p – 2q)2 + 3(p – 2q) – 10].
Exercise 10.1
Factorise the following (1 to 9) polynomials:
(ii) 15ax3 – 9ax2
(iii) 20l2m + 30alm
1. (i) 8xy3 + 12x2y2 (ii) 4x3 – 6x2
(iii) 5x2y – 15xy2
2. (i) 21py2 – 56py
3. (i) 2πr2 – 4πr
(ii) 18m + 16n. (iii) –16z + 20z3
4. (i) 25abc2 – 15a2b2c
(ii) 28p2q2r – 42pq2r2.
(iii) x2yz + xy2z + xyz2
5. (i) 8x3 – 6x2 + 10x
(ii) 14mn + 22m – 62p
(iii) – 4a2 + 4ab – 4ca
(ii)27a3b3 – 18a2b3 + 75a3b2
6. (i)18p2q2 – 24pq2 + 30p2q
(iii)ax2y + bxy2 + cxyz
7. (i)15a(2p – 3q) – 10b(2p – 3q)
(ii)3a(x2 + y2) + 6b(x2 + y2)
(ii)14(a – 3b)3 – 21p(a – 3b)
8. (i)6(x + 2y)3 + 8(x + 2y)2
9. 10a(2p + q)3 – 15b(2p + q)2 + 35(2p + q)
Factorising by grouping of terms
When the grouping of terms of the given polynomial gives rise to common factor, then the given
polynomial can be factorised by the following procedure:
(i)Arrange the terms of the given polynomial in groups in such a way that each group has a common
factor.
(ii)Factorise each group.
(iii)Take out the factor which is common to each group.
Note. Factorisation by grouping is possible only if the given polynomial contains an even number of
terms.
Example 1. Factorise the following polynomials:
(i)ax – ay + bx – by
(ii) 4x2 – 10xy – 6xz + 15yz.
Solution.
(i)ax – ay + bx – by = (ax – ay) + (bx – by)
= a(x – y) + b(x – y)
= (x – y)(a + b)
(ii)4x2 – 10xy – 6xz + 15yz = (4x2 – 10xy) – (6xz – 15yz)
= 2x(2x – 5y) – 3z(2x – 5y)
= (2x – 5y)(2x – 3z).