]> Factorization< algebra< mathematics< high school< ICSE CBSE< mentorials.com
Top Box
Section Navigator
Author: S.LAL
Created: 20 Sep, 2010; Last Modified: 20 Sep, 2010

Algebra II - 05

Factorisation

Factorisation of an algebraic expression implies writing the expression as a product of simpler algebraic expressions, each of which is called a factor. Factorisation is also termed resolution into factors.

Principle Rules of Factorisation

Taking out the common factor in polynomials

Simple factor

When each of the terms in a polynomial contains a common factor, which is basically a monomial, it can be easily separated out. Such a common factor is called a simple factor. Simple factors are easily spotted by inspection.

Factorise (i) 3a26ab (ii) 5a2bx315abx220b3x2. (Hall & Knight, Elementary Algebra, p. 125 Example 1, 2)

(i) The terms of the expression have 3a as the common factor, which can be separated out. Thus,

3a26ab=3a(a2b)

(ii) The common factor is 5bx2. So,

5a2bx315abx220b3x2=5bx2(a2x3a4b2)

Compound factor

Sometimes, the common factor itself is a polynomial, and can be identified and fished out only after rearranging the terms of the original polynomial into suitable groups. Such a common factor is called a compound factor.

Resolve into factors (i) x2ax+bxab (ii)6x29ax+4bx6ab. (Hall & Knight, Elementary Algebra, p. 126 Example 1, 2)

(i) Noticing that the first two terms contain a common factor x, and the last two terms a common factor b, we group the terms as follows and continue:

x2ax+bxab=(x2ax)+(bxab)=x(xa)+b(xa)=(x+b)(xa)

(ii) Using a similar thinking process, we have

6x29ax+4bx6ab=(6x29ax)+(4bx6ab)=3x(2x3a)+2b(2x3a)=(3x+2b)(2x3a)

Figuring out the compound factors in complex expressions becomes easier with practice.

Splitting the middle term of trinomials of degree 2

Factorisation of (x2+bx+c)

Let the factors of x2+bx+c be of the form (x+p)(x+q). Then,

x2+bx+c=(x+p)(x+q)=x2+px+qx+pq=x2+(p+q)x+pq

Comparing the coefficients of like terms on both sides, we get (p+q)=b and pq=c.

Thus, to factorise a trinomial of the form x2+bx+c,

  1. Find two number p and q such that (p+q)=b and pq=c
  2. The trinomial now stands factorised as (x+p)(x+q).

Resolve into factors (i) x2+11x+24 (ii) x210x+24. (Hall & Knight, Elementary Algebra, p. 127 Example 1, 2)

(i) We need to find p and q such that (p+q)=11 and pq=24. Since the product is positive, as well as the sum, both the numbers must be positive. Clearly, p = 8 and q = 3 satisfy the conditions. Thus,

x2+11x+24=(x+8)(x+3)

(ii) We need to find p and q such that (p+q)=10 and pq=24. This implies that both must be negative, since their sum is negative but the product is positive. Clearly, p = -6 and q = -4 satisfy the conditions. Thus,

x210x+24=(x4)(x6)

Resolve into factors (i) x2+2x35 (ii) x23x54. (Hall & Knight, Elementary Algebra, p. 129 Example 1, 2)

(i) For (p+q)=2 and pq=35 to be true, the numbers must have opposite signs, and the greater must be positive for the sum to be positive. Clearly, p = 7 and q = -5 satisfy the conditions.

x2+2x35=(x+7)(x5)

(ii) For (p+q)=3 and pq=54 to be true, the numbers must have opposite signs, and the greater must be negative for the sum to be negative. p = 7 and q = -5 fit the bill.

x23x54=(x9)(x+6)

Factorisation of (ax2+bx+c)

Let factors of (ax2+bx+c) be of the form (rx+s)(tx+u). Then,

ax2+bx+c=(rx+s)(tx+u)=rtx2+rux+stx+su=rtx2+(ru+st)x+su

Comparing the coefficients of like terms on both sides, we get a = rt, b = ru + st and c = su.

If we let p = ru and q = st, then

pq = rtsu = ac and (p+q)=ru+st=b

Thus, we should find p and q such that (p+q)=b and pq = ac.

So, to factorise a trinomial of the form (ax2+bx+c),

  1. Find two number p and q such that

    (p+q)=b and pq = ac.

  2. Rewrite the expression (ax2+bx+c) with the middle term split as:

    (ax2+px+qx+c) or (ax2+px+qx+pqa)

  3. In this form, the compound factors can then be taken out as follows:

    ax2+px+qx+pqa=x(ax+p)+q(x+pa)=x(ax+p)+qa(ax+p)=(ax+p)(x+qa)=1a(ax+p)(ax+q)

Resolve into factors (i) 6x2+7x3 (ii) 15x226x+8. (Aggarwal & Aggarwal, Mathematics Class IX, p. 35 Ex.2 1, 2)

(i) We need to find p and q such that (p+q)=7 and pq = -18. Clearly, one of them is negative, while the positive number has a larger absolute value. p = 9 and q = -2 satisfy the conditions. Thus,

6x2+7x3=6x2+9x2x3=3x(2x+3)1(2x+3)=(3x1)(2x+3)

(ii) We need to find p and q such that (p+q)=26 and pq = 120. Clearly, both of them are negative. p = -20 and q = -6 satisfy the conditions. Thus,

15x226x+8=15x220x6x+8 =5x(3x4)2(3x4) =(5x2)(3x4)

The difference of two squares

As we have seen in the monograph on Expansions for the simplification of (a+b)(ab):

(a+b)(ab)=a2ab+bab2=a2b2

Thus,

a2b2=(a+b)(ab)     .......[1]

In words, the difference of the squares of two numbers is equal to the product of the sum and the difference of the two numbers.

Resolve into factors (i) 25x216y2 (ii) 149c6. (Hall & Knight, Elementary Algebra, p. 133 Example 1, 2)

(i) 25x216y2=(5x)2(4y)2=(5x+4y)(5x4y)

(ii) 149c6=(1)2(7c3)2=(1+7c3)(17c3)

Resolve into factors (i) x2+xy2y4 (ii) (1a2)(1b2)+4ab. ICSE

(i) x2+xy2y4=(x24)+xy2y(Rearranging the terms)=(x+2)(x2)+y(x2)=(x+y+2)(x2)

(ii) (1a2)(1b2)+4ab=1b2a2+a2b2+4ab=1(a2+b2)+a2b2+2ab+2ab=1+2ab+a2b2(a2+b22ab)=(1+ab)2(ab)2=((1+ab)+(ab))((1+ab)(ab))=(1+ab+ab)(1a+b+ab)

The sum and difference of two cubes

Dividing a3+b3 by (a+b) gives the quotient a2ab+b2, while dividing a3b3 by (a-b) gives the quotient a2+ab+b2. This leads us to the following identities:

a3+b3=(a+b)(a2ab+b2)     .......[2]

a3b3=(ab)(a2+ab+b2)     .......[3]

Hence, any expression which can be written as a sum or a difference of two cubes can be resolved into factors as above.

Resolve 64a3+1 into factors. (Hall & Knight, Elementary Algebra, p. 136 Example 2)

64a3+1=(4a)3+(1)3=(4a+1)((4a)24a1+(1)2)=(4a+1)(16a24a+1)

Feedback

List of References

Aggarwal, RS & Aggarwal, V, Secondary School Mathematics for Class IX, Patna, India: Bharati Bhawan, 1999.
Hall, HS & Knight, SR, Elementary Algebra for Schools, metric edn, Agra, India: AK Publications, 1966.

Bibliography

Aggarwal, RS & Aggarwal, V, Secondary School Mathematics for Class IX, Patna, India: Bharati Bhawan, 1999.
Gupta, SD & Banerjee, A, ICSE Mathematics for Class 9, Patna, India: Bharati Bhawan, 2003.
Hall, HS & Knight, SR, Elementary Algebra for Schools, metric edn, Agra, India: AK Publications, 1966.
Schultze, A, Elements of Algebra, NY, USA: The Macmillan Company, 1910.