how to solve factorization questions
Answers
Solve factorization by
1.taking out the common factor.
Eg:5a(b+c)-7b(b+c)
=(b+c)(5a-7b)
2.Sometimes in a expression it's not possible to take out a common factor directly.However the terms of the expression are grouped in such a manner that we may have a common factor.
Eg:ab+bc+ax+cx
=(a+b)+(ax+cx)
=b(a+c)+x(a+c)
=(a+c)(b+x)
Answer:
Simple Factorization :
1. Factorize 36x2y2 – 15xy
We see that the HCF of both the terms is found. The HCF of the terms 36x2y2 and 15xy is 3xy.
Each term of the given expression is multiplied and divided by the HCF.
3xy(36x2y23xy−15xy3xy)
= 3xy(12xy – 5)
2.Find the HCF of 15a3b2c3 and 12a4bc4.
The HCF of 15 and 12 = 3, common literal numbers are a, b and c. The lowest power of a = 3, b = 1 and c = 3.
Therefore, HCF of 15a3b2c3 and 12a4bc4 = 3a3bc3.
3. Find the HCF of 4a2bc, 14a3b and 2ac.
The HCF of 4, 14 and 2 = 2, common literal number is a. The lowest power of a = 1.
Therefore, HCF of 4a2bc, 14a3b and 2ac = 2a.
Remember:
(i) HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
(ii) HCF of literal coefficients = product of each common literal raised to the lowest power.
Some solved examples:
1. Factorize 8x3 - 32x5
The HCF of the terms 8x3 and 32x5 = 8x3 (common terms in both)
Therefore, 8x3 - 32x5 = 8x3(1 - 4x2).
2. Factorize 14m2n5 - 28mn2 + 7m5n
The HCF of the terms 14m2n5, 28mn2 and 7m5n = 7mn (common terms in all)
Therefore, 14m2n5 - 28mn2 + 7m5n = 7mn (2mn4 - 4n + m4).
3. Factorize 5a(b + 3c) - 5m(b + 3c)
The HCF of the terms 5a(b + 3c) and 5m(b + 3c) = 5(b + 3c)
Therefore, 5a(b + 3c) - 5m(b + 3c) = 5(b + 3c) (a - m).
Note:
Thus, factorization is the method of expressing an algebraic expression as a product of two or more expressions.