Factorization is nothing but the breaking down an entity (a number, a matrix, or a polynomial) into the product of another entity or factors. When you multiply the product of the factors, you will get the original number or matrix. Difficult algebraic or quadratic equations can reduce into small forms using the Factorization method. The factors of equations can be a variable, an integer, or an algebraic expression itself.
In simple words, Factorization can be explained as the reverse process of multiplication.
Examples:
(i) Product: 4x (3x – 6y) = 12x² – 24xy; Factorization: 12x² – 24xy = 4x (3x – 6y)
(ii) Product: (x + 4)(x – 3) = x² + x – 12; Factorization: x² + x – 12 = (x + 4)(x – 3
(iii) Product: (4a + 6b)(4a – 6b) = 16a2 – 36b2; Factorization: 16a2 – 36b2 = (4a + 6b)(4a – 6b)
Maths Factorization
Find out different Factorization concepts and concern links below. You can learn each individual concept with a clear explanation with the help of the below links. We have explained every topic details separately with the solved examples. Therefore, students can easily get a grip on Factorization concepts by referring to our detailed concepts.
- Factors of Algebraic Expressions
- Monomial is a Common Factor
- Factorization when Monomial is Common
- Binomial is a Common Factor
- Factorization when Binomial is Common
- Factorization by Grouping
- Factorize by Grouping The Terms
- Factoring Terms by Grouping
- Factorization by Regrouping
- Factorize by Regrouping The Terms
- Factoring Terms by Regrouping
- Factorization by Using Identities
- Factorization of Perfect Square
- Factorization of Perfect Square Trinomials
- Difference of Two Squares
- Factoring Differences of Squares
- Factorize the Difference of Two Squares
- Evaluate the Difference of Two Squares
- Factorize the Trinomial x^2 + px + q
- Factorize the Trinomial ax^2 + bx + c
- Factorization of Quadratic Trinomials
Simple Factorization
Simple Factorization can be easily understood by the below examples.
(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.
Factor
The factor is something that is to be multiplied.
Sum = term + term
Product = factor × factor
For example, x = m(n + 1); m and n + 1 are the factors.
Examples:
1. Factorize 4x²y² – 2xy
Solution:
Firstly, find the HCF of both given terms.
HCF of their numerical coefficients 4 and 2 is 2.
HCF of literal coefficients:
The lowest power of x is 1
The lowest power of y is 1
Therefore, the HCF of literal coefficients is xy.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 4x²y² – 2xy is 2xy.
Multiply and divide each term of the given expression 4x²y² – 2xy with 2xy
2xy((4x²y²/2xy) – (2xy/2xy)) = 2xy (2xy – 1)
The final answer is 2xy (2xy – 1)
2. Find the HCF of 24a3b2c3 and 27a4bc4.
Solution:
Firstly, find the HCF of both given terms.
HCF of their numerical coefficients 24 and 27 is 3.
HCF of literal coefficients:
The lowest power of a is 3.
The lowest power of b is 1.
And, the lowest power of c is 3.
Therefore, the HCF of literal coefficients is a³bc³.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 24a3b2c3 and 27a4bc4 is 3a³bc³.
The final answer is 3a³bc³
3. Find the HCF of 2a2bc, 4a3b and 14bc.
Solution:
Firstly, find the HCF of both given terms.
HCF of their numerical coefficients 2, 4, and 14 is 2.
HCF of literal coefficients:
The lowest power of a is 0.
The lowest power of b is 1.
And, the lowest power of c is 0.
Therefore, the HCF of literal coefficients is b.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 2a2bc, 4a3b and 14bc is 2b.
The final answer is 2b.
Factorization Solved Examples
1. Factorize 4y3 – 24y5
Solution:
Firstly, find the HCF of both given terms.
HCF of their numerical coefficients 4 and 24 is 4.
HCF of literal coefficients:
The lowest power of y is 3
Therefore, the HCF of literal coefficients is y³.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 4y3 – 24y5 is 4y³.
Multiply and divide each term of the given expression 4y3 – 24y5 with 4y³
4y³((4y3/4y³) – (24y5 /4y³) = 4y³(1 – 6y²)
The final answer is 4y³(1 – 6y²)
2. Factorize 21m2n5 – 7mn2 + 28m5n
Solution:
Firstly, find the HCF of both given terms.
HCF of their numerical coefficients 21, 7, and 28 is 7.
HCF of literal coefficients:
The lowest power of m is 1
The lowest power of n is 1
Therefore, the HCF of literal coefficients is mn.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 21m2n5 – 7mn2 + 28m5n is 7mn.
Multiply and divide each term of the given expression 21m2n5 – 7mn2 + 28m5n with 7mn
7mn((21m2n5 /7mn) – (7mn2 /7mn) + (28m5n/7mn)) = 7mn(3mn4 – n + 4m4)
The final answer is 7mn(3mn4 – n + 4m4)
3. Factorize 3m(d + 5e) – 3n(d + 5e)
Solution:
Firstly, find the HCF of both given terms.
HCF of their numerical coefficients 3 and 3 is 3.
HCF of literal coefficients:
The lowest power of m is 0
The lowest power of n is 0
Therefore, the HCF of literal coefficients is d+ 5e.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 3m(d + 5e) – 3n(d + 5e) is 3(d + 5e).
Multiply and divide each term of the given expression 3m(d + 5e) – 3n(d + 5e) with 3(d + 5e)
3(d + 5e)((3m(d + 5e)/3(d + 5e)) – (3n(d + 5e)/3(d + 5e))) = 3(d + 5e)(m – n)
The final answer is 3(d + 5e)(m – n)
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