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Factorization of algebraic expressions refers to the process of determining the factors of a given expression, which involves finding two or more expressions whose product is the given expression.
- A factor is a number that divides another number without leaving any remainder.
- Simply put, it implies representing a number as the product of two other numbers.
- Similarly, in Algebra, algebraic formulas are written as a product of their factors.
- The only difference is that an algebraic expression includes numbers and variables with an arithmetic operation such as addition or subtraction.
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Key Terms: Algebraic factorization, Factors, Algebraic expression, Product of two numbers, Highest common factor, Multiplication
Factorization of Algebraic Expressions
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Factorization of an algebraic expression refers to the method of identifying two or more expressions whose product is the given expression.
- Hence, factorization is the multiplication of algebraic statements in reverse.
- For example, the factors of 10 are 1, 2, 5, and 10.
- In the same way, we are able to factorize an algebraic expression.
- When the factors are multiplied they end in the first number or an expression that's factorized.
- For example, consider the expression (2x2 + 8x).
- It can be factorized as 2x(x + 4).
- When we multiply (2x) and (x + 4), we get the original expression (2x2 + 8x).
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Methods of Factorization of Algebraic Expressions
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An algebraic expression is composed of terms, as we all know.
- The expression 5xy, for example, can be factorized as 5 × x × y.
- There is no way to factorize this expression anymore.
Let’s look at the methods for factoring algebraic expressions now.
Factorization of Algebraic Expressions By Grouping of Terms
All of the terms in some algebraic formulas may not have the same factor.
Take the algebraic statement 15x + y - xy - 15 as an example.
Step 1:
Look for terms that have a lot of the same things in them.
- The first and last terms are the only ones with a common factor of 15.
- A common factor y exists in both the second and third terms.
Step 2:
The terms can now be regrouped as 15x + y - xy - 15 = 15x - 15 + y - xy
Step 3:
Eliminate common variables.
15x - 15 - xy - y = 15(x - 1) - y(x -1)
(x - 1) is clearly a common factor
Step 4:
As a result, the factorization of the given formula 15x - 15 - xy - y is (x -1) (15 - y)
We can factorize an algebraic expression by regrouping the terms in that algebraic statement.
Factorization of Algebraic Expressions Using the Common Factor Method
Follow the steps to find the factors of the expression: x2 + 4x
Step 1:
x2 can be broken down into x × x as well as 4x can be broken down into 4 × x.
Step 2:
Determine which of the two terms has the highest common factor.
- We can observe in this example that x is the most prevalent factor.
- This factor is kept outside the brackets, the polynomial terms are divided by this factor, and the remaining expression is written inside the brackets.
Step 3:
The expression is factored as follows: x(x + 4)
Factorization of algebraic expressions using the common factor method
Factorization of Algebraic Expressions Using Identities
For factorization, this method uses algebraic expression formulae.
Example: x2 + 6x + 9
We can see that the three terms in the equation have no common components. However, we can see in the expression that 9 is a square that is perfectly square.
In this example, we use algebraic identities to easily factorize the expression.
This expression appears to be the same as the identity:
(a + b)2 = a2 + 2ab + b2
When the provided expression is compared to the identity, we get
a = x, b = 3
As a result, the elements are (x+3)2 or (x + 3)(x + 3)
List of Identities to Factorize Algebraic Expressions
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The formulae for algebraic expressions are listed below.
- (a + b)2= a2 + 2ab + b2
- (a – b)2= a2 - 2ab + b2
- a2 – b2= (a + b)(a – b)
- a3 + b3= (a + b)(a2 - ab + b2)
- a3 – b3 = (a – b)(a2 + ab + b2)
- (a + b)3 = a3 + 3a2b + 3ab2 + b3
- (a – b)3= a3 – 3a2b + 3ab2 – b3
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| Other Maths Concepts | ||
|---|---|---|
| Degree of polynomial | Difference of Squares Formula | Factorial Formula |
| Integrals | Remainder Theorem | Like and Unlike Algebraic Terms |
Things to Remember
- An algebraic expression is factored when it is written as a product of factors.
- To see if the factors are valid, multiply them and see if the original algebraic equation appears.
- The common factor approach, regrouping like terms together, and algebraic identities can all be used to factorize algebraic expressions.
- It's the same as finding the factors of a whole number to discover the factors of a monomial.
- Check if any algebraic expressions are in the form of identity while factoring them.
- In this scenario, we can use the identity formula to obtain the factors.
Sample Questions
Ques. How to factorize algebraic expressions? (2 Marks)
Ans. Many methods are available for factorizing algebraic expressions. The following are the most common methods for factoring algebraic expressions:
- Factorization by regrouping terms.
- Factorization using common factors.
- Factorization using identities.
Ques. How do you factorize algebraic expressions using the common factor method? (2 Marks)
Ans. We take the highest common factor and insert it before the brackets to factorize using the common factor method. The expression inside the brackets is then obtained by dividing the polynomial terms by the highest common factor.
Ques. Factorize the following polynomials: (4 Marks)
i) ax-ay+bx-by
ii) 4x2-10xy-6xy+15yz
Ans. i) ax-ay+bx-by = (ax-ay)+(bx-by)
a(x-y)+b(x-y)
= (x-y)(a+b)
ii) 4x2-10xy-6xy+15yz = (4x2-10xy)-(6xy+15yz)
2x(2x-5y)-3z(2x-5y)
= (2x-5y)(2x-3z)
Ques. Factorize the following expression: (4 Marks)
i) xy-pq+qy-px
ii) a2+bc+ab+ca
Ans. i) To get something in common to be able to solve. We tend to interchange -pq and -px
Therefore, we write it as
xy-pq+qy-px = (xy-px)+(qy-pq)
x(y-p)+q(y-p) = (y-p)(x+q)
ii) a2+bc+ab+ca = a2+ab+bc+ca
a(a+b)+c(b+a) = a(a+b)+c(a+b)
= (a+b)(a+c)
Ques. Factorize 25a2- 64b2 (3 Marks)
Ans. 25a2-64b2
= (5a)2-(8b)2
= (5a+8b)(5a-8b)
Ques. Factorize using H.C.F. method: 6x2y2 and 8xy3 (3 Marks)
Ans. 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 is =xy2
Therefore, H.C.F. of 6x2y2 and 8xy3 = 2 × xy2 = 2xy2
Ques. Factorize the following polynomials: (4 Marks)
i) 24x3- 32x2
ii) 15ab2- 21a2b
Ans. Factorizing algebraic expressions of the following:
i) H.C.F. of 24x3 and 32x2 is 8x2
24x3-32x2 = 8x2(3x-4)
ii) H.C.F. of 15ab2 and 21a2b is 3ab
15ab2-21a2b = 3ab(5b-7a)
Ques. Factorize the following: (4 Marks)
i) 3x(y+2z)+5a(y+2z)
ii) 10(p - 2q)3+ 6(p - 2q)2- 20(p - 2q)
Ans. i) H.C.F. of the expressions
3x(y+2z) and 5a(y+2z) is y+2z
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)
Therefore, 10(p-2q)3+6(p-2q)2- 20(p-2q) = 2(p-2q){5(p−2q)2+3(p−2q)−10}
Ques. Factorize x2 − 10x + 25. (3 Marks)
Ans. The algebraic expression given has no common terms for all three terms.
However, there is an identity that we know of, that matches the formula of the given algebraic expression,
(a-b)2= a2- 2ab+b2
So, on Comparing the identity to this equation we get, a=x, b=5.
Therefore, the factors are (x−5)2 or (x−5)(x−5).
Therefore, the factors of the algebraic expression x2−10x+25 are (x−5)2 or (x−5)(x−5)
Ques. Factorize 5z3 − 10z2. (3 Marks)
Ans. The first term of the given expression 5z3 can be factorized as 5×z×z×z and the second term of it that is, 10z2 can be factorized as 10×z×z
The common factor in both terms is 5z2
Taking out the common factor,
we get 5z2(z−2) as factors.
Therefore, the factors of 5z3−10z2 are 5z2 is (z−2)
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