Factorisation of Algebraic Expressions: Definition, Types, Examples

Factorisation of algebraic expressions is the process of finding two or more expressions whose product is the given expression. A factor is a number that divides a given integer entirely without leaving any remainder. We write algebraic expressions as a product of their factors in algebra. The only distinction is that an algebraic expression contains numbers and variables and an arithmetic operation like addition or subtraction.

This article will learn about techniques we use to find an algebraic expression factor. Also, we have mentioned standard algebraic identities, which are used to find the factors of an algebraic expression. Read on to find more

Factorisation of Algebraic Expressions Definition

Writing a given algebraic expression as the product of two or more factors is called factorisation.

Factors

If an algebraic expression is written as the product of algebraic expressions, then each of these expressions is called the factors of the given algebraic expression.

Standard Algebraic Identities

The standard algebraic identities play an important role in finding the factors of a given algebraic expression. The important algebraic identities are listed below.
(i) \({(a + b)^2} = {a^2} + {b^2} + 2ab\)
(ii) \({(a – b)^2} = {a^2} + {b^2} – 2ab\)
(iii) \((a + b)(a – b) = {a^2} – {b^2}\)
(iv) \({(a + b)^3} = {a^3} + {b^3} + 3ab(a + b)\)
(v) \({(a – b)^3} = {a^3} – {b^3} – 3ab(a – b)\)
(vi) \({a^3} + {b^3} = {(a + b)^3} – 3ab(a + b) = (a + b)\left( {{a^2} – ab + {b^2}} \right)\)
(vii) \({a^3} – {b^3} = {(a – b)^3} + 3ab(a – b) = (a – b)\left( {{a^2} + ab + {b^2}} \right)\)
(viii) \({(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca\)

Types

There are several types to find an algebraic expression factor. They are,

1. Factorisation by taking out the common factors
2. Factorisation by grouping the terms.
3. Factorisation of Quadratic Polynomials by Splitting the Middle term
4. Factorisation using standard algebraic identities.
5. Factorisation by making a perfect square

Factorisation by Taking Out the Common Factors

To factorise an algebraic expression consisting of common factors, we use the following steps:

Check the given algebraic expression

1. Find the highest common factor (GCF/HCF) of its terms.
2. Express each term in the given expression as the product of the GCF and the quotient when it is divided by the GCF.
3. Use the distributive property of multiplication to express the given algebraic expression as the product of the GCF and the quotient of the given expression by the GCF.

Example: Factorise the expression \(2x+8\)

Solution: The GCF of the two terms \(2x\) and \(8\) is \(2\). 
Also, \(2x=2×x\) and \(8=2×4\).
So, \(2x + 8 = 2\left( {x + 4} \right)\)

Factorisation by Grouping the Terms

To find the factors of an algebraic expression by grouping the terms, we need to observe the terms of the expression; if we do not find the common terms, then we need to group the like terms in the given expression and start factorising.

Example: Factorise the expression \(15xy-6x+10y-4\)

Solution: There are four terms in the given algebraic expression. We observe that there are no common factors among all terms. So, let us group the like terms in pairs so that there are some factors common to the terms in each pair.
In the given expression \(15xy-6x+10y-4\), the first two terms have \(3x\) as a common factor. Taking \(3x\) common from the first two terms, we get
\(3x\left( {5y – 2} \right)\)

Also, in the next two terms, we can notice \(2\) as the common factor. So, taking \(2\) as a common factor from the next two terms, we get
\(10y – 4 = 2\left( {5y – 2} \right)\)
So, \(15xy-6x+10y-4=3x(5y-2)+2(5y-2)\)
Again we can observe that, \((5y-2\)) is a common factor. So taking \((5y-2\)) as a common factor, we get
\((5y – 2)(3x + 2)\)

Therefore, factors of \(15xy – 6x + 10y – 4\) are \((5y – 2)(3x + 2)\)

Factorisation of Quadratic Polynomials by Splitting the Middle Term

The algebraic expression is of the form \(a{x^2} + bx + c\), where \(a \ne 0\) is called a quadratic polynomial. To factorise the quadratic polynomial of the type \({x^2} + bx + c\), we have to find the numbers say \(p\) and \(q\), such that \(p + q = b\) and \(pq = c\) i.e. two numbers whose sum is equal to the coefficient of \(x\) and product is equal to the constant term. After finding \(p\) and \(q\), we split the middle term in the quadratic as \(px + qx\) and get the factors by grouping the terms.

Example: Factorise \({x^2} + 6x + 8\)

Solution: To factorise \({x^2} + 6x + 8\), we find two numbers \(p\) and \(q\), such that \(p + q = 6\) and \(pq=8\). We know, \(2 + 4 = 6\) and \(2 \times 4 = 8\)
So, splitting the middle term \(6x\) as \(2x+4x\), we get
\({x^2} + 6x + 8 = {x^2} + 2x + 4x + 8\)
\( = x(x + 2) + 4(x + 2)\)
\( = (x + 2)(x + 4)\)

So, factors of \({x^2} + 6x + 8\) are \((x + 2)(x + 4)\)

Factorisation Using Standard Algebraic Identities

Factorisation using standard algebraic identities is one of the most common method. In this method, we use some of the standard identities of algebra to find the factors of given algebraic expressions. The most used algebraic identities are discussed before.

Example: Factorise \(4{p^2} – 9{q^2}\)

Solution: Given expression \(4{p^2} – 9{q^2}\) can be written as \({(2p)^2} – {(3q)^2}\), this is of the form \({a^2} – {b^2}\) which is equal to \((a + b)(a – b)\)
So, \({(2p)^2} – {(3q)^2} = (2p + 3q)(2p – 3q)\)

Therefore, factors of \(4{p^2} – 9{q^2},(2p + 3q)(2p – 3q)\).

Factorisation by Making a Perfect Square

The ability to recognise exceptional polynomials cases that we can easily factor in is essential for solving any algebraic expressions involving polynomials.

One of these effortless to factor polynomials is the perfect square trinomial. We recall that a trinomial is an algebraic expression of three terms attached by addition or subtraction. Similarly, a binomial is an expression made of two times. Therefore, a perfect square trinomial can be explained as an expression that is obtained by squaring a binomial

The following are the steps on how to recognise a perfect square trinomial:

1. Identify whether the first and last terms of the trinomial are perfect squares.
2. Multiply the roots of the first and third terms altogether.
3. Compare to the middle terms with the result as in the above step.
4. If the first term and last term are perfect squares, and the mid term’s coefficient is twice the product of the square roots of the first and last terms, then the expression is a perfect square trinomial.
5. Most commonly used algebraic identities here are
   \({(a + b)^2} = {a^2} + {b^2} + 2ab\)
   \({(a – b)^2} = {a^2} + {b^2} – 2ab\)

Example: Factorise \({x^2} + 8x + 16\)

Solution: The given expression \({x^2} + 8x + 16\) can be written as \({x^2} + 2 \times x \times 4 + {4^2}\), which is of the form \({a^2} + {b^2} + 2ab\), which is equal to \({(a + b)^2}\)
So, \({x^2} + 2 \times x \times 4 + {4^2} = {(x + 4)^2}\)

Therefore, factors of \({x^2} + 8x + 16\) are \((x + 4)(x + 4)\).

Solved Examples

Q.1. Factorise \(12{a^2}b + 15a{b^2}\)
Ans: We know that, \(12{a^2}b = 2 \times 2 \times 3 \times a \times a \times b\)
\(15a{b^2} = 3 \times 5 \times a \times b \times b\)
The two terms have \(3\), \(a\) and \(b\). 
Therefore, \(12{a^2}b + 15a{b^2} = (2 \times 2 \times 3 \times a \times a \times b) + (3 \times 5 \times a \times b \times b)\)
\( = 3 \times a \times b \times [(2 \times 2 \times a) + (5 \times b)]\)
\( = 3ab \times (4a + 5b)\)
Therefore, factors of \(12{a^2}b + 15a{b^2}\) are \(3ab(4a + 5b)\)

Q.2. Factorise \(6xy – 4y + 6 – 9x\)
Ans: Given \(6xy – 4y + 6 – 9x\)
Taking \(2y\) as common in the first two terms and \(3\) as common in the next two terms, we get
\(2y(3x – 2) + 3(2 – 3x) \Rightarrow 2y(3x – 2) – 3(3x – 2)\)
\( = (2y – 3)(3x – 2)\)
Therefore, factors of \(6xy – 4y + 6 – 9x\) are \((2y – 3)(3x – 2)\)

Q.3 Factorise \({a^2} – 2ab + {b^2} – {c^2}\)
Ans: Given \({a^2} – 2ab + {b^2} – {c^2}\)
We know that, \({a^2} – 2ab + {b^2} = {(a – b)^2}\)
So, \({a^2} – 2ab + {b^2} – {c^2} = {(a – b)^2} – {c^2}\)
Also, we know that \((a + b)(a – b) = {a^2} – {b^2}\)
So, \({(a – b)^2} – {c^2} = (a – b + c)(a – b – c)\)
Therefore, factors of \({a^2} – 2ab + {b^2} – {c^2}\) are \((a – b + c)(a – b – c)\)

Q.4. Factorise \(3\,{m^2} + 9\,m + 6\)
Ans: Given \(3\,{m^2} + 9\,m + 6\)
Here, \(3\) is a common factor in all three terms. So,
\(3\,{m^2} + 9\,m + 6 = 3\,\left( {{m^2} + 3\,m + 2} \right)\)
Consider \({{m^2} + 3\,m + 2}\)
We know that, \(2 + 1 = 3\) and \(2 \times 1 = 2\)
So, \(3\,m = 2\,m + m\)
Therefore, by splitting the middle term of \({m^2} + 3\,m + 2\) we get
\({m^2} + 3\,m + 2 = {m^2} + m + 2\,m + 2\)
\( = m\left( {m + 1} \right) + 2\left( {m + 1} \right)\)
\( = \left( {m + 2} \right)\left( {m + 1} \right)\)
Therefore, \(3\left( {{m^2} + 3\,m + 2} \right) = 3\left( {m + 2} \right)\left( {m + 1} \right)\)
So, factors of \(3\,{m^2} + 9\,m + 6\) are \(3\left( {m + 2} \right)\left( {m + 1} \right)\)

Q.5. Factorise \(4{a^2} + 12ab + 9{b^2} – 8a – 12b\)
Ans: We have, \(4{a^2} + 12ab + 9{b^2} – 8a – 12b\)
\(= {(2a)^2} + 2 \times 2a \times 3b + {(3b)^2} – 4(2a + 3b)\)
\( = {(2a + 3b)^2} – 4(2a + 3b)\)
\( = (2a + 3b)(2a + 3b – 4)\)
Therefore, factors of \(4{a^2} + 12ab + 9{b^2} – 8a – 12b\) are \((2a + 3b)(2a + 3b – 4)\)

Q.6. Factorise \({x^2} + 4x – 21\)
Ans: Given \({x^2} + 4x – 21\)
We know that, \(7 + ( – 3) = 4\) and \(7 \times – 3 = – 21\)
Therefore, by splitting the middle term of \({x^2} + 4x – 21\), we get
\({x^2} + 4x – 21 = {x^2} + 7x – 3x – 21\)
\( = x(x + 7) – 3(x + 7)\)
\( = (x – 3)(x + 7)\)
Therefore, factors of \({x^2} + 4x – 21\) are \((x – 3)(x + 7)\)

Factorisation of algebraic expressions: Summary

In this article, we have learned the factorisation of an algebraic expression. Different methods include finding common factors, regrouping the terms, splitting the middle term, perfect square, and factorising algebraic identities. Also, we have solved some example problems on the factorisation of an algebraic expression.

FAQs

Q.1. Explain factorisation of algebraic expressions with example?
Ans: Writing a given algebraic expression as the product of two or more factors is called factorisation. If an algebraic expression is written as the product of algebraic expressions, then each of these expressions is called the factors of the given algebraic expression.
Example: Factors of \(a{x^2} + bx\) is \(x(ax + b)\)

Q.2. How do you factorise algebraic expressions?
Ans: We have several methods to find an algebraic expression factor or factorise the algebraic expressions. They are:
a. Factorisation by taking out the common factors
b. Factorisation by grouping the terms.
c. Factorisation of Quadratic Polynomials by Splitting the Middle term
d. Factorisation using standard algebraic identities.
e. Factorisation by making a perfect square
f. We use any of the methods based on the given algebraic expression.

Q.3. What does it mean to factorise an algebraic expression?
Ans: The meaning of factorisation of an algebraic expression is to find the factors of the given algebraic expression. In other words, the method to find the factors or write the factors of an algebraic expression is known as factorisation of algebraic expression.

Q.4. What is the formula for factorisation?
Ans: There is no particular formula for factorisation of the given algebraic expression. We use different methods to find the factors of the given algebraic expression.

Q.5. What is the formula for \({a^3} + {b^3}\)?
Ans: \({a^3} + {b^3} = {(a + b)^3} – 3ab(a + b) = (a + b)\left( {{a^2} – ab + {b^2}} \right)\)

Q.6. What is the formula of \({a^3} – {b^3}\)?
Ans: \({a^3} – {b^3} = {(a – b)^3} + 3ab(a – b) = (a – b)\left( {{a^2} + ab + {b^2}} \right)\)

Q.7. What is the formula of \({a^2} + {b^2}\)?
Ans: \({a^2} + {b^2} = {(a + b)^2} – 2ab = {(a – b)^2} + 2ab\)

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