Factorisation in Maths: Identities and Solved Examples

A complete guide to factorisation: the HCF of monomials, fifteen identities from common factors to the three-cube identity, worked examples of each, splitting the middle term, and dividing algebraic expressions.

Factorisation means writing an algebraic expression as a product of simpler expressions, called its factors. It is the reverse of expanding brackets: expanding turns $3ab(4a + 5b)$ into $12a^2b + 15ab^2$ , and factorising turns it back. You will use it to simplify fractions, divide polynomials, solve quadratic equations in Class 10 and cancel terms in almost every later chapter. This lesson starts with the highest common factor of monomials, lists the fifteen identities and methods used for factorising, works through examples of each, and ends with division of algebraic expressions.

Highest common factor (HCF) of monomials

Definition

The HCF of two or more monomials is the common factor with the greatest numerical coefficient and the highest power of each variable that divides all of them. In practice this means: the HCF of the coefficients, multiplied by every variable common to all the monomials, each raised to its smallest power.

Method

Write each monomial as a product of prime numbers and variables.
Pick out the prime factors and variables that appear in every monomial.
For each one, take the lowest power that appears.
Multiply these together.

Worked examples

Example 1. Find the HCF of $6x^3y$ and $18x^2y^3$ .

$\begin{aligned} 6x^3y &= 2 \times 3 \times x^3 \times y \\ 18x^2y^3 &= 2 \times 3^2 \times x^2 \times y^3 \\ \text{HCF} &= 2 \times 3 \times x^2 \times y = 6x^2y \end{aligned}$

Example 2. Find the HCF of $5xy$ and $10x$ .

$\begin{aligned} 5xy &= 5 \times x \times y \\ 10x &= 2 \times 5 \times x \\ \text{HCF} &= 5 \times x = 5x \end{aligned}$

The variable $y$ is not in $10x$ , so it cannot be part of the HCF.

Example 3. The HCF of $12a^2b$ and $15ab^2$ is $3ab$ , because $12a^2b = 2^2 \times 3 \times a^2 \times b$ and $15ab^2 = 3 \times 5 \times a \times b^2$ : the only common prime is 3, and the lowest powers of $a$ and $b$ are $a^1$ and $b^1$ .

Example 4. The HCF of $2x$ and $4$ is $2$ . The number 4 has no $x$ , so only the numbers share a factor.

Example 5. The HCF of $12x$ and $36$ is $12$ , since $12 = 2^2 \times 3$ divides $36 = 2^2 \times 3^2$ .

Example 6. The HCF of $14pq$ and $35pqr$ is $7pq$ : $\text{HCF}(14, 35) = 7$ , and $p$ and $q$ are common, while $r$ is not.

What factorisation is, and the fifteen identities

Factorisation is the process of expressing an algebraic expression as a product of two or more factors. The methods and identities below are the tools for doing it. Each one is simply an expansion read from right to left.

Basic methods (common factor and grouping)

$ab + ac = a(b + c)$
$ab - ac = a(b - c)$
$ac + ad + bc + bd = (a + b)(c + d)$

Square-based identities

$a^2 - b^2 = (a + b)(a - b)$
$a^2 + 2ab + b^2 = (a + b)^2$
$a^2 - 2ab + b^2 = (a - b)^2$
$x^2 + (a + b)x + ab = (x + a)(x + b)$
$a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2$
$a^2 + b^2 + c^2 + 2ab - 2bc - 2ca = (a + b - c)^2$

Cubic and advanced identities

$a^3 + 3a^2b + 3ab^2 + b^3 = (a + b)^3$
$a^3 - 3a^2b + 3ab^2 - b^3 = (a - b)^3$
$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
If $a + b + c = 0$ , then $a^3 + b^3 + c^3 = 3abc$ .

A general order of attack

Take out the HCF of all the terms first.
Count the terms. Two terms: look for a difference of squares or a sum or difference of cubes. Three terms: look for a perfect square or split the middle term. Four terms: try grouping, or a perfect cube.
Check each factor you get: can it be factorised again?
Expand your answer mentally to confirm it gives the original expression.

Basic methods in factorisation

1. Common factor (sum): $ab + ac = a(b + c)$

Example 1. Factorise $2x + 4$ .

$2x + 4 = 2 \times x + 2 \times 2 = 2(x + 2)$

Example 2. Factorise $5xy + 10x$ .

$5xy + 10x = 5x \times y + 5x \times 2 = 5x(y + 2)$

Example 3. Factorise $12a^2b + 15ab^2$ .

$\begin{aligned} 12a^2b + 15ab^2 &= 3 \times 4 \times a^2 \times b + 3 \times 5 \times a \times b^2 \\ &= 3ab(4a + 5b) \end{aligned}$

2. Common factor (difference): $ab - ac = a(b - c)$

Example 1. Factorise $22y - 33z$ .

$\begin{aligned} 22y - 33z &= 11 \times 2 \times y - 11 \times 3 \times z \\ &= 11(2y - 3z) \end{aligned}$

Example 2. Factorise $10x^2 - 18x^3 + 14x^4$ .

The HCF of 10, 18 and 14 is 2, and the lowest power of $x$ is $x^2$ , so the HCF of the terms is $2x^2$ .

$10x^2 - 18x^3 + 14x^4 = 2x^2(5 - 9x + 7x^2)$

3. Grouping: $ac + ad + bc + bd = (a + b)(c + d)$

Example 1. Factorise $2xy + 2y + 3x + 3$ .

$\begin{aligned} 2xy + 2y + 3x + 3 &= 2y(x + 1) + 3(x + 1) \\ &= (2y + 3)(x + 1) \end{aligned}$

Example 2. Factorise $6xy - 4y + 6 - 9x$ .

Rearrange so that each pair shares a factor, then take out $2y$ from the first pair and $-3$ from the second.

$\begin{aligned} 6xy - 4y + 6 - 9x &= 6xy - 4y - 9x + 6 \\ &= 2y(3x - 2) - 3(3x - 2) \\ &= (2y - 3)(3x - 2) \end{aligned}$

Square-based identities in factorisation

4. Difference of two squares: $a^2 - b^2 = (a + b)(a - b)$

Left: square of side a plus b split into a squared, two ab rectangles and b squared. Right: a b by b corner cut from an a by a square, re-laid as an a plus b by a minus b rectangle. — Area models: $(a+b)^2$ splits into $a^2 + 2ab + b^2$ , and $a^2 - b^2$ rearranges into a rectangle $(a+b)$ by $(a-b)$ .

Example 1.

$49y^2 - 36 = (7y)^2 - 6^2 = (7y + 6)(7y - 6)$

Example 2. Here the identity is used twice, because $p^2 - 9$ is itself a difference of squares.

$p^4 - 81 = (p^2)^2 - 9^2 = (p^2 + 9)(p^2 - 9) = (p^2 + 9)(p + 3)(p - 3)$

Example 3 (mixed). Factorise $a^2 - 2ab + b^2 - c^2$ . The first three terms form a perfect square.

$\begin{aligned} a^2 - 2ab + b^2 - c^2 &= (a - b)^2 - c^2 \\ &= [(a - b) + c][(a - b) - c] \\ &= (a - b + c)(a - b - c) \end{aligned}$

5. and 6. Perfect square trinomials

These come from squaring a binomial sum or difference: $a^2 + 2ab + b^2 = (a + b)^2$ and $a^2 - 2ab + b^2 = (a - b)^2$ . To spot one, check that the first and last terms are perfect squares and the middle term is twice the product of their square roots.

Example 1. $x^2 + 8x + 16 = x^2 + 2(x)(4) + 4^2 = (x + 4)^2$

Example 2. $25m^2 + 30m + 9 = (5m)^2 + 2(5m)(3) + 3^2 = (5m + 3)^2$

Example 3. $4y^2 - 12y + 9 = (2y)^2 - 2(2y)(3) + 3^2 = (2y - 3)^2$

7. Splitting the middle term: $x^2 + (a + b)x + ab = (x + a)(x + b)$

Method.

For $x^2 + px + q$ , find two numbers whose sum is $p$ and whose product is $q$ .
For $ax^2 + bx + c$ with $a \neq 1$ , find two numbers whose sum is $b$ and whose product is $ac$ .
Split the middle term using those numbers and factorise by grouping.

Example 1. $x^2 + 5x + 6$ : the numbers 3 and 2 add to 5 and multiply to 6, so $x^2 + 5x + 6 = (x + 3)(x + 2)$ .

Example 2. $y^2 - 7y + 12$ : the numbers $-4$ and $-3$ add to $-7$ and multiply to 12, so $y^2 - 7y + 12 = (y - 4)(y - 3)$ .

Example 3. $3x^2 + 10x + 3$ : here $ac = 9$ , and $9 + 1 = 10$ .

$\begin{aligned} 3x^2 + 10x + 3 &= 3x^2 + 9x + x + 3 \\ &= 3x(x + 3) + 1(x + 3) \\ &= (x + 3)(3x + 1) \end{aligned}$

Example 4. $6x^2 + 5x - 6$ : here $ac = -36$ , and $9 + (-4) = 5$ with $9 \times (-4) = -36$ .

$\begin{aligned} 6x^2 + 5x - 6 &= 6x^2 + 9x - 4x - 6 \\ &= 3x(2x + 3) - 2(2x + 3) \\ &= (2x + 3)(3x - 2) \end{aligned}$

Two by two grid with rows 2x and +3 and columns 3x and -2; cells read 6x squared, -4x, +9x and -6, showing 6x squared + 5x - 6 = (2x + 3)(3x - 2). — The box method for $6x^2 + 5x - 6$ : the two middle cells, $9x$ and $-4x$ , are the split of $5x$ .

Example 5 (substitution). Factorise $(2a - b)^2 + 2(2a - b) - 8$ . Put $t = 2a - b$ ; then $t^2 + 2t - 8 = (t + 4)(t - 2)$ , since $4 \times (-2) = -8$ and $4 + (-2) = 2$ .

$(2a - b)^2 + 2(2a - b) - 8 = (2a - b + 4)(2a - b - 2)$

8. Square of a trinomial: $a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2$

Example 1. $4a^2 + b^2 + c^2 + 4ab + 2bc + 4ac$ has square terms $(2a)^2$ , $b^2$ , $c^2$ and cross terms $2(2a)(b) = 4ab$ , $2(b)(c) = 2bc$ , $2(c)(2a) = 4ac$ .

$4a^2 + b^2 + c^2 + 4ab + 2bc + 4ac = (2a + b + c)^2$

Example 2. $9a^2 + 4b^2 + 16 + 12ab + 16b + 24a$ has square terms $(3a)^2$ , $(2b)^2$ , $4^2$ and cross terms $2(3a)(2b) = 12ab$ , $2(2b)(4) = 16b$ , $2(4)(3a) = 24a$ .

$9a^2 + 4b^2 + 16 + 12ab + 16b + 24a = (3a + 2b + 4)^2$

9. Square of a trinomial with two negative cross terms: $a^2 + b^2 + c^2 + 2ab - 2bc - 2ca = (a + b - c)^2$

Use this when the two cross terms that contain one particular variable are both negative. That variable takes the minus sign.

Example. Factorise $25x^2 + y^2 + 4z^2 - 10xy - 4yz + 20zx$ .

The negative cross terms $-10xy$ and $-4yz$ both contain $y$ , so $y$ plays the role of $c$ . Take $a = 5x$ , $b = 2z$ , $c = y$ : then $2ab = 20zx$ , $2bc = 4yz$ and $2ca = 10xy$ .

$\begin{aligned} 25x^2 + y^2 + 4z^2 - 10xy - 4yz + 20zx &= (5x)^2 + (2z)^2 + y^2 + 2(5x)(2z) - 2(2z)(y) - 2(y)(5x) \\ &= (5x + 2z - y)^2 = (5x - y + 2z)^2 \end{aligned}$

You get the same answer from identity 8 with the terms $5x$ , $-y$ and $2z$ .

Cubic and advanced identities in factorisation

10. Cube of a sum: $a^3 + 3a^2b + 3ab^2 + b^3 = (a + b)^3$

Example. Factorise $8a^3 + b^3 + 12a^2b + 6ab^2$ .

$\begin{aligned} 8a^3 + b^3 + 12a^2b + 6ab^2 &= (2a)^3 + b^3 + 3(2a)^2(b) + 3(2a)(b)^2 \\ &= (2a + b)^3 \end{aligned}$

11. Cube of a difference: $a^3 - 3a^2b + 3ab^2 - b^3 = (a - b)^3$

Example. Factorise $27 - 125a^3 - 135a + 225a^2$ .

$\begin{aligned} 27 - 125a^3 - 135a + 225a^2 &= 3^3 - (5a)^3 - 3(3)^2(5a) + 3(3)(5a)^2 \\ &= (3 - 5a)^3 \end{aligned}$

12. Sum of two cubes: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

Example. Factorise $y^3 + 125$ .

$\begin{aligned} y^3 + 125 &= y^3 + 5^3 \\ &= (y + 5)(y^2 - y \times 5 + 5^2) \\ &= (y + 5)(y^2 - 5y + 25) \end{aligned}$

13. Difference of two cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Example. Factorise $\displaystyle \frac{x^3}{216} - 8y^3$ .

$\displaystyle \begin{aligned} \frac{x^3}{216} - 8y^3 &= \left(\frac{x}{6}\right)^3 - (2y)^3 \\ &= \left(\frac{x}{6} - 2y\right)\left[\left(\frac{x}{6}\right)^2 + \frac{x}{6}(2y) + (2y)^2\right] \\ &= \left(\frac{x}{6} - 2y\right)\left(\frac{x^2}{36} + \frac{xy}{3} + 4y^2\right) \end{aligned}$

14. The three-cube identity

$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$

Example. Factorise $a^3 - b^3 + 1 + 3ab$ .

Write the terms as cubes of $a$ , $-b$ and $1$ . Then $-3(a)(-b)(1) = 3ab$ , which matches the last term.

$\begin{aligned} a^3 - b^3 + 1 + 3ab &= a^3 + (-b)^3 + 1^3 - 3(a)(-b)(1) \\ &= (a - b + 1)\left[a^2 + (-b)^2 + 1^2 - a(-b) - (-b)(1) - (1)(a)\right] \\ &= (a - b + 1)(a^2 + b^2 + 1 + ab + b - a) \end{aligned}$

15. The conditional cube property

If $a + b + c = 0$ , the first factor of identity 14 is zero, so $a^3 + b^3 + c^3 - 3abc = 0$ , that is, $a^3 + b^3 + c^3 = 3abc$ .

Example. Factorise $(x - y)^3 + (y - z)^3 + (z - x)^3$ .

Let $a = x - y$ , $b = y - z$ , $c = z - x$ . Then $a + b + c = x - y + y - z + z - x = 0$ .

$(x - y)^3 + (y - z)^3 + (z - x)^3 = 3abc = 3(x - y)(y - z)(z - x)$

Division of algebraic expressions

Dividing a monomial by a monomial

Factorise both, then cancel the common factors.

Example 1.

$\displaystyle 6x^3 \div 2x = \frac{2 \times 3 \times x \times x \times x}{2 \times x} = 3x^2$

Example 2.

$\displaystyle -20x^4 \div 10x^2 = \frac{-2 \times 10 \times x^2 \times x^2}{10 \times x^2} = -2x^2$

Example 3.

$\displaystyle 7x^2y^2z^2 \div 14xyz = \frac{7 \times xyz \times xyz}{2 \times 7 \times xyz} = \frac{xyz}{2}$

Dividing a polynomial by a monomial

Take the HCF out of the polynomial first, so that the divisor cancels in one step.

Example 1. Divide $24(x^2yz + xy^2z + xyz^2)$ by $8xyz$ .

$\displaystyle \begin{aligned} \frac{24(x^2yz + xy^2z + xyz^2)}{8xyz} &= \frac{24xyz(x + y + z)}{8xyz} \\ &= 3(x + y + z) \end{aligned}$

Example 2. Divide $3y^8 - 4y^6 + 5y^4$ by $y^4$ .

$\displaystyle \begin{aligned} \frac{3y^8 - 4y^6 + 5y^4}{y^4} &= \frac{y^4(3y^4 - 4y^2 + 5)}{y^4} \\ &= 3y^4 - 4y^2 + 5 \end{aligned}$

Common mistakes

Taking out a common factor from only one term: $22y - 33z$ is $11(2y - 3z)$ , not $11(2y - 33z)$ . Divide every term by the factor.
Stopping too early: $(p^2 + 9)(p^2 - 9)$ is not finished, because $p^2 - 9 = (p + 3)(p - 3)$ .
Trying to factorise $a^2 + b^2$ as $(a + b)(a + b)$ . A sum of two squares has no real factors of this kind; $(a + b)^2 = a^2 + 2ab + b^2$ .
Mixing up the signs in the cube identities: $a^3 + b^3$ has $-ab$ in the second factor, and $a^3 - b^3$ has $+ab$ .
Choosing numbers whose product is $c$ instead of $ac$ when splitting the middle term of $ax^2 + bx + c$ with $a \neq 1$ .
Including a variable in the HCF that is missing from one of the monomials, as in writing the HCF of $5xy$ and $10x$ as $5xy$ .

Try these

Find the HCF of $8a^3b^2$ and $20a^2b^4$ . Answer: $4a^2b^2$
Factorise $x^2 + 7x + 12$ . Answer: $(x + 3)(x + 4)$
Factorise $2x^2 - 7x + 3$ . Answer: $(2x - 1)(x - 3)$
Factorise $16a^2 - 81b^2$ . Answer: $(4a + 9b)(4a - 9b)$
Factorise $27x^3 - 8$ . Answer: $(3x - 2)(9x^2 + 6x + 4)$
Divide $15x^3y^2 - 10x^2y^3$ by $5x^2y^2$ . Answer: $3x - 2y$

Key terms

Factor: An expression that divides another expression exactly; in $3ab(4a + 5b)$ , both $3ab$ and $4a + 5b$ are factors.
Factorisation: Writing an expression as a product of its factors.
Monomial: An expression with a single term, such as $6x^3y$ .
Highest common factor (HCF): The largest expression that divides each of the given expressions exactly.
Identity: An equality that holds for every value of its variables, such as $a^2 - b^2 = (a + b)(a - b)$ .
Perfect square trinomial: A three-term expression that equals the square of a binomial, such as $x^2 + 8x + 16 = (x + 4)^2$ .
Splitting the middle term: Rewriting the middle term of a quadratic as two terms so that the expression can be factorised by grouping.
Grouping: Arranging terms in pairs that share a common factor, then taking out the common bracket.

Common questions

How do I know which identity to use?

Take out the HCF first, then count the terms. Two terms suggest a difference of squares or a sum or difference of cubes; three terms suggest a perfect square or splitting the middle term; four terms suggest grouping or a perfect cube; six terms with three squares suggest the square of a trinomial.

How can I check that a factorisation is right?

Multiply the factors back out. If the expansion gives the original expression exactly, the factorisation is correct. Also check that no factor can be broken down further.

Is the HCF of monomials found the same way as the HCF of numbers?

Yes. Find the HCF of the numerical coefficients as usual, and for the variables take only those common to every monomial, each with its lowest power.

Can every quadratic be factorised by splitting the middle term?

No. If no two integers have the required sum and product, the quadratic has no factors with integer coefficients. For example, $x^2 + x + 1$ cannot be split this way. In Class 10 you will learn the quadratic formula for such cases.

Why does $a^3 + b^3 + c^3 = 3abc$ when $a + b + c = 0$ ?

Identity 14 says $a^3 + b^3 + c^3 - 3abc$ equals $(a + b + c)$ times another factor. When $a + b + c = 0$ the product is zero, so $a^3 + b^3 + c^3 = 3abc$ .

References

National Council of Educational Research and Training. Mathematics: Textbook for Class X. NCERT, New Delhi.
Hall, H. S. and Knight, S. R. Higher Algebra. Macmillan.
Gelfand, I. M. and Shen, A. Algebra. Birkhäuser.

Menu

High­est com­mon fac­tor (HCF) of mono­mi­als#

Def­i­n­i­tion#

Method#

Worked exam­ples#

What fac­tori­sa­tion is, and the fif­teen iden­ti­ties#

Basic meth­ods (com­mon fac­tor and group­ing)#

Square-based iden­ti­ties#

Cubic and advanced iden­ti­ties#

A gen­eral order of attack#

Basic meth­ods in fac­tori­sa­tion#

1. Com­mon fac­tor (sum): ab+ac=a(b+c)ab + ac = a(b + c)ab+ac=a(b+c)#

2. Com­mon fac­tor (dif­fer­ence): ab−ac=a(b−c)ab - ac = a(b - c)ab−ac=a(b−c)#

3. Group­ing: ac+ad+bc+bd=(a+b)(c+d)ac + ad + bc + bd = (a + b)(c + d)ac+ad+bc+bd=(a+b)(c+d)#

Square-based iden­ti­ties in fac­tori­sa­tion#

4. Dif­fer­ence of two squares: a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b)a2−b2=(a+b)(a−b)#

5. and 6. Per­fect square tri­no­mi­als#

7. Split­ting the mid­dle term: x2+(a+b)x+ab=(x+a)(x+b)x^2 + (a + b)x + ab = (x + a)(x + b)x2+(a+b)x+ab=(x+a)(x+b)#

8. Square of a tri­no­mial: a2+b2+c2+2ab+2bc+2ca=(a+b+c)2a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2a2+b2+c2+2ab+2bc+2ca=(a+b+c)2#

9. Square of a tri­no­mial with two neg­a­tive cross terms: a2+b2+c2+2ab−2bc−2ca=(a+b−c)2a^2 + b^2 + c^2 + 2ab - 2bc - 2ca = (a + b - c)^2a2+b2+c2+2ab−2bc−2ca=(a+b−c)2#

Cubic and advanced iden­ti­ties in fac­tori­sa­tion#

10. Cube of a sum: a3+3a2b+3ab2+b3=(a+b)3a^3 + 3a^2b + 3ab^2 + b^3 = (a + b)^3a3+3a2b+3ab2+b3=(a+b)3#

11. Cube of a dif­fer­ence: a3−3a2b+3ab2−b3=(a−b)3a^3 - 3a^2b + 3ab^2 - b^3 = (a - b)^3a3−3a2b+3ab2−b3=(a−b)3#

12. Sum of two cubes: a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2 - ab + b^2)a3+b3=(a+b)(a2−ab+b2)#

13. Dif­fer­ence of two cubes: a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2)a3−b3=(a−b)(a2+ab+b2)#

14. The three-cube iden­tity#

15. The con­di­tional cube prop­erty#

Divi­sion of alge­braic expres­sions#

Divid­ing a mono­mial by a mono­mial#

Divid­ing a poly­no­mial by a mono­mial#

Com­mon mis­takes#

Try these#

Key terms#

Com­mon ques­tions#

How do I know which iden­tity to use?#

How can I check that a fac­tori­sa­tion is right?#

Is the HCF of mono­mi­als found the same way as the HCF of num­bers?#

Can every qua­dratic be fac­torised by split­ting the mid­dle term?#

Why does a3+b3+c3=3abca^3 + b^3 + c^3 = 3abca3+b3+c3=3abc when a+b+c=0a + b + c = 0a+b+c=0?#

Ref­er­ences#

Related read­ing#