Factorization in Maths: Methods, Identities and Solved Examples

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

1. Write each mono­mial as a prod­uct of prime num­bers and vari­ables.
2. Pick out the prime fac­tors and vari­ables that appear in every mono­mial.
3. For each one, take the low­est power that appears.
4. 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)

1. $ab + ac = a(b + c)$
2. $ab - ac = a(b - c)$
3. $ac + ad + bc + bd = (a + b)(c + d)$

Square-based iden­ti­ties

4. $a^2 - b^2 = (a + b)(a - b)$
5. $a^2 + 2ab + b^2 = (a + b)^2$
6. $a^2 - 2ab + b^2 = (a - b)^2$
7. $x^2 + (a + b)x + ab = (x + a)(x + b)$
8. $a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2$
9. $a^2 + b^2 + c^2 + 2ab - 2bc - 2ca = (a + b - c)^2$

Cubic and advanced iden­ti­ties

10. $a^3 + 3a^2b + 3ab^2 + b^3 = (a + b)^3$
11. $a^3 - 3a^2b + 3ab^2 - b^3 = (a - b)^3$
12. $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
13. $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
14. $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
15. If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

A gen­eral order of attack

1. Take out the HCF of all the terms first.
2. 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.
3. Check each fac­tor you get: can it be fac­torised again?
4. 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: $a^2 - b^2 = (a + b)(a - b)$

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.

1. For $x^2 + px + q$, find two num­bers whose sum is $p$ and whose prod­uct is $q$.
2. For $ax^2 + bx + c$ with $a \neq 1$, find two num­bers whose sum is $b$ and whose prod­uct is $ac$.
3. 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 $9 \times (-4) = -36$.

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

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 $\displaystyle \dfrac{x^3}{216} - 8y^3$.

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

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

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

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

Related read­ing

• Lat­est Ncert Maths Text­book Class 10 pdf Down­load For Free
• Expo­nent For­mu­las Detailed Expla­na­tion
• Fac­tor­iza­tion course