Common factors and grouping

When you mul­ti­ply, you start with fac­tors and fin­ish with a prod­uct: $2(x + 2) = 2x + 4$. Fac­tori­sa­tion runs the other way. You are given the prod­uct, $2x + 4$, and asked for the fac­tors that make it. This les­son defines fac­tori­sa­tion and teaches the two meth­ods every­thing else is built on: tak­ing out a com­mon fac­tor, and group­ing.

What fac­tori­sa­tion means#

A fac­tor of an expres­sion is some­thing that divides it exactly. The process of writ­ing an alge­braic expres­sion as the prod­uct of two or more fac­tors is called fac­tori­sa­tion.

With num­bers, $12 = 2 \times 2 \times 3$ writes $12$ as a prod­uct of fac­tors that can­not be bro­ken down fur­ther. With expres­sions the aim is the same: break the expres­sion into fac­tors until no fac­tor can be split any more.

Fac­tori­sa­tion and expan­sion are oppo­sites:

$$\underbrace{2(x + 2)}_{\text{factorised}} \;\; \overset{\text{expand}}{\underset{\text{factorise}}{\rightleftharpoons}} \;\; \underbrace{2x + 4}_{\text{expanded}}$$

This gives you a free check on every answer: mul­ti­ply your fac­tors back out, and you must get the expres­sion you started with.

Method 1: take out the com­mon fac­tor#

If every term has a com­mon fac­tor, write that fac­tor once out­side a bracket, and inside the bracket write what is left of each term after divid­ing by it. This is the dis­trib­u­tive law, $ab + ac = a(b + c)$, read from right to left.

Always take out the H.C.F of the terms, found as in the last les­son. Tak­ing out a smaller com­mon fac­tor leaves the job unfin­ished.

Exam­ple 1#

Fac­torise $2x + 4$.

The H.C.F of $2x$ and $4$ is $2$. Divid­ing, $2x \div 2 = x$ and $4 \div 2 = 2$.

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

Exam­ple 2#

Fac­torise $6x^{3}y + 18x^{2}y^{3}$.

From the last les­son, the H.C.F is $6x^{2}y$. Divide each term by it: $6x^{3}y \div 6x^{2}y = x$ and $18x^{2}y^{3} \div 6x^{2}y = 3y^{2}$.

$$6x^{3}y + 18x^{2}y^{3} = 6x^{2}y(x + 3y^{2})$$

Exam­ple 3: three terms#

Fac­torise $5a^{2}b - 10ab^{2} + 15ab$.

The H.C.F of the coef­fi­cients $5$, $10$ and $15$ is $5$. Both $a$ and $b$ appear in every term, each at least to power $1$. So the H.C.F is $5ab$.

$$5a^{2}b - 10ab^{2} + 15ab = 5ab(a - 2b + 3)$$

Check: $5ab \times a = 5a^{2}b$, $5ab \times (-2b) = -10ab^{2}$ and $5ab \times 3 = 15ab$. If a term is exactly equal to the H.C.F, it leaves $1$ in the bracket, never noth­ing: for exam­ple, $4x + 2 = 2(2x + 1)$.

Exam­ple 4: a bracket as the com­mon fac­tor#

Fac­torise $3x(a - b) + 4y(a - b)$.

The com­mon fac­tor here is the whole bracket $(a - b)$. Take it out just as you would a sin­gle let­ter.

$$3x(a - b) + 4y(a - b) = (a - b)(3x + 4y)$$

Exam­ple 5: brack­ets that are oppo­sites#

Fac­torise $2x(a - b) - 5(b - a)$.

The brack­ets $(a - b)$ and $(b - a)$ look dif­fer­ent, but $b - a = -(a - b)$. So $-5(b - a) = +5(a - b)$.

$$\begin{aligned} 2x(a - b) - 5(b - a) &= 2x(a - b) + 5(a - b) \\ &= (a - b)(2x + 5) \end{aligned}$$

Method 2: group­ing#

Some­times the terms have no fac­tor com­mon to all of them, but they fall into groups that do. Take

$$ax + bx + ay + by$$

No let­ter is in all four terms. But the first two share $x$ and the last two share $y$:

$$\begin{aligned} ax + bx + ay + by &= x(a + b) + y(a + b) \\ &= (a + b)(x + y) \end{aligned}$$

After the first step a com­mon bracket, $(a + b)$, appears, and tak­ing it out fin­ishes the job. This method, fac­toris­ing by group­ing, is used again in the les­son on split­ting the mid­dle term, so prac­tise it well.

Exam­ple 6#

Fac­torise $x^{2} + 3x + 2x + 6$.

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

Exam­ple 7: rear­rang­ing first#

Fac­torise $6xy - 4y + 6 - 9x$.

Group the first two terms and the last two terms:

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

The brack­ets are oppo­sites. Take out $-3$ instead of $3$ from the sec­ond group, so that the brack­ets match:

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

If a group­ing does not pro­duce a com­mon bracket, try a dif­fer­ent group­ing or take out the neg­a­tive of the fac­tor. Check: $(3x - 2)(2y - 3) = 6xy - 9x - 4y + 6$, the same four terms.

Prac­tice#

Fac­torise.

1. $7x + 21$
2. $12a^{2}b + 18ab^{2}$
3. $4x^{3} - 8x^{2} + 12x$
4. $10p^{2}q^{3} - 15p^{3}q^{2}$
5. $x(y + 2) + 5(y + 2)$
6. $3a(x - y) - 2b(y - x)$
7. $ab + 3a + 2b + 6$
8. $x^{2} - 4x + 3x - 12$
9. $10xy + 6 + 4y + 15x$
10. $a^{3} - a^{2}b + ab - b^{2}$

Answers#