Highest Common Factor (H.C.F) of Monomials

You already know how to find the H.C.F of two num­bers such as $12$ and $18$. In alge­bra the same ques­tion is asked about terms that con­tain let­ters, like $6x^{3}y$ and $18x^{2}y^{3}$. The answer is again the biggest thing that divides every term exactly. You need this skill the moment you start fac­toris­ing: to fac­torise $6x^{3}y + 18x^{2}y^{3}$ you first take out the H.C.F of the two terms.

What a mono­mial is

A mono­mial is a sin­gle alge­braic term: a num­ber mul­ti­plied by let­ters, with whole-num­ber pow­ers. $5xy$, $-7a^{2}$ and $12$ are mono­mi­als. $3x + 2$ is not, because it has two terms joined by a plus sign.

Every mono­mial has two parts. The coef­fi­cient is the num­ber in front. The vari­able part is the let­ters with their pow­ers. In $18x^{2}y^{3}$ the coef­fi­cient is $18$ and the vari­able part is $x^{2}y^{3}$.

Remem­ber what a power means. $x^{3}$ is $x \times x \times x$, three $x$s mul­ti­plied together. So a power sim­ply counts how many copies of a let­ter the term holds. That count­ing is the whole secret of this les­son.

The rule

The H.C.F of given mono­mi­als is the com­mon fac­tor hav­ing great­est coef­fi­cient and high­est pow­ers of the vari­ables.

Remem­ber. The rule above works in two halves. For the num­bers, find the H.C.F of the coef­fi­cients. For each let­ter, take it only if it appears in every term, and take the small­est power that appears. The phrase "high­est pow­ers" means the high­est power that still divides every term, and that is always the small­est power you can see.

Why the small­est power? Take $x^{3}$ and $x^{2}$. The first holds three copies of $x$, the sec­ond only two. A com­mon fac­tor can use only as many copies as both terms can spare, so it can use two. $x^{3}$ would not divide $x^{2}$.

The method, step by step

1. Write each mono­mial as its coef­fi­cient bro­ken into primes, times its let­ters.
2. Find the H.C.F of the coef­fi­cients.
3. List the let­ters that appear in every mono­mial. A let­ter miss­ing from even one term is left out.
4. For each of those let­ters, take the small­est power.
5. Mul­ti­ply the results together.

The acad­e­my's exam­ples

Ex: 1) Find H.C.F of $6x^{3}y$ and $18x^{2}y^{3}$.

Sol: $6x^{3}y = 2 \times 3 \times x^{3} \times y$

$\quad \quad \ 18x^{2}y^{3} = 2 \times 3^{2} \times x^{2} \times y^{3}$

$\quad \quad \ \text{H.C.F} = 2 \times 3 \times x^{2} \times y$

$\quad \quad \quad \quad \quad \quad \ = 6x^{2}y$

Here is what hap­pened in Exam­ple 1. The coef­fi­cients $6$ and $18$ share $2 \times 3 = 6$. The pow­ers of $x$ are $3$ and $2$, so we take $x^{2}$. The pow­ers of $y$ are $1$ and $3$, so we take $y$. The pic­ture lines the fac­tors up in columns so you can see this.

Check it by divid­ing: $6x^{3}y \div 6x^{2}y = x$ and $18x^{2}y^{3} \div 6x^{2}y = 3y^{2}$. Both divi­sions come out with noth­ing left over, and $x$ and $3y^{2}$ share noth­ing more, so $6x^{2}y$ really is the high­est.

Ex: 2) Find H.C.F of $5xy$ and $10x$.

Sol: $5xy = 5 \times x \times y$

$\quad \quad \ 10x = 2 \times 5 \times x$

$\quad \quad \ \text{H.C.F} = 5x$

In Exam­ple 2 the let­ter $y$ appears in $5xy$ but not in $10x$. So $y$ can­not be part of the H.C.F. The coef­fi­cients $5$ and $10$ have H.C.F $5$, and $x$ appears once in each, giv­ing $5x$.

The next four are given with their answers. Try each one your­self before read­ing the rea­sons below.

Ex: 3) H.C.F of $12a^{2}b$ and $15ab^{2}$ is 3ab

Ex: 4) H.C.F of $2x$ and $4$ is 2

Ex: 5) H.C.F of $12x$ and $36$ is 12

Ex: 6) H.C.F of $14pq$ and $35pqr$ is 7pq

Why those four answers are right

• $12a^{2}b$ and $15ab^{2}$: the H.C.F of $12$ and $15$ is $3$; the smaller power of $a$ is $a$; the smaller power of $b$ is $b$. So $3ab$.
• $2x$ and $4$: the H.C.F of $2$ and $4$ is $2$. The term $4$ has no $x$, so no let­ter is taken. So $2$.
• $12x$ and $36$: $12$ divides $36$, so the H.C.F of the num­bers is $12$, and again $36$ has no $x$. So $12$.
• $14pq$ and $35pqr$: the H.C.F of $14$ and $35$ is $7$; $p$ and $q$ are in both; $r$ is in only one. So $7pq$.

More worked exam­ples

Exam­ple 7: three mono­mi­als

Find the H.C.F of $8a^{3}b^{2}$, $12a^{2}b^{4}$ and $20a^{4}b^{3}$.

$$\begin{aligned}8a^{3}b^{2} &= 2^{3} \times a^{3} \times b^{2} \\ 12a^{2}b^{4} &= 2^{2} \times 3 \times a^{2} \times b^{4} \\ 20a^{4}b^{3} &= 2^{2} \times 5 \times a^{4} \times b^{3} \\ \text{H.C.F} &= 2^{2} \times a^{2} \times b^{2} = 4a^{2}b^{2}\end{aligned}$$

The only prime in all three coef­fi­cients is $2$, and the small­est count of it is two. The small­est power of $a$ is $2$, and the small­est power of $b$ is $2$.

Exam­ple 8: a let­ter in one term only

Find the H.C.F of $9x^{2}y$ and $15xy^{2}z$.

$$\begin{aligned}9x^{2}y &= 3^{2} \times x^{2} \times y \\ 15xy^{2}z &= 3 \times 5 \times x \times y^{2} \times z \\ \text{H.C.F} &= 3 \times x \times y = 3xy\end{aligned}$$

The let­ter $z$ is left out, because $9x^{2}y$ has none.

Exam­ple 9: noth­ing in com­mon

Find the H.C.F of $7m^{2}$ and $11n$. The coef­fi­cients $7$ and $11$ are dif­fer­ent primes, so their H.C.F is $1$. The two terms share no let­ter. So the H.C.F is $1$. That is a per­fectly good answer: it tells you the terms have no com­mon fac­tor except $1$.

Where this is used

The H.C.F is exactly what you take out when you fac­torise. Because $6x^{2}y$ is the H.C.F of $6x^{3}y$ and $18x^{2}y^{3}$, you can write

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

Mul­ti­ply the bracket back out to check: $6x^{2}y \times x = 6x^{3}y$ and $6x^{2}y \times 3y^{2} = 18x^{2}y^{3}$.

When a coef­fi­cient is neg­a­tive

Signs do not change the method. Find the H.C.F of $-6x^{2}y$ and $21xy^{3}$ by ignor­ing the minus sign while you work. The H.C.F of $6$ and $21$ is $3$; the smaller power of $x$ is $x$; the smaller power of $y$ is $y$. So the H.C.F is $3xy$. By cus­tom the H.C.F is writ­ten as a pos­i­tive term. When you fac­torise, the minus sign sim­ply stays inside the bracket: $-6x^{2}y + 21xy^{3} = 3xy(-2x + 7y^{2})$.

How to check any answer

A good habit is to test your H.C.F in two ways before you move on.

1. Does it divide every term? Divide each mono­mial by your answer. Each result must be a mono­mial with a whole-num­ber coef­fi­cient and no neg­a­tive pow­ers.
2. Is it the high­est? Look at the results of those divi­sions. If they still share a num­ber big­ger than $1$, or a let­ter, your answer was too small.

For Exam­ple 7, divid­ing by $4a^{2}b^{2}$ gives $2a$, $3b^{2}$ and $5a^{2}b$. No num­ber other than $1$ divides $2$, $3$ and $5$, and no let­ter is in all three. So $4a^{2}b^{2}$ passes both tests.

Sup­pose a stu­dent had writ­ten $2ab$ instead. It does divide all three terms, so it passes the first test. But divid­ing gives $4a^{2}b$, $6ab^{3}$ and $10a^{3}b^{2}$, which still share $2ab$. So $2ab$ is a com­mon fac­tor, but not the high­est one.

Your turn

Find the H.C.F of each set of mono­mi­als.

1. $4x^{2}$ and $6x$
2. $9ab$ and $12b^{2}$
3. $10p^{3}q$ and $25p^{2}q^{2}$
4. $16xyz$ and $24x^{2}z$
5. $3m^{2}n$, $6mn^{2}$ and $9m^{3}n^{3}$
6. $18a^{2}b^{3}c$ and $27a^{3}b^{2}$
7. $5x^{2}$ and $7y^{2}$
8. $-6x^{2}y$ and $21xy^{3}$

Com­mon mis­takes

• Tak­ing the largest power instead of the small­est. The H.C.F of $x^{3}$ and $x^{2}$ is $x^{2}$, not $x^{3}$.
• Includ­ing a let­ter that is miss­ing from one of the terms, such as putting $y$ in the H.C.F of $5xy$ and $10x$.
• Find­ing the L.C.M of the coef­fi­cients by mis­take, for exam­ple giv­ing $18$ instead of $6$ for $6$ and $18$.
• For­get­ting the coef­fi­cient alto­gether and writ­ing only the let­ters.
• Think­ing an answer of $1$ is wrong. When terms share noth­ing, the H.C.F is $1$.

Key terms

Mono­mial

A sin­gle alge­braic term, such as $6x^{3}y$.

Coef­fi­cient

The num­ber that mul­ti­plies the let­ters in a term.

Vari­able

A let­ter that stands for a num­ber.

Power (expo­nent)

The small raised num­ber that counts how many copies of a let­ter are mul­ti­plied.

Com­mon fac­tor

Some­thing that divides every given term exactly.

H.C.F of mono­mi­als

The com­mon fac­tor with the great­est coef­fi­cient and the high­est pow­ers that divide every term.

Answers

1. $2x$
2. $3b$
3. $5p^{2}q$
4. $8xz$
5. $3mn$
6. $9a^{2}b^{2}$
7. $1$, because the terms share no fac­tor.
8. $3xy$. The minus sign belongs to one term only, so the H.C.F is taken as pos­i­tive.