Lesson 5: Dividing terms (with indices)

Introduction

With multiplying terms, you had a lesson on 'the basics' before you had to deal with indices. With division, there aren't really any basics - it's pretty much all about indices.

There is one basic thing - how to write division. You'll see below that I've used a  \(\div\)  sign. Not the best way to write division in algebra! But in my defense, it helps me with my explanation! And you might get a question written like this in an exam so you do need to be familiar with it. The proper way to write division, though, is as a fraction!

SECTION A

So, you're going to learn how to simplify something like  \(a^5\div a\) .

If we open it up, we get  \(a\times a\times a\times a\times a\div a\)

Focus on the last two terms. If you multiply by  \(a\)  then divide by  \(a\) , you end up back where you started ( \(3\times 5\div 5~=~3\) ). The last two operations cancel each other out!

\(\require{cancel} a\times a\times a \times a \times \cancel{a}\div \cancel{a}~=~a\times a\times a\times a~=~a^4\)

We call this cancelling terms.

Now we'll try  \(a^5\div a^2\)  but let's write this one as a fraction.

\(\Large\frac{a^5}{a^2}\normalsize~=~\Large\frac{a\times a\times a\times \cancel{a}\times \cancel{a}}{\cancel{a}\times \cancel{a}}\normalsize~=~a\times a\times a~=~a^3\)

You just did 5 take away 2, right? I feel a formal rule coming on...

\(x^a\div x^b~=~\Large\frac{x^a}{x^b}\normalsize~=~x^{a-b}\)

Examples

1. Simplify  \(x^7\div x^2\)

       Answer:  \(x^{7-2}~=~x^5\)

2. Simplify  \(\Large\frac{b^3}{b}\)

       Answer: \(b^{3-1}~=~b^2\)      Please don't forget the invisible power of 1!

Practise to master

SECTION A

Simplify each of these expressions using your knowledge of indices.

01)  \(a^6\div a^2\)

\(a^{6-2}~=~a^4\)

02)  \(\Large\frac{m^5}{m^4}\)

\(m^{5-4}~=~m^1~=~m\)

03)  \(v^4\div v^2\)

\(v^{4-2}~=~v^2\)

04)  \(\Large\frac{c^5}{c}\)

\(c^{5-1}~=~c^4\)

05)  \(y^9\div y^5\)

\(y^{9-5}~=~y^4\)

06)  \(\Large\frac{x^8}{x^3}\)

\(x^{8-3}~=~x^5\)

07)  \(w^7\div w\)

\(w^{7-1}~=~w^6\)

08)  \(\Large\frac{n^6}{n^5}\)

\(n^{6-5}~=~n^1~=~n\)

SECTION B

If there's more than one term on the top and/or bottom, just deal with them one matching pair at a time. Start by simplifying any numbers the way you'd simplify a 'normal' fraction.

Example 1

Simplify  \(\Large\frac{6x^5}{2x^2}\)

       Answer:  \(3x^{5-2}~=~3x^3\)

Since 6 divided by 2 is exactly 3, the denominator (the bottom bit) disappears.

Example 2

Simplify  \(12x^7y^3\div 8x^2y^2\)

       Answer: \(\Large\frac{12x^7y^3}{8x^2y^2}\normalsize~~=~\Large\frac{3x^{7-2}y^{3-2}}{2}\normalsize~~=~\Large\frac{3x^5y}{2}\)

Since 8 does NOT divide exactly into 12, the denominator does not disappear and the answer is written as a fraction. So I thought I might as well switch to fraction form straight away, rather than use the  \(\div\)  sign.

Practise to master

SECTION B

Simplify these expressions using your knowledge of indices.

01)  \(8b^6\div 2b^2\)

\(4b^{6-2}~=~4b^4\)

02)  \(6y^7\div 4y^3\)

\(\Large\frac{6y^7}{4y^3}\normalsize~=~\Large\frac{3y^{7-3}}{2}\normalsize~=~\Large\frac{3y^4}{2}\)

03)  \(\Large\frac{9e^5}{3e^2}\)

\(3e^{5-2}~=~3e^3\)

04)  \(\Large\frac{14x^4}{6x}\)

\(\Large\frac{7x^{4-1}}{3}\normalsize~=~\Large\frac{7x^3}{3}\)

05)  \(4x^6y^2\div 2x^3y\)

\(2x^{6-3}y^{2-1}~=~2x^3y\)

06)  \(20a^9b^2\div 12a^7b\)

\(\Large\frac{20a^9b^2}{12a^7b}\normalsize~=~\Large\frac{5a^{9-7}b^{2-1}}{3}\normalsize~=~\Large\frac{5a^2b}{3}\)

07)  \(\Large\frac{12x^7y^4}{6x^6y^2}\)

\(2x^{7-6}y^{4-2}~=~2xy^2\)

08)  \(\Large\frac{18m^5n^4}{4mn^3}\)

\(\Large\frac{9m^{5-1}n^{4-3}}{2}\normalsize~=~\Large\frac{9m^4n}{2}\)

09)  \(16u^2v^7\div uv\)

\(16u^{2-1}v^{7-1}~=~16uv^6\)

10)  \(8x^5y^9\div 6x^2y^8\)

\(\Large\frac{8x^5y^9}{6x^2y^8}\normalsize~=~\Large\frac{4x^{5-2}y^{9-8}}{3}\normalsize~=~\Large\frac{4x^3y}{3}\)

11)  \(\Large\frac{10g^5h^4}{5g^3h}\)

\(2g^{5-3}h^{4-1}~=~2g^2h^3\)

12)  \(\Large\frac{7x^5y^6}{6xy^5}\)

\(\Large\frac{7x^{5-1}y^{6-5}}{6}\normalsize~=~\Large\frac{7x^4y}{6}\)

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