Multiplying Monomials
Monomials can be multiplied in the same manner as numbers.
To multiply monomials, we use commutative property of multiplication and properties of exponents.
Let's start from a simple example, involving only one variable.
Example 1. Simplify $$${\left({2}{{x}}^{{3}}\right)}{\left({7}{{x}}^{{4}}\right)}$$$ (parenthesis are written to separate monomials).
$$${2}{{x}}^{{3}}\cdot{7}{{x}}^{{4}}=$$$
$$$={2}\cdot{7}\cdot{{x}}^{{3}}\cdot{{x}}^{{4}}=$$$ (commutative property of multiplication, multiple times)
$$$={14}{{x}}^{{3}}\cdot{{x}}^{{4}}=$$$ (simplify coefficient)
$$$={14}{{x}}^{{{3}+{4}}}=$$$ (sum rule for exponents)
$$$={14}{{x}}^{{7}}$$$ (simplify).
Answer: $$${\left({2}{{x}}^{{3}}\right)}{\left({7}{{x}}^{{4}}\right)}={14}{{x}}^{{7}}$$$.
Let's see how to multiply, if there are more than one variable (in fact technique is same).
Example 2. Simplify $$${\left(\frac{{1}}{{3}}{{x}}^{{5}}{{y}}^{{7}}{{z}}^{{2}}\right)}{\left(-{7}{{z}}^{{3}}{x}{{y}}^{{3}}\right)}$$$.
$$$\frac{{1}}{{3}}{{x}}^{{5}}{{y}}^{{7}}{{z}}^{{2}}\cdot{\left(-{7}\right)}{{z}}^{{3}}{x}{{y}}^{{3}}=$$$
$$$=\frac{{1}}{{3}}\cdot{\left(-{7}\right)}\cdot{{x}}^{{5}}\cdot{x}\cdot{{y}}^{{7}}\cdot{{y}}^{{3}}\cdot{{z}}^{{2}}\cdot{{z}}^{{3}}=$$$ (we apply commutative property of multiplication multiple times)
$$$=-\frac{{7}}{{3}}\cdot{{x}}^{{{5}+{1}}}\cdot{{y}}^{{{7}+{3}}}\cdot{{z}}^{{{2}+{3}}}=$$$ (sum rule for exponents).
$$$=-\frac{{7}}{{3}}{{x}}^{{6}}{{y}}^{{10}}{{z}}^{{5}}$$$ (simplify).
Answer: $$${\left(\frac{{1}}{{3}}{{x}}^{{5}}{{y}}^{{7}}{{z}}^{{2}}\right)}{\left(-{7}{{z}}^{{3}}{x}{{y}}^{{3}}\right)}=-\frac{{7}}{{3}}{{x}}^{{6}}{{y}}^{{10}}{{z}}^{{5}}$$$.
We can even multiple more than two monomials!
Example 3. Multiply $$${\left({4}{{x}}^{{2}}{y}\right)}{\left({3}{z}{{y}}^{{2}}{{x}}^{{3}}\right)}{\left({2}{y}{z}{x}\right)}$$$.
$$${4}\cdot{{x}}^{{2}}\cdot{y}\cdot{3}\cdot{{y}}^{{2}}\cdot{{x}}^{{3}}\cdot{2}\cdot{y}\cdot{z}\cdot{x}=$$$
$$$={4}\cdot{3}\cdot{2}\cdot{{x}}^{{2}}\cdot{{x}}^{{3}}\cdot{x}\cdot{y}\cdot{{y}}^{{2}}\cdot{y}\cdot{z}\cdot{z}=$$$ (commutative property of multiplication)
$$$={24}\cdot{{x}}^{{{2}+{3}+{1}}}\cdot{{y}}^{{{1}+{2}+{1}}}\cdot{{z}}^{{{1}+{1}}}=$$$ (sum rule for exponents)
$$$={24}{{x}}^{{6}}{{y}}^{{4}}{{z}}^{{2}}$$$ (simplify).
Answer: $$${\left({4}{{x}}^{{2}}{y}\right)}{\left({3}{z}{{y}}^{{2}}{{x}}^{{3}}\right)}{\left({2}{y}{z}{x}\right)}={24}{{x}}^{{6}}{{y}}^{{4}}{{z}}^{{2}}$$$.
Note: we explicitly wrote commutative property, but, after some practice, you will want to skip that step and just multiple "like" variables immediately, making commutations in your head.
Finally, let's work through a slightly harder example.
Example 4. Simplify the following: $$${{\left({2}{{x}}^{{3}}{{y}}^{{2}}\right)}}^{{3}}{\left({{x}}^{{3}}{{y}}^{{5}}\right)}$$$.
What outer exponent is doing here??? It appears, that we raise monomial to 3rd power.
But recall, that exponent is just multiplication, so we, actually, multiply here.
First method is to rewrite expression as $$${{\left({2}{{x}}^{{3}}{{y}}^{{2}}\right)}}^{{3}}{\left({{x}}^{{3}}{{y}}^{{5}}\right)}={\left({2}{{x}}^{{3}}{{y}}^{{2}}\right)}{\left({2}{{x}}^{{3}}{{y}}^{{2}}\right)}{\left({2}{{x}}^{{3}}{{y}}^{{2}}\right)}{\left({{x}}^{{3}}{{y}}^{{5}}\right)}$$$, and proceed as follows. But what if exponent is a large number? It can take much time and efforts to perform such multiplication.
Solution is to use another method, namely, other properties of exponents.
$$${{\left({2}{{x}}^{{3}}{{y}}^{{2}}\right)}}^{{3}}{\left({{x}}^{{3}}{{y}}^{{5}}\right)}=$$$
$$$={{2}}^{{3}}{{\left({{x}}^{{3}}\right)}}^{{3}}{{\left({{y}}^{{2}}\right)}}^{{3}}{\left({{x}}^{{3}}{{y}}^{{5}}\right)}=$$$ (power of a product: $$${{\left({a}{b}{c}\right)}}^{{n}}={{a}}^{{n}}{{b}}^{{n}}{{c}}^{{n}}$$$)
$$$={8}{{x}}^{{{3}\cdot{3}}}{{y}}^{{{2}\cdot{3}}}{\left({{x}}^{{3}}{{y}}^{{5}}\right)}=$$$ (rule for multiplying exponents)
$$$={8}{{x}}^{{9}}{{y}}^{{6}}{\left({{x}}^{{3}}{{y}}^{{5}}\right)}=$$$ (simplify)
$$$={8}{{x}}^{{12}}{{y}}^{{11}}$$$ (multiply)
Both methods gave the same answer, but, as stated above, second method is better, because it is faster and requires less writing.
Answer: $$${{\left({2}{{x}}^{{3}}{{y}}^{{2}}\right)}}^{{3}}{\left({{x}}^{{3}}{{y}}^{{5}}\right)}={8}{{x}}^{{12}}{{y}}^{{11}}$$$.
Now, it is time to exercise.
Exercise 1. Multiply $$${\left({2}{{a}}^{{3}}\right)}{\left({3}{{a}}^{{4}}\right)}$$$.
Answer: $$${6}{{a}}^{{7}}$$$.
Exercise 2. Simplify $$${\left(-\frac{{1}}{{3}}{{x}}^{{2}}{{y}}^{{5}}{{z}}^{{2}}\right)}{\left({3}{{y}}^{{7}}{z}{{x}}^{{2}}\right)}$$$.
Answer: $$$-{{x}}^{{4}}{{y}}^{{12}}{{z}}^{{3}}$$$.
Exercise 3. Multiply the following: $$${\left(-\frac{{3}}{{5}}{{c}}^{{2}}{{a}}^{{3}}{{b}}^{{7}}\right)}\cdot{\left(-\frac{{2}}{{7}}{{b}}^{{4}}{a}{{c}}^{{3}}\right)}\cdot{\left({a}{b}{c}\right)}$$$.
Answer: $$$\frac{{6}}{{35}}{{a}}^{{5}}{{b}}^{{12}}{{c}}^{{6}}$$$.
Exercise 4. Simplify $$${{\left({3}{{x}}^{{2}}{y}\right)}}^{{3}}{\left(-{{x}}^{{2}}\right)}$$$.
Answer: $$$-{27}{{x}}^{{8}}{{y}}^{{3}}$$$.
Exercise 5. Simplify the following: $$${{\left({{a}}^{{2}}{{z}}^{{3}}{{b}}^{{2}}{c}\right)}}^{{4}}{{\left(-{3}{{a}}^{{2}}{b}{{z}}^{{5}}\right)}}^{{3}}\cdot{{z}}^{{3}}{a}$$$.
Answer: $$$-{27}{{a}}^{{15}}{{b}}^{{11}}{{c}}^{{4}}{{z}}^{{30}}$$$.