# Operations with monomials

Two monomials are called similar when they have the same literal part. For example:

$$4x^4y$$ and $$\dfrac{1}{5}x^4y$$

Other examples of similar monomials are:

$$5x^2yh$$ and $$\dfrac{6}{7}x^2yh$$

$$x^2$$ and $$3x^2$$

We have to remark that the order in which the variables appear is not relevant.

## Sum of monomials

The first rule we have to remember is that we can only add similar monomials. In this case, the independent term remains equal, and the coefficients must be added.

$$3x^5y+2x^5y=(3+2)x^5y=5x^5y$$

$$4x^3+6x^3=(4+6)x^3=10x^3$$

$$3xyh+11xyh=(3+11)xyh=14xyh$$

## Subtraction of monomials

As we have seen in the case of the sums, we can only reduce similar monomials. In this case, the independent term remains equal, and the coefficients must be reduced.

$$3x^5y-2x^5y=(3-2)x^5y=1x^5y=x^5y$$

$$4x^3-6x^3=(4-6)x^3=-2x^3$$

$$3xyh-11xyh=(3-11)xyh=-8xyh$$

## Product of monomials

The independent terms of the product is the product of independent terms, and the coefficient of the product is the product of the coefficients.

If we multiply the monomials $$3x^2y$$, $$\dfrac{3}{4}zy$$

The product of the coefficients is $$3\cdot\dfrac{3}{4}=\dfrac{9}{4}$$$And that of the independent terms are $$(x^2y)\cdot(zy)=x^2yzy=x^2y^2z$$$

So, the final result is $$\dfrac{9}{4}x^2y^2z$$$In the same way, if we multiply $$\dfrac{3}{4}x^6z$$, $$\dfrac{16}{7}z^2y$$ The product of the coefficients is $$\dfrac{3}{4}\cdot\dfrac{16}{7}=\dfrac{12}{7}$$$

And that of the independent terms are $$(x^6z)\cdot(z^2y)=x^6zz^2y=x^6z^3y$$$So, the final result is $$\dfrac{12}{7}x^6z^3y$$$

## Division of monominals

The independent term of the division is the quotient between the independent term of the numerator and the independent term of the denominator.

The coefficient of the division is the quotient between the coefficient of the numerator and the coefficient of the denominator.

$$\dfrac{3x^2y}{2xy}=\dfrac{3}{2}\cdot\dfrac{x^2y}{xy}=\dfrac{3}{2}x$$

$$\dfrac{3x^2y}{2x^4y}=\dfrac{3}{2}\cdot\dfrac{x^2y}{x^4y}=\dfrac{3}{2}\dfrac{1}{x^2}$$

$$\dfrac{3x^2}{2xz}=\dfrac{3}{2}\cdot\dfrac{x^2}{xz}=\dfrac{3}{2}\dfrac{x}{z}$$

As we can see in the examples, the result of a division of monomials is not always a monomial. Sometimes, as we can see in the second and the third example, we have an unknown in the denominator that cannot be simplified.

Generally, we will find two kinds of results, according to the variables and their exponents.

So:

• if we have the same variables, and the exponents of every numerator variable are bigger than or equal to those of the denominator: the result is a monomial.

$$\dfrac{x^5z}{5x^3z}=\dfrac{1}{5}\cdot\dfrac{x^5z}{x^3z}=\dfrac{1}{5}x^2$$

$$\dfrac{7x^3z}{5x^3z}=\dfrac{7}{5}\cdot\dfrac{x^3z}{x^3z}=\dfrac{7}{5}\cdot1=\dfrac{7}{5}$$

$$\dfrac{2h^7}{7h^3}=\dfrac{2}{7}\cdot\dfrac{h^7}{h^3}=\dfrac{2}{7}h^4$$

• In other words: the result is a rational fraction which, as we have already seen, is a quotient between two monomials that cannot be reduced. Let's see examples of two situations where a rational fraction can appear.

Identical variables in the numerator and the denominator, some of them with a superior degree in the denominator.

$$\dfrac{xz}{2x^3z}=\dfrac{1}{2}\cdot\dfrac{xz}{x^3z}=\dfrac{1}{2}\dfrac{1}{x^2}=\dfrac{1}{2x^2}$$

$$\dfrac{3x^3y}{x^3y^4}=\dfrac{3}{1}\cdot\dfrac{x^3y}{x^3y^4}=3\cdot\dfrac{1}{y^3}=\dfrac{3}{y^3}$$

$$\dfrac{4x}{x^3}=\dfrac{4}{1}\cdot\dfrac{x}{x^3}=4\cdot\dfrac{1}{x^2}=\dfrac{4}{x^2}$$

Cases with different variable in the denominator.

$$\dfrac{4x^4y^2}{3h}=\dfrac{4}{3}\cdot\dfrac{x^4y^2}{h}$$

$$\dfrac{2x^5}{3x^6}=\dfrac{2}{3}\cdot\dfrac{x^5}{x^6}=\dfrac{2}{3}\cdot\dfrac{1}{x}=\dfrac{2}{3x}$$

$$\dfrac{1}{t}$$

It must be said that in the last example the degree of the monomial of the numerator is not relevant at all: it does not matter how big it is since if a new variable appears in the denominator, the result will always be a rational fraction.

## Power of monomials

The independent term and the quotient are both the result of raising the variables and the coefficient of the original monomial to the given exponent. If the coefficient has more than one variable, we should remember that the power of a product is the product of the elements raised to the mentioned potency.

$$(2x)^2=2^2\cdot x^2=4x^2$$

$$(3xy^2)^3=3^3\cdot(xy^2)^3=27\cdot x^3 \cdot (y^2)^3=27x^3y^6$$

$$(\dfrac{1}{2}xyz^3)^4=(\dfrac{1}{2})^4\cdot(xyz^3)^4=\dfrac{1}{2^4}\cdot x^4\cdot y^4\cdot(z^3)^4=\dfrac{1}{16}\cdot x^4\cdot y^4\cdot z^{12}$$

In the first example of all, we must be aware that:

$$(-2x)^2=(-2)^2\cdot x^2=4x^2$$

But

$$(-2x)^3=(-2)^3\cdot x^3=-8x^3$$

So, if a monomial with negative coefficient is raised to an even exponent, the result will be positive; if it is raised to an odd exponent, it will be negative.

Let's see some more examples:

$$(-\dfrac{1}{2}xy^2)^5=(-\dfrac{1}{2})^5\cdot(xy^2)^5=-\dfrac{1}{2^5}\cdot x^5\cdot y^{10}=-\dfrac{1}{32}\cdot x^5\cdot y^{10}$$

$$(-\dfrac{1}{2}xy^2)^4=(-\dfrac{1}{2})^4\cdot(xy^2)^4=\dfrac{1}{2^4}\cdot x^4\cdot y^8=\dfrac{1}{16}\cdot x^4\cdot y^8$$

$$(-\dfrac{1}{3}x^3h)^3=(-\dfrac{1}{3})^3\cdot(x^3h)^3=-\dfrac{1}{3^3}\cdot x^9\cdot h^3=-\dfrac{1}{27}\cdot x^9\cdot h^3$$

$$(-\dfrac{1}{3}x^3h)^2=(-\dfrac{1}{3})^2\cdot(x^3h)^2=\dfrac{1}{3^2}\cdot x^6\cdot h^2=\dfrac{1}{9}\cdot x^6\cdot h^2$$