Polynomials are expressions consisting of indeterminates and coefficients connected with additions and subtractions. [[Quadratics]] are a restricted case of polynomials where the highest degree (exponent) is $2$.
$
\begin{align}
f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{2}x^{2}+a_{1}x^{1}+a_{0}
\end{align}
$
The subscript part associated with a coefficient always matches the power of the variable in the same term. This is really just a way of labelling coefficients so we can use them in various formulas later on.
$
\begin{align}
f(x)=7x^4+3x^3-5x^2+x+9
\end{align}
$
In the above polynomial, for instance, $a_4=7$, $a_3=3$, $a_2 = -5$, $a_1=1$, $a_0=9$.
A polynomial can be as simple as one term, even though they rarely are. All exponents must be non-negative integers; we cannot have rational (fractional) exponents in a polynomial. The domain of polynomials is all real numbers.
## Standard Form
In standard form, a polynomial is sorted in descending order of power.
This allows us to identify immediately the leading term, which provides us with lots of information about the given polynomial, like its degree, or what it looks like graphically.
The **degree** $n$ (largest exponent in the polynomial) informs us about end behavior and how many zeros we can expect the function to have if we include complex numbers.
Additionally, we can immediately identify the **constant** $a_0$, if any. If there is none, we can immediately factor out as many $x$ as the least power, which makes solving that much more convenient.
Considering for instance the function $f(x)=7x^4-3x^3-5x^2$.
$
\begin{align}
7x^4-3x^3-5x^2=0
\\
x^2(7x^2-3x-5)=0
\end{align}
$
Applying the zero product property, we see that $x=0$ is a root and are left with solving $7x^2-3x-5$. We find that the second real root is $x=\frac{3\pm \sqrt{149}}{14}$.
This matters because starting from the fifth degree there are no formulas to find the zeros of the function; up to the fourth degree there is but they are quite time-consuming.
We do, however, have the **rational zero theorem**, which allows us to leverage the leading term $a_nx^n$ and the constant $a_0$ to find the rational zeros.
## Factored form
Putting a polynomial in factored form sometimes makes it easier to reason about them.
Consider for instance the function $g(x)=3(x+2)^5-3$.
If we were to distribute it to obtain a polynomial in standard form, it would give us $3x^{5}+30x^{4}+120x^{3}+240x^{2}+240x+93$.
However, $g(x)=3(x+2)^5-3$ allows us to better understand what is happening and what the shape of the graph will look like. Leveraging our knowledge of [[Functions#Transformations|transformations]], we can immediately tell the shape of the graph will be that of the function $x^5$, with a down shift, $-3$, and a left shift, $+2$, applied. Lastly, the coefficient, $3$, at the beginning of the expression applies a vertical stretch to the overall function $x^5$.
Meanwhile, in standard form we would not have been able to understand all this.
While useful in many situation, standard polynomial form is not *always* the best tool for the job.
## Identifying polynomials
### $f(x)=5x^2+4x^4$
All exponents are non-negative integers.
We can rewrite it using the standard form.
$
\begin{align}
f(x)=4x^4+0x^3+5x^2+0x+0
\end{align}
$
Indeed, this is a polynomial of the fourth degree. The leading term is $4x^4$ and the constant $a_0=0$.
### $g(x)=3-\frac{1}{2}x$
Here, we find a fraction. But it is not an exponent, and all exponents are positive integers. This is a polynomial as well, which we can write in standard form.
$
\begin{align}
g(x)=-\frac{1}{2}x+3x^0
\end{align}
$
>For any non-zero value of $x$, $x^0=1$.
The leading coefficient is $-\frac{1}{2}$ and the constant is $3$.
### $h(x)=9$
This is a polynomial, $h(x)=9x^0$. Its degree is $0$, even though it doesn't really have a variable. Because it only has a constant term, this function is called a constant.
### $f(x)=0$
If a function is a constant equal to zero, we say it has no degree. Indeed, no matter the exponent, we get $0$ back. Since we don't have any information about the exponent, we can't assign a degree to this polynomial.
### $F(x)=x^{\frac{3}{2}}+2x-1$
This can be rewritten as $F(x)=\sqrt{x^3}+2x-1$. Since there is a radical in this expression, this is not a polynomial.
### $G(x)=1-\frac{4}{x}$
This expression has a fraction with a variable on the denominator, and we cannot get rid of it. This leads to domain issues. This is not a polynomial.
This can also be written as $G(x)=1-4x^{-1}$. Polynomials can't have negative exponents.
## $f(x)=9x^4-\pi x^2+\frac{1}{2}$
Expressions depend on their structure and exponents in order to qualify as polynomials. The presence of irrational numbers in the coefficients does not affect this. This is a polynomial, already in order.
### $g(x)=\frac{x^2-5}{x^4}$
This is a rational function, which divides a polynomial by another polynomial.
We can rewrite this to make it obvious why this expression is not a polynomial.
$
\begin{align}
\frac{x^2}{x^4}-\frac{5}{x^4}=\frac{1}{x^2}-\frac{5}{x^4}=x^{-2}-5x^{-4}
\end{align}
$
We can't have negative powers in a polynomial.
### $h(x)=\frac{3x^2-1}{7}$
Since there is no indeterminate on the denominator, this is a valid polynomial.
$
\begin{align}
h(x)=\frac{3}{7}x^2-\frac{1}{7}
\end{align}
$
The leading term is $\frac{3}{7}x^2$ and the constant is $-\frac{1}{7}$.
### $G(x)=2\sqrt{x}(\sqrt{x}+1)$
We can rewrite it to try and get rid of the square roots.
$
\begin{align}
G(x)=2\sqrt{x}(\sqrt{x}+1)
\\
G(x)=2x+2\sqrt{x}
\\
G(x)=2x+2x^{\frac{1}{2}}
\end{align}
$
>⚠️ The domain must be determined *before* manipulating an expression; it will not change for the better as we do so.
The domain of polynomials is all real numbers, they're not compatible with domain issues.
Expressions manipulated into polynomial form can *sometimes* be treated as polynomials despite such domain issues. Here, we're not able to get rid of all the square roots / rational exponents anyway. This is not a polynomial.
### $F(x)=2x^3(x^2-1)(x-1)^2$
Often, we get polynomials that are factored. It's possible to find the degree of the polynomial without factoring, though. We can perform a *fake distribution* of sorts, distributing only the factors.
This way we see this is a polynomial of degree $3 + 2 + 2 = 7$, and that the leading coefficient is $2$.
>Multiplying variables with exponents in them such as $x^4 \times x^3 \times x^2$ is solved by *adding* the exponents. Thus, $x^4 \times x^3 \times x^2 = x^9$.
## Multiplicity
Multiplicity stands for *multiple roots*. It refers to the repetition of a root (aka zero or $x$-intercept) in a polynomial (or any other function for that matter).
If you can factor a polynomial, and you have the exponent, the exponent is going to determine the multiplicity of that particular zero (how many times it's repeated).
Consider the following polynomial.
$
\begin{align}
f(x)=7 \cdot x^2 \cdot (x-2) \cdot (x+\frac{1}{2})^4 \cdot (x+3)^5
\end{align}
$
We can take all of the factors and set then equal to zero. It follows from the zero product property that we set this expression equal to zero, we will find as many zeros as there are factors: $x^2=0$, $x-2=0$, $(x+\frac{1}{2})^4=0$ and $(x+3)^5=0$.
In each of the factor, multiplicity is determined by the exponent which tells us whether that zero is going to touch the $x$-axis if the exponent is even, or cross it if it's odd.
From our example, the zeros obtained from the factors $x^2=0$ and $(x+\frac{1}{2})^4=0$ will bounce off the $x$-axis. Meanwhile, the zeros coming from the factors $x-2=0$ and $(x+3)^5=0$ will cross the $x$-axis.
>👉 To find the zeros, you want to isolate the factors that have a variable in it. Namely, you don't want or need to solve $7x^2=0$, just $x^2=0$. You can think about it in two ways. First, you can break it down further following the zero product property, but it wouldn't make sense to solve $7=0$. Second, in the expression $7x^2$, all what the $7$ coefficient really does is to perform a vertical stretch / horizontal compression, but it doesn't change the $x$-intercept itself, even if the coefficient is negative. Wrapping all factors in parentheses beforehand is a good way of making your life simpler by identifying small trivialities like this.
Let us determine the zeros given by the factors.
$
\begin{align}
x=0
\\
x-2=0
\\
x+\frac{1}{2}=0
\\
x+3=0
\end{align}
$
The zeros from our polynomial are found at:
- $x=0$. We know from its power is a double root (the graph touches the $x$-axis).
- $x=2$. From its odd power we know the graph crosses the $x$-axis at that point.
- $x=-\frac{1}{2}$. The graph bounces off the $x$-axis at that point.
- $x=-3$. The graph crosses the $x$-axis at that point.
A look at the graph of this function confirms our findings.
But why don't we find 12 zeros, since this is a polynomial of degree $12$? This is because some are *repeated* zeros. This is the whole idea behind multiplicity. To better understand it, imagine we had written our polynomial *without* the powers in the following form:
$
\begin{align}
7 \cdot x \cdot x \cdot (x-2) \cdot (x+\frac{1}{2}) \cdot (x+\frac{1}{2}) \cdot (x+\frac{1}{2})...
\end{align}
$
It follows from the zero product property that each of the factors corresponds to a zero, but we can now easily see how $(x+\frac{1}{2})$, $(x+\frac{1}{2})$ and $(x+\frac{1}{2})$ are going to yield the exact same solution multiple times.
>👉 Remember that a positive integer exponent is merely a way of repeating a factor without having to write it down. It acts as an iterator for multiplication.
Coming back to our zeros, we see that $x=0$ has a multiplicity of $2$, $x=2$ has a multiplicity of $1$, $x=-\frac{1}{2}$ has a multiplicity of of $4$ and $x=-3$ has a multiplicity of $5$. All that adds up to the $12$ solutions we can expect for of polynomial of degree $12$.
## End Behavior
End behavior is determined by the leading term of a polynomial.
Indeed, as $x$ approaches $\infty$, the shape of the function's graph will resemble more and more that of its leading term, while it might look very different locally, for smaller values of $x$.
A polynomial of degree $n$ can have at most $n-1$ turning points (local minimum / maximum).
All polynomials create continuous curves, with no cusps (sharp edges / points).
### $f(x)=3x^5-2x^2+1$
Here, the leading term is $3x^5$. This means that at a large scale, the graph of $3x^5-2x^2+1$ will be very close to that of $3x^5$. Below is a representation of the two graphs superposed. You see that locally, the behaviors differ, while the overall slopes grow nearly identical as $x$ approaches $\infty$.
![[polynomial-leading-term-end-behavior-example 1.png]]
We say that $3x^5-2x^2+1\approx 3x^5$.
### $g(x)=-2x^3-5x^4+7$
Here, the leading term is $-5x^4$ (the above polynomial is not in standard form).
Again, this means that $-2x^3-5x^4+7 \approx -5x^4$ as $x$ approaches $\infty$.
## Creating Polynomials from Real Zeros
We can build a function from its graph and $x$-intercepts.
If we have an $x$-intercept, you can create a factor. Using the zero product property, you can take the factors corresponding to the graph's roots and build a polynomial from it.
If you can find some zeros for a polynomial, it means you can factor it.
### $f(x)=x^2+2x-8$
This is a simple [[Quadratics|quadratic function]].
$
\begin{align}
x^2+2x-8=0
\\
(x+4)(x-2)=0
\end{align}
$
Using the zero product property, we find two real solutions, $x=-4$ and $x=2$.
Now imagine you had been given only the zeros to start with. How can we go backwards and build the function form it?
First, we need to pick the adequate exponents with respect to multiplicity, and that we match the degree of the polynomial.
Second, if we have a point given to us, we can find a specific polynomial whose graph passes through this point. If we don't, we find a general polynomial, a family of curves that have the same $x$-intercepts, and we usually mark it upfront with the letter $a$ to indicate we're referring to a *family* of polynomials. Think of $a$ as a vertical stretch factor; for the same roots, changing $a$ stretches or compresses the graph vertically, or flips it across the $x$-axis. It changes the steepness of the curve without affecting its zeros.
Here, knowing we are dealing with a polynomial of degree two and that its roots are $x=-4$ and $x=2$, we can factor backwards to the following polynomial $(x+4)(x-2)=0$. We only have to distribute it to get it into standard form.
### $-3,4,1,0$
Imagine you're given the above zeros. Let us rewrite them.
$
\begin{align}
x=-3 \Rightarrow x+3=0
\\
x=4 \Rightarrow x-4=0
\\
x=1 \Rightarrow x-1=0
\\
x=0 \Rightarrow x=0
\end{align}
$
Following the zero product property, we are now able to write the following polynomial.
$
\begin{align}
a \cdot x \cdot (x+3) \cdot (x-4) \cdot (x-1) = 0
\end{align}
$
>👉 We typically put individual terms in the front out of convenience.
We write $a$ in front of the polynomial, because this coefficient doesn't affect our roots.
We also need to remember that each of the factored terms could be raised to *any* power without affecting our zeros. Indeed we have no information regarding the multiplicity of the roots.
Below are examples that all are equally true and belong to the set of polynomials that fit those zeros.
$
\begin{align}
f(x) = 3 \cdot x \cdot (x+3) \cdot (x-4) \cdot (x-1)
\\
f(x) = 2 \cdot x \cdot (x+3)^2 \cdot (x-4) \cdot (x-1)
\\
f(x) = -13 \cdot (x+3)^{11}\cdot(x-4)^{5}\cdot(x-1)^{3}\cdot x
\end{align}
$
This is the case because we don't have a point on the graph; without more information this is all we're able to determine about this polynomial.
### $-9,-1,0,4$ - Degree $4$
$
\begin{align}
x = -9 \Rightarrow x+9=0
\\
x=-1 \Rightarrow x+1=0
\\
x=0 \Rightarrow x=0
\\
x=4 \Rightarrow x-4 =0
\end{align}
$
We can now write a polynomial.
$
\begin{align}
a \cdot x \cdot (x+9) \cdot (x+1) \cdot (x-4)
\end{align}
$
This time, however, we have more information at hand, and know this is a polynomial of degree 4. If we 'fake distribute' this expression, we see that our polynomial here already satisfies this requirement. None of the terms will be further raised by any power. The only part that remains undetermined is $a$.
Because the degree was given to us, we know there can't be things like *irreducible quadratics* in the polynomial that wouldn't result in any more zeros and be transparent in regard to the zeros we were given.
Consider for instance the irreducible quadratic $x^2+9$, which cannot be reduced any further over the real number system, as it would yield a complex solution $x=\pm i\sqrt{9}$.
### $1$ (mult $1$), $2$ (mult $2$) - Degree $4$
$
\begin{align}
x=1 \Rightarrow x-1=0
\\
x=2 \Rightarrow x-2=0
\end{align}
$
Because we know the multiplicity of each root, we can write the following polynomial:
$
\begin{align}
a \cdot (x-2)^2 \cdot (x-1)
\end{align}
$
### $-2$ (mult $2$), $5$ (mult $1$), $0$ (mult $3$), $\frac{1}{2}$ (mult $3$) - Degree $9$
$
\begin{align}
x=-2 \Rightarrow x+2=0
\\
x=5 \Rightarrow x-5=0
\\
x=0 \Rightarrow x=0
\\
x=\frac{1}{2} \Rightarrow x-\frac{1}{2}=0
\end{align}
$
We can now write the polynomial.
$
\begin{align}
a \cdot x^3 \cdot (x+2)^2 \cdot (x-5) \cdot (x-\frac{1}{2})^3
\end{align}
$
We see this is a polynomial of degree $9$, which satisfies the information we were given.
## Sketching Polynomials
Here, we concern ourselves with the *sketching* of polynomials, rather than graphing. The goal is to capture the essential features of the function with its leading term, degree, end behavior, intercepts and general shape, without plotting the exact points.
Indeed, it's generally hard to infer from the polynomial itself how dynamic the graph is going to be and what are its turning points (local maxima / minima). To find those, however, we need derivatives, which are already in the domain of Calculus and not the matter at hand.
But it's often helpful to build a general understanding of what a graph is going to be like without the need for complete accuracy, which only comes at the cost of time.
To do so, we focus on a few key informations about polynomials that inform us about their behavior: leading term, degree, maximum number of turning points, end behavior, $y$-intercept and $x$-intercept. Additionally, we can approximate where the turning points are going to be by plugging in the average of two $x$-intercepts.
Let us start with a somewhat convoluted polynomial to better understand this.
### $f(x)=-2(x-7)(x+\frac{1}{2})^2(x+4)^3(x^2+9)^2$
We will not try to write it in standard form; as it is factored, it already gives most of the information we need to understand what's going on.
The first question we must ask of this polynomial is what the leading term is. Looking at the powers at play, we see this is a polynomial of degree ten. Since the leading coefficient is $-2$, we see the leading term must be $-2x^{10}$, without even bothering with distribution.
Knowing this gives us more information still. The degree $10$ tells us we'll have a maximum of $9$ turning point. Meanwhile, the leading coefficient being negative associated with the fact $10$ is an *even* power informs us about the end behavior. We know the outside of the polynomial will look like a downward opening parabola, with something going on in the middle.
We also can find the $y$-intercept of the function by solving $f(0)$. This is trivial, so we'll skip over the calculations here. We find that $f(0)=18144$. Seeing that, we see how unrealistic it would be to graph this, and that the best we can do is try to get an understanding what it must look like.
Next, we turn toward $x$-intercepts. From the zero product property, we know we can solve each of the factor that has a variable in it by setting it up equal to zero.
$
\begin{align}
(x-7) = 0 \Rightarrow x=7
\\
(x+\frac{1}{2})^2=0 \Rightarrow x=-\frac{1}{2}
\\
(x+4)^3=0 \Rightarrow x = -4
\\
(x^2+9)^2=0=> x=\pm 3i
\end{align}
$
We derive the multiplicity of each solution from the term's power. Remember, if the power is even, then this is a *bounce*, the graph touches the $x$-axis for this value of $x$ and bounces off. This is the case of $x=-\frac{1}{2}$.
While $(x^2+9)^2$ is an even power, we also see it has no real solution. This is an irreducible quadratic because it can't be factored over the real numbers to get rid of the power inside the parentheses, which would yield a complex solution, $x=\pm3i$. A complex root usually correspond to a turning point on the graph that neither touches nor crosses the $x$-axis.
Had it been written $(x^2-9)^2$, this would have been easily reduced into linear terms, $(x-3)^2(x+3)^2$, yielding $x=\pm3$. Had it been written a cubic power, this would have not been an issue either since odd roots preserve signs.
The rest of the zeros are odd, meaning at those points the graph crosses the $x$-axis.
This is really all the information we can infer from this polynomial, but also all we need to get an understanding about its behavior.
### $h(x)=x^2(x-3)$
We see this is a polynomial of degree $3$, with a coefficient of $1$.
Thus, the leading term must be $x^3$.
We therefore know the graph must have at most $2$ turning points.
Its end behavior is going to be that of an odd function, ranging from $-\infty$ to $\infty$.
The $y$-intersect is $(0,0)$, since $h(0)=0$.
From the zero product theorem, we find two $x$-intercepts, $x=0$ of multiplicity $2$, making it a bounce, and $x=3$, of multiplicity $1$, which means the graph crosses the $x$-axis at that point.
Unlike the previous polynomial, it's much clearer what the graph must look like here.
To check our work, we can take a look at the graph of the function, which corroborates our findings.
![[factored-cubic-example.png]]
### $g(x)=(x+4)^2(1-x)$
Let us rewrite this factored polynomial.
$
\begin{align}
g(x)=(x+4)^2(1-x)
\\
g(x)=(x+4)^2(-x+1)
\\
g(x)=(x+4)^2 \times -(x-1)
\\
g(x)=-(x+4)^2(x-1)
\end{align}
$
This help us figure out end behavior and $x$-intercepts easier while avoiding sign mistakes.
Now we see this is a polynomial of degree $3$, and that the leading term must be $-x^3$.
The graph's shape will be determined by this cubic function, going from $\infty$ towards $-\infty$.
It will have a maximum of $2$ turning points.
The $y$-intercept will be located at $(0,16)$, since $g(0)=16$.
We find two real roots, $x=-4$ of multiplicity $2$ and $x=1$ of multiplicity $1$.
### $f(x)=-5x(x^2-4)(x+3)$
We first need to make sure the polynomial is factored all the way up.
$(x^2-4)$ is a reducible quadratic and can be rewritten as $(x-2)(x+2)$.
This leaves us with $f(x)=-5x(x+2)(x-2)(x+3)$.
This is a polynomial of degree $4$, and our leading term is going to be $-5x^4$.
There will be at most three turning points in our graph, whose end behavior will be that of a downward opening curve.
Solving $f(0)=0$ gives us a $y$-intercept at $y=0$.
We find four real roots, $x=-3$, $x=-2$, $x=0$ and $x=2$, each of multiplicity $1$.
## Division of Polynomials
We already learned how to create polynomials from real zeros. To be more precise, we learned to narrow down families of polynomials from real zeros, asking what would be the simplest possible form that could correspond to those.
Even in that simple form, the leading coefficient was missing. It could also be the case the graph we were trying to write a polynomial for had an irreducible quadratic factor, which we could not infer from the information we had.
This was a first step in building an understanding of polynomials leveraging our knowledges of real roots and of the zero product property.
Here, we go further and ask how we can factor a polynomial in standard form so that they not only share the same roots but are exactly the same.
To achieve this, we need to understand how to divide polynomials.
>👉 The **remainder theorem** states that when you divide a polynomial $P(x)$ by $(x−c)$, the remainder is exactly $P(c)$.
>👉 The **factor theorem** is a direct consequence of the remainder theorem. It states that if the remainder $P(c)$ is $0$, then $(x−c)$ must be a factor. In plain English, if you get a remainder of $0$, the divisor is a factor of the dividend. For instance, $6$ goes $8$ times into $48$, leaving a remainder of $0$. Therefore, the divisor $6$ is a factor of the dividend $48$.
The first method we'll review is the **long division of polynomials**, as it is a robust method we can use unconditionally.
We'll take a look at **synthetic division** right after, and we'll see we can use it to speed up the factoring process a little under some circumstances.
### Long Division
The long division algorithm for polynomials follows these steps:
1. Divide the first term of the dividend by the highest term of the divisor, placing the result above the bar
2. Multiply all divisor's terms by the first term of the quotient, writing the result under the like terms of the dividend
3. Subtract the product obtained in the above step from the appropriate terms of the original dividend, writing the result underneath
4. Bring down the next term
5. Repeat the process for the resulting polynomial, adding to the quotient with each iteration until the polynomial can't be divided further
>⚠️ It's common to make sign mistakes when dividing polynomials. Pay attention to the cases where you subtract a negative number, which is really an addition.
#### $f(x)=2x^3-x^2+2x-3$
Here, we don't have a quadratic and there's nothing to group that would help us factor.
In order to factor, we need the roots of the polynomial. This can be done in a few ways, one of which is the rational roots theorem, which we'll get to below. This will give a us a list of numbers to test for potential zeros by plugging them into the polynomial. Once we have a zero, we can create a factor from it, as we did below when we created polynomials from real roots.
For now, say we we were given $x=1$. We see that $f(1)=0$, which means that we can go ahead divide our polynomial by $(x-1)$ using long division and expect a remainder of $0$.
The first thing we need to do is to write our polynomial in standard form, making sure we include all powers all the way down to $0$ even if the coefficient is $0$, like $-2x^3+5x^2-0x+11$.
Applying the long division algorithm to $f(x)$, we find a remainder of $0$.
$
\begin{array}{r}
2x^2+x+3
\\
x-1\enclose{longdiv}{2x^3-x^2+2x-3}
\\
\underline{-(2x^3-2x^2) \phantom{00000^{0}}}
\\
x^2+2x-3
\\
\underline{-(x^2-x) \phantom{000^0}}
\\
3x-3
\\
\underline{-(3x-3)}
\\
0
\end{array}
$
Remember, it cannot be the case the remainder is not $0$ if you're dividing the polynomial by a real root. If this is not the case, you either the zero is wrong or you made a mistake in the division process.
Our polynomial can now be rewritten as $(x-1)(2x^2+x+3)$.
We could try and factor this further, isolating $(2x^2+x+3)$. However, this is an irreducible quadratic and cannot be reduced further as it would not yield a real zero.
We can prove it by showing that $b^2-4ac$, the part under the square root from the quadratic formula, yields a negative number, and therefore a complex root.
$
\begin{align}
b^2-4ac=(1)^2-4(2)(3)=1-24=-23
\end{align}
$
Therefore our polynomial is completely factored over the real numbers.
### $g(x)=6x^4-9x^2+18$
Above, we divided a polynomial $f(x)$ using a divisor derived from one of the function's roots.
This isn't always the case, which we'll illustrate here, to show how you would write the remainder if there was one.
We'll divide $g(x)$ by $x-3$.
Let us rewrite the polynomial in standard form first, $6x^4-9x^2+0x+18$.
$
\begin{array}{r}
6x^3+18x^2+45x+135
\\
x-3\enclose{longdiv}{6x^4-9x^2+0x+18 \phantom{0000000^0}}
\\
\underline{-(6x^4-18x^3)} \phantom{00000000000000^{0}}
\\
18x^3-9x^2+0x+18 \phantom{00}
\\
\underline{-(18x^3-54x^2)} \phantom{000000000}
\\
45x^2+0x+18
\\
\underline{-(45x^2-135x)} \phantom{0^0}
\\
135x+18
\\
\underline{-(135x-405)}
\\
423
\end{array}
$
We're left with a remainder of $423$.
There are two standard ways to write the final result to include the remainder.
We can write a mixed fraction, which is common in Precalculus.
$
\begin{align}
\frac{6x^4-9x^2+18}{x-3}=6x^3+18x^2+45x+135+\frac{423}{x-3}
\end{align}
$
We can also write our solution based on the long division algorithm $P(x)=D(x)⋅Q(x)+R(x)$.
$
\begin{align}
6x^4-9x^2+18=(x-3)(6x^3+18x^2+45x+135)+423
\end{align}
$
Both forms are equivalent and show the relationship between the dividend, divisor, quotient and remainder.
### Synthetic Division
## Descartes Rule of Signs
## Rational Roots Theorem