## Introduction: How to Fully Solve Polynomials- Finding Roots of Polynomials

A polynomial, if you don't already know, is an expression that can be written in the form *a**sub(n) x^n + a sub(n-1) x^(n-1) + . . . + a sub(2) x^2 + a sub(1)x + a sub(0).* An expression is only a polynomial when it meets the following criteria:

1. There are no fractions that contain variables in the denominator (5/x).

2. There are only whole number exponents (0, 1, 2, . . .)- no fractions or negative exponents (x^-2).

3. There are no radical expressions that contain variables (sqrt(2x)).

4. All coefficients are real, though they may be fractions, radicals, or irrational numbers (4ix).

## Step 1: The History of Polynomials and Personal Interest

**History**

Polynomials were not always around, but the earliest forms of "polynomials" were found as far back as in Mesopotamia during the Babylonian period. They did not always take the practical form as they do today, and were often seen as word problems or another form. After algebra in general started to pick up speed, mathematicians developed a geometrical approach, which later helped in finding what is now the root of a quadratic equation. In the Whetstone of Witte, by Robert Recorde, the first appearances of addition, subtraction, and equal signs were made, as well as a letter variable for an unknown value (note that other variables were used in times before, but may not have been a letter like we use today). Mathematics continued to develop when Rene Descartes, an important mathematician in the 1500s, discovered the graph of a polynomial equation. He also created Descartes' Rule, a rule that is quite useful when solving polynomials.

**Personal Interest**

I never used to like math. I despised it in elementary school, but upon starting junior high, I entered into algebra straightaway. It was a huge problem for me, since I had never really done pre-algebra, but I worked hard and ended up passing the class with an A. From there on, I hadn't really noticed my passion for math until I started algebra II and then it hit me that I found math fun and enjoyable. I find the process for solving polynomials tedious, and not very fun compared to math theory or understanding paradoxes, but it has many steps and can branch off depending on the circumstances. That makes it interesting and something I can write about in a step-by-step instructional. Rather than saying I'm completely immersed and interested in mathematics, a more specific idea is that I love the theory of math and paradoxes that involve mathematics. They're just so fun and interesting that it completely grabs me and holds me in a trance. In fact, astronomy and space gives me the same feeling, so I decided that in the future, I'd want to go into astrophysics :)

## Step 2: What You Need to Know (sort Of)

This is written assuming that the reader has basic knowledge in the following:

Polynomial identities

Synthetic division

Relative extrema

Upper bounds and lower bounds

Fundamental Theorem of Algebra

Conjugate Radical Theorem

Complex Conjugate Theorem

End behavior

Multiplicity

If the reader understands these concepts, feel free to skip the next section.

## Step 3: What You Need to Know (sort Of) (cont.)

You're most likely here because you did not understand at least one of the concepts in the previous section. Here's a quick overview of each. If you'd like to dig in further on any of these concepts, I suggest you Google it :)

**Polynomial Identities-**

"Shortcuts to getting where you want to go without the messy details." There are different kinds of identities so I suggest you search them up.

Ex. *a +b = (a + c) + (b - c)*, with *c* as the common element

**Synthetic division-**

A form of division where you take the coefficients of the polynomial and put them where the dividend typically goes and the *a *from *( x - a ) *where the divisor typically goes, *( x - a ) *being a plausible factor of the polynomial.

**Descartes' Rule of Signs (was not mentioned previously)-**

"A rule that states that the number of positive roots is less than or equal to the number of sign changes for the terms of a polynomial function, *f(x)*. Also, the number of negative roots is less than or equal to the number of sign changes for the terms of a polynomial function, *f(-x)*. In each case, if the number of roots is less than the number of sign changes by a multiple of 2."

**Relative extrema-**

Relative maximums or minimums of the graph of a polynomial.

**Upper bounds and lower bounds-**

"When using synthetic division, an upper bound is a number that is greater than or equal to the real-number roots for a polynomial. Similarly, a lower bound is a number that is less than or equal to all of the real-number roots for a polynomial. They can be found when using synthetic division."

*Upper bound-* "for a positive number *c, *if *f(x)* is divided by *( x - c ) *and the resulting quotient polynomial and remainder have no changes in sign, then *f(x)* has no real roots greater than *c."*

*Lower bound- "*for a negative number *c,* if *f(x) *is divided by *( x - c ) *and the resulting quotient and remainder have alternating signs, then *f(x) *has no real roots less than *c."*

**Fundamental Theorem of Algebra-**

"A polynomial function *f(x)* of degree *n *has exactly *n *roots, or zeros, as long as you permit complex numbers to be considered zeros. Note that real numbers are a subset of the complex numbers."

**Conjugate Radical Theorem-**

"If *a + sqrt(b) *is a root, then *a - sqrt(b) *is also a root."

**Complex Conjugate Theorem-**

"If *a + bi *is a root, then *a - bi *is also a root."

**End behavior-**

The way a graph's behavior is depending on the degree of the function and the leading coefficient. More specifically, the way the ends of the graph face when they continue infinitely.

**Multiplicity-**

"Two identical roots. They must be included when you say that a polynomial of degree *n *has *n* roots."

## Step 4: Use the Fundamental Theorem of Algebra...

"Use the Fundamental Theorem of Algebra to identify the total number of roots in a polynomial."

(If you need to go back a section to review what the Fundamental Theorem of Algebra is, go ahead).

Take the equation 10x^3-10x^2-32, for example. The degree of the function is the highest degree, and the degree of the first term when put in standard form. The Fundamental Theorem of Algebra ultimately says that the degree of the polynomial, *n,* is how many roots the polynomial will have as long as you are counting complex numbers (which we are). That means that this polynomial will have 3 roots, whether complex or not.

## Step 5: Use the Rational Root Theorem...

"Use the Rational Root Theorem to create a list of possible rational roots."

The Rational Root Theorem is a theorem that says all possible combinations of *(p/q) *are all the possible roots for a function, *p *being the factors of the constant term and *q *the factors of the leading coefficient.

Take for example the same equation, 10x^3-10x^2-32. The *p* are the factors of the constant term, or in other words, -32. The factors of -32 are 1, 2, 4, 8, 16, and 32 (all of them being either positive of negative marked by the symbol +-). The factors of the leading coefficient, *q, *are the factors of 10 since the coefficient of 10x^3 is 10. That means that all possible combinations are:

+-1/1, +-1/2, +-1/5, +-1/10, +-2/1, +-2/2, +-2/5, +-2/10, +-4/1, +-4/2, +-4/5, +-4/10, +-8/1, +-8/2, +-8/5, +-8/10, +-16/1, +-16/2, +-16/5, +-16/10, +-32/1, +-32/2, +-32/5, +-32/10

## Step 6: Use Descartes' Rule of Signs...

"Use Descartes' Rule of Signs to determine the numbers of positive and negative roots. Remember that multiplicity applies."

So looking at the equation 10x^3-10x^2-32, we can look at the signs of each of the terms. It starts with a positive 10x^3, so a +, and then has two negative terms following. That would make the sequence + - -. There is one sign change, so there is one positive real root.

To find the number of negative roots, we multiply *x* by -1. The new equation would be -10x^3-10x^2-32. That makes the sequence to be - - -, and no sign changes, so no negative real roots.

But what about the complex roots?

Well we know we're going to have 1 real root and no negative roots, but we need to have 3 roots in total, whether complex or not. That means that there will be 2 complex roots to fill in for the other roots to make a total of three; one positive real root, and two complex roots.

## Step 7: Use Synthetic Division...

"Use synthetic division to try to find roots."

Using the list of possible roots we found before (*p/q)*, and use those as the divisor for synthetic division, and the polynomial 10x^3-10x^2-32 for the dividend.

When solving using synthetic division, the goal is to have no remainder, so the space all the way to the right must be a zero. Once there is no remainder, we know the divisor in that equation is a root of the dividend because the remainder is always the corresponding y-coordinate to the x-coordinate (divisor). In other words, *f(x) = f(a) *(remember, the possible roots that we found before would go in as ( x - a )).

Once testing the roots (also while looking for the upper bound and lower bound since they help make things faster), you should find that *nothing works.*

At least from that set of numbers.

Remember that the whole *(p/q) *thing only looks for all *possible roots.* Normally, one would assume that the actual root is a possible root and ultimately included in the set, but that is not always true. Instead, for this equation, you'd have to plug 0 in for *y* and solve for *x*, but know that typically using *(p/q) *works for most polynomials.

**Why?**

You may be wondering why *p/q* doesn't work for everything, and the reason that probably is is because of derivatives and the shifting of the graph at any point in time. The possible roots found when using the Rational Root Theorem are only pertaining to the change in the graph's motion or direction and would be used, along with sigma, to find derivatives and solve them as well. That's why the actual root does not have to be included in the Rational Root Theorem's method, even if it is a rational root.

## Step 8: Factor the Polynomial...

"Factor the polynomial using synthetic division, factoring rules, or the quadratic formula. Remember that the conjugate radical theorem and complex conjugate theorem apply."

Let's say that you could use synthetic division to find the roots of a polynomial unlike the last equation. If we had used a different equation (one that worked with synthetic division), we would factor out the factor of the polynomial, and most likely end up with a quadratic equation. Then we can use factoring rules or the quadratic formula to finish solving the polynomial.

(I assume you know how to use the quadratic formula and know the factoring rules since that's basic algebra).

To make things easier, keep in mind that the conjugate radical theorem and complex conjugate theorem apply, so while you're going through the steps to factor, see if you can find any times that the two may be useful.

## Step 9: Identify the Roots...

"Identify the roots that are solutions."

Basically, all you need to do is identify them. You can write them in the form *( x - a )( x - a )( x - a )* or just identify the roots. For example:

10x^3-10x^2-32

( x - (1/15)((5+cbrt(5525-300sqrt(399))+5^(2/3)cbrt(221+12sqrt(339))) )( x - (1/3)+(1/6i)(sqrt(3)+i)cbrt((1/5)(221-12sqrt(339)))-(1/6)(1+isqrt(3))cbrt((1/5)(221+12sqrt(339))) )( x - (1/3)-(1/30)(1+isqrt(3))cbrt(5525-300sqrt(339))+(1/6i)(sqrt(3)+i)cbrt((1/5)(221+12sqrt(339))) )

or

Roots:

(1/15)((5+cbrt(5525-300sqrt(399))+5^(2/3)cbrt(221+12sqrt(339))), (1/3)+(1/6i)(sqrt(3)+i)cbrt((1/5)(221-12sqrt(339)))-(1/6)(1+isqrt(3))cbrt((1/5)(221+12sqrt(339))), and (1/3)-(1/30)(1+isqrt(3))cbrt(5525-300sqrt(339))+(1/6i)(sqrt(3)+i)cbrt((1/5)(221+12sqrt(339)))

## Step 10: Graphing the Polynomial

I can only really help you with the steps to graph it, since this isn't supposed to be primarily on how to graph polynomials, but you can use the roots and relative extremas to *sketch* the graph out. You know that the roots are where the graph will be crossing the x-axis, so we can mark those on our graph. Then using end behavior rules and the idea that *n-1* is the number of times the graph will be switching directions, you can sketch out an approximation of what the actual graph would look like.

If you wanted exact measurements, you'd have to do a bit more, but it's tedious and practically pointless to me, so I typically wouldn't do it unless you've specifically been told to.

(Being honest here, I don't fully understand how to graph them completely either so I'm not going to).

## Step 11: Thank You! and Bibliography

That's basically it. It's probably hard to understand for a lot of people, but I'm just a student doing this for her language arts project :) Thank you for reading and I hope you learned something.

Made by Kelly Oh ㅎㅅㅎ

**Bibliography**

Apex Learning, Algebra II, Semester 1, Unit 5, 1997, course.apexlearning.com. Accessed 31 Jan 2019.

IXL Learning, 1998, https://www.ixl.com/math/algebra2. Accessed 31 Jan 2019.

## 5 Comments

4 years ago

It's a lot easier to use a graphics calculator like the Casio CFX-9850GB ! Your explanation is quite clear, however. Well done.

Reply 4 years ago

thank you ^^

4 years ago

to everyone who even reads my instructable (and sees this): IM SO SORRY. it's so terribly explained (especially why p/q doesn't work all the time) and i didn't expect so many people to look at this..if anyone can elaborate on why the rational root theorem doesn't work all the time or on how to fully a graph a polynomial (other than just plugging in numbers), i'd appreciate comments :) i also take constructive criticism but please don't be mean. thank you so much and i apologize again <3

4 years ago

Very nicely explained :)

Reply 4 years ago

thank you so much ! i feel like it could've definitely been better, but it's for a language arts project which demands a limited number of steps and some other criteria which needed to be met. i really appreciate the feedback <3