# Polynomial Calculator

**
Instructions:
**
Use this polynomial equation calculator to solve any polynomial equation, showing all the steps. Please type in the polynomial equation you
want to solve.

Note that some equations may have complex roots and higher order equations may not be solved with elementary methods).

## Polynomial Equation Calculator

This polynomial equation solver will assist you at solving polynomials equations that you provide, such as for example '3x^2 - 2/3 x + 1/4 = 0' , which is a simple quadratic equation, or polynomial equations of higher order such as 'x^5 - x^2 + 1 = 0', etc.

If you don't add an equality sign "=" to the expression provided, the calculator will automatically add a " = 0" to it so to convert it to an equation..

Once a valid polynomial equation has been provided, you can then click on the "Calculate" button, and then you will be presented with the step-by-step calculation of the solutions of the equation solutions.

A polynomial equation is a type of algebra equation, and one of the simplest kinds, barred from linear equations. The fact that polynomial equations are simple, does not mean they are EASY to solve, and in fact, at times they take quite a long to be solved, if they can be solved at all.

## How can I solve polynomial?

Though polynomials are simple expressions, solving polynomial equations can be really complicated, especially for polynomial degree greater than 2.

For quadratic equations, the solutions are simply found using a quadratic formula. Sure, you may think it is hard to memorize the formulas, but at least there is a formula.

For cubic (degree 3) and quartic (degree 4), there are some very clever equations to be used, but they are by no means easy to use or remember. For poly equations of degree 5 and above there is no formula.

That does not mean we cannot find the polynomial roots for those equations , but we don't have a formula for it, and a formula does not exist (if you are curious about that, such conclusions was one of the main breakthroughs of modern mathematics in the late 18th century.

## Steps for Finding Solutions to a Polynomial Equation

There are number of systematic steps you can follow to have the best chances at finding the solutions to a polynomial equation, but be aware that you may end up not finding any solutions, specially for higher degree equations.

**Step 1:**Be aware that in theory there are \(n\) solutions to an polynomial equation of degree \(n\). But those solutions may be real or complex, and beyond degree 4, there is no formula for them**Step 2:**Try to factor the polynomial terms. Put all the terms in one side of the equation and look a way to factor the polynomial expression. By factoring you can try to find solutions to each factor, reducing the problem to lower degrees**Step 3:**Try to find rational/integer solutions first using the rational zero theorem. This is achieved by finding integer factors of the constant term, dividing them by factors of the leading term (the one that goes with the highest power)**Step 4:**Using these rational candidates, you test them one by one (there could be many of them), in hopes you find solutions. If by chance you found \(n\) solutions to an equation of degree \(n\), then you finish**Step 5:**If you found one or more rational roots, but not all of them, you construct a multiplication of the terms \(x - \alpha\), where \(\alpha\) is a rational root found. Multiply those terms, form a polynomial and then DIVIDE the polynomial of the original equation by this product consisting of the terms \(x - \alpha\). To find the remaining roots, you need to find roots of the result of the division (which will have a lower degree than the original polynomial.

It sounds difficult, and honestly, it is. It is a cumbersome process, that requires lots of calculations, very likely. That is why you should use an equation calculator that will show you steps, because you will save a lot of time and will minimize your chances of making an error in the calculation .

## How do you find the equation of a polynomial?

Solving polynomial equations is definitely not a trivial task. You won't be able to do it in general, as there is no general equation to solve ALL polynomials. We do know, by virtue of the Fundamental Theorem of Algebra that there are \(n\) solutions to an polynomial equation of degree \(n\).

As the name suggests it, these result is a major accomplishment because it tells us exactly HOW MANY solutions we are on the lookout. For example, if we have the equation \(x^4 = x^6\), what we have is an equation of degree 6 (because it is the highest polynomial power that can be found there). Hence, by the Fundamental Theorem of Algebra, we KNOW there are 6 solutions.

Now, it can be tricky because not all solutions will be real, some could be complex, and some could be repeated. If we had say a polynomial of degree \(n\), then we know there are \(n\) solutions, and another notable thing stated by this theorem, is that the polynomial part can be written as

\[\displaystyle p(x) = (x - \alpha_1) (x - \alpha_2) (x - \alpha_3) \cdot\cdot\cdot (x - \alpha_n) \]where \(\alpha_1\), ..., \(\alpha_n\) are the solutions. But it can happen that not all the solutions are different. In fact, we could have something like

\[ p = (x - \alpha)^n\]indicating that all n solutions are the same.

## What are the rules for polynomials?

**Step 1:**Polynomials are linear combinations of expressions of the form \(x^k\)**Step 2:**The polynomials we are interested in are those with terms \(x^k\), only with \(k\) integers**Step 3:**Polynomials are a simple type of functions that can be added, subtracted, multiplied and divided.

Observe that polynomial operations are not closed. Notice that when adding, subtracting and multiplying polynomials, the result will always be a polynomial. But when dividing polynomials, the result will not necessarily be a polynomial, though the division and remainder will be polynomials. Check the polynomial long division algorithm.

## What is a polynomial equation and how do we solve it?

A polynomial equation, put simply, is a math equation in which the terms in the left and right hand side of the equation are polynomials. Usually, these equations are given with a constant on the right hand side, but that is not always the case.

For example, \(x^2 + 3x = 2\) is a polynomial equation, because the terms in both sides of the equation are polynomials (the constant '2' is a polynomial of order 0).

But, \(x^2 + \sin(x) = 2x\) is NOT a polynomial equation, because the terms in the left hand side is not a polynomial (because of the presence of the \(\sin(x)\) term.

### Example: Calculating solutions to polynomial equations

Calculate the solution to: \(x^2 = x^4\)

Solution:

We need to solve the following given polynomial equation:

\[x^2=x^4\]The equation we need to solve has only one variable, which is \(x\), so the objective is to solve for it.

Observe that the degree of the given polynomial is \(\displaystyle deg(p) = 4\), its leading coefficient is \(\displaystyle a_{4} = -1\) and its constant coefficient is \(\displaystyle a_0 = 0\).

Since the first term with a non-zero coefficient in \(p(x)\) is \(x^2\), we can factor this term out to get:

\[\displaystyle p(x) = -x^4+x^2 = x^2 \left(\displaystyle -x^2+1 \right) \]but the term in parenthesis has degree 2, and we need to see if it can be further factorized.

We need to solve the following given quadratic equation \(\displaystyle -x^2+1=0\).

For a quadratic equation of the form \(a x^2 + bx + c = 0\), the roots are computed using the following formula:

\[x = \displaystyle \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]In this case, we have that the equation we need to solve is \(\displaystyle -x^2+1 = 0\), which implies that corresponding coefficients are:

\[a = -1\] \[b = 0\] \[c = 1\]First, we will compute the discriminant to assess the nature of the roots. The discriminating is computed as:

\[\Delta = b^2 - 4ac = \displaystyle \left( 0\right)^2 - 4 \cdot \left(-1\right)\cdot \left(1\right) = 4\]Since in this case we get the discriminant is \(\Delta = \displaystyle 4 > 0\), which is positive, we know that the equation has two different real roots.

Now, plugging these values into the formula for the roots we get:

\[x = \displaystyle \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \displaystyle \frac{0 \pm \sqrt{\left(0\right)^2-4\left(-1\right)\left(1\right)}}{2\cdot -1} = \displaystyle \frac{0 \pm \sqrt{4}}{-2}\]so then, we find that:

\[ {x}_1 = \frac{0}{-2}-\frac{1}{-2}\sqrt{4}=\frac{0}{-2}-\left(-1\right)=1 \] \[{x}_2 = \frac{0}{-2}+\frac{1}{-2}\sqrt{4}=\frac{0}{-2}-1=-1\]In this case, the quadratic equation \( \displaystyle -x^2+1 = 0 \), has two real roots, so then:

\[\displaystyle -x^2+1 = - \left(x-1\right)\left(x+1\right)\]so then the original polynomial is factored as \(\displaystyle p(x) = -x^4+x^2 = - x^2 \left(x-1\right)\left(x+1\right) \), which completes the factorization.

**Conclusion**: Therefore, the final factorization that we obtain is:

The roots found using the factorization process are \(0\), \(1\), and \(-1\) .

## Other useful equation calculators

Equation solvers are really important in math, as equations are usually the way we express the association between related quantities. Being able to solve equations will uncover some special points that satisfy some specific equality.

General calculators are difficult to achieve as different equation structures will require different solving strategies. A trigonometric equation calculator will usually exploit the relationship between different trigonometric functions in order to find solutions, the same way as exponential equations and logarithmic equations will have their own approaches, based on key properties held by exponents and logarithms, respectively. .

Most algebra problems can be represented, so then by solving equations we are finding the key to those algebra problems, those special points that satisfy specific properties of interest.

Solving equations is not easy in general. You may follow certain useful strategies, like rearranging equations, factoring, or simplifying expressions. But ultimately, each type of equation will give you a type of structure which will unveil the path to it solution

For example, for radical equations you most certainly need to solve for the term that has a root, and use a power to eliminate the root, turning it into a polynomial equation. But that route, which work perfectly for a radical equation, may not work for a trigonometric equation, for example.