Polynomial Factoring

This polynomial type of calculator is a type of polynomial calculator that will allow you to put an expression as a multiplication of irreducible factors.

All you need to do is to provide a polynomial that you want to factor. It can be a polynomial of lower degree that comes already simplified, like x^2 - 2x + 3, of you can provide higher order polynomials that require simplification, like x^4 - x + 2x^4 - x^3 + 1.

Once you provide a valid polynomial expression, what you need to do next is to click on the "Calculate" button, and then you will get all the steps of the process shown to you .

Though they are among the simplest expression to be factored, polynomials still are hard to deal with in general, especially with polynomials of degree higher than 5.

How to factor polynomials

There one and only systematic way of factoring polynomials is finding its roots or zeros. Knowing its roots you will be able to find its factors, due to the Fundamental Theorem of Algebra.

For example, for a polynomial of degree 3, if there are three roots \(x_1\), \(x_2\) and \(x_3\), Fundamental Theorem of Algebra says that the polynomial can be written as:

\[\displaystyle p(x) = a (x-x_1)(x-x_2)(x-x_3) \]

for a constant \(a\), and the same would happen for a polynomial of degree \(n\), with \(n\) roots \(x_1\), \(x_2\), ...., \(x_{n-1}\) and \(x_n\), which can be written as:

\[\displaystyle p(x) = a (x-x_1)(x-x_2)(x-x_3)....(x-x_n) \]

What are the steps for factoring polynomials

• Step 1: Identify the polynomial you need to factor, and make any obvious expression simplifications if any
• Step 2: Find the polynomial roots, using a suitable method, depending on the degree of the polynomial
• Step 3: If the polynomial has degree 2, use the quadratic formula, otherwise, use the rational zero theorem
• Step 4: Once you have found all the roots, you can express the final factorization as \(\displaystyle p(x) = a (x-x_1)(x-x_2)(x-x_3)....(x-x_n) \)

The good thing about finding roots of polynomials is that you can go finding one root at time and making the problem progressively simple. Let me show how:

Suppose that you have one polynomial \(P(x)\) which you want to find all its roots. Say that the polynomial has degree 5, so you are expecting 5 roots, some of them not real (complex).

Say that by mere luck you found one root, say we call it \(x_1\). Then, by the Fundamental Theorem of Algebra you know that \(x-x_1\) divides \(P(x)\), so then \(P(x) = Q(x)(x-x_1)\), where \(Q(x)\) is a polynomial of degree 4.

You may be wondering "How do I get Q(x)??". Simple \(Q(x)\) is the obtained by using long division to divide \(P(x)\) by \(x-x_1\). We know that the remainder is zero because \(x_1\) is a root.

Do not forget you are trying to solve \(P(x) = 0\), so then we have now to solve \(Q(x)(x-x_1)\), which is reduced to solve \(Q(x) = 0\). So now you have another polynomial equation, only that simpler than the original. And then you proceed with this one trying to find one solution and repeating then the process.

Is there a simpler way to completely factor polynomials?

Not really. Anecdotally, you may factor by exploding some specific structures, you can do factor by grouping if possible, or you can exploit some obvious factor opportunities, like for example an expression like \(x^4 + x^2\) obviously lends itself to factor \(x^2\) out.

But all of these tricks are structure-dependent, which means, they need a specific simplified structure to work out, and they are by no means general ways to approach the problem.

For polynomials, the factored form equation and the actual roots provide the same information, except for a constant, which is the constant that goes with the leading term (the term with the highest exponent).

Why factor polynomials

Very simple, because it is the way to solve equations. We cannot skip the process of factoring polynomials because it is tightly connected with the process of solving polynomial equations.

The same happens with more general equations, where factoring can help to break down a complicated equation into simpler ones. Solving equations gets broken down into simpler problems if you are able to effectively factor and reduce expressions.

Example: Using polynomial factoring to solve equations

Solve the following equation: \(x^5 = -x^3\)

Solution:The usual approach consists of putting everything on one side of the equation. If your first reflex is to cancel x^2 from both sides of the equation, please refrain because you will lose solutions doing so. You will see why. So we start like this

\[x^5 = -x^3 \Rightarrow x^5 + x^3 = 0\]

and now we can factor \(x^2\) out:

\[x^5 = -x^3 \Rightarrow x^5 + x^3 = 0 \Rightarrow x^2(x^3 + 1)\]

Now, we use the old trick that tells us that \(x^3+1 = (x+1)(x^2-x+1)\), which means that

\[x^2(x^3+1) = x^2 (x+1)(x^2-x+1)\]

Now that you have the left hand side of the equation completely factored, we see that we need to solve

\[x^5 = -x^3 \Rightarrow x^2(x^3+1) = x^2 (x+1)(x^2-x+1) = 0\]

so we need to solve:

\[x^2 (x+1)(x^2-x+1) = 0\]

Now we use its factors to find the solutions, all we need to do is to set the factors equal to zero. The solutions of the equation are \(x = 0\), \(x = -1\) and \(x = \frac{-1 \pm i\sqrt 3}{2}\).

More polynomial calculators

Polynomials are very useful objects in Algebra, Calculus in Physics, and they are simple enough to have some very general and useful theorems, such as the Fundamental Theorem of Algebra (which by the states that all polynomial equations have has many complex solutions as its degree).

Yet, polynomials are hard enough to provide us with some polynomial equations and polynomial inequalities that cannot be solved with elementary methods, and you will need to try to reduce the degree of the polynomial by using the polynomial division and the Remainder Theorem.

So when dealing with objects that are more complex than polynomials, it is only reasonable to think that you will need a factor calculator that can detect complex structures and apply diverse identities to reach a proper factoring, and ultimately, it is not always possible.