Question

Factor each polynomial completely. If the polynomial cannot be factored, say it is prime. $$ x^{4}-1 $$

Short answer

\[x^4 - 1 = (x - 1)(x + 1)(x^2 + 1)\]

Step-by-step solution

- Recognize the Polynomial Type

The polynomial given is a difference of squares, which can be identified by the form \[x^{4} - 1^{4}\].

- Apply the Difference of Squares Formula

Use the difference of squares formula, \[a^2 - b^2 = (a - b)(a + b)\]. Here, \[a = x^2\] and \[b = 1\]. Therefore, \[x^4 - 1 = (x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1)\].

- Factor Further, if Possible

Notice that \[x^2 - 1\] can be factored further because it is also a difference of squares:\[x^2 - 1 = (x - 1)(x + 1)\]. The term \[x^2 + 1\] cannot be factored further using real numbers since it does not fit the difference of squares or any other common factorization formula.

- Write the Complete Factorization

Combine the factors to get the complete factorization of the original polynomial: \[x^4 - 1 = (x - 1)(x + 1)(x^2 + 1)\].

Key concepts

Difference of Squares

To start understanding the problem, we first need to recognize the concept of the difference of squares. This happens when we have two squared terms subtracted from each other. It looks like this: \[a^2 - b^2\].
The difference of squares is handy because it always factors into \[(a - b)(a + b)\].

Why does this happen? It's based on a property of multiplication where the middle terms cancel out. For example, \[(x - 1)(x + 1) = x^2 - 1\]. The same principle can be applied again if the conditions are met.
In our problem, \[x^4 - 1\] fits this pattern. We can identify \[x^4\] as \[(x^2)^2\] and \[1\] as \[1^2\], which transforms our expression into \[(x^2)^2 - 1^2\].

Polynomial Factorization

With the difference of squares identified, we can now apply the polynomial factorization method. Polynomial factorization involves breaking down a polynomial into simpler factors that, when multiplied together, give back the original polynomial.

Starting with \[x^4 - 1\], we use the difference of squares formula: \[a^2 - b^2 = (a - b)(a + b)\]. Here, \[a\] is \[x^2\] and \[b\] is \[1\]. Thus, \[x^4 - 1\] becomes \[(x^2 - 1)(x^2 + 1)\].

Next, we see if these factors can be broken down further. \[x^2 - 1\] is another difference of squares (\[x^2 - 1^2\]), so it factors to \[(x - 1)(x + 1)\].
However, \[x^2 + 1\] doesn't fit the difference of squares or any other simple factorization pattern within real numbers, so it stays as it is.

Complete Factorization

Now, let’s summarize our complete factorization. We started with \[x^4 - 1\], factored it as \[(x^2 - 1)(x^2 + 1)\], and then factored \[x^2 - 1\] further into \[(x - 1)(x + 1)\].
Thus, the complete factorization of \[x^4 - 1\] is:

• \[(x - 1)\]
• \[(x + 1)\]
• \[(x^2 + 1)\]

Combining these, we get the fully factored form: \[x^4 - 1 = (x - 1)(x + 1)(x^2 + 1)\].
Complete factorization means that we can no longer factorize any part of the polynomial into simpler polynomial factors. This method helps us simplify and solve polynomial expressions efficiently.