Problem 89 - Elementary Algebra

Question

Factor completely. Identify any prime polynomials. $$ x^{13}-x^{5} y^{2} $$

Step-by-Step Solution

Answer

\( x^5 (x^4 - y) (x^4 + y) \)

Step 1: Identify common factors

Look at the given polynomial: \( x^{13}-x^{5} y^{2} \). Both terms have a common factor involving \( x \). Find the greatest common factor (GCF) of the powers of \( x \), which is \( x^5 \).

Step 2: Factor out the GCF

Factor out \( x^5 \) from the polynomial \( x^{13}-x^{5} y^{2} \). This yields: \[ x^5 (x^{8} - y^{2}) \]

Step 3: Recognize difference of squares

Notice that \( x^{8} - y^{2} \) is a difference of squares since \( x^{8} = (x^{4})^2 \) and \( y^{2} = (y)^2 \). Use the difference of squares formula: \( a^2 - b^2 = (a - b)(a + b) \), with \( a = x^4 \) and \( b = y \).

Step 4: Apply the difference of squares formula

Applying the difference of squares to \( x^8 - y^2 \), we have: \[ x^8 - y^2 = (x^4 - y)(x^4 + y) \]Therefore, the factored form is: \[ x^5 (x^4 - y)(x^4 + y) \]

Key Concepts

Greatest Common Factor

The greatest common factor (GCF) is the highest number or terms that can divide each term of the polynomial without leaving a remainder.

To find the GCF, we look for the largest factor common to each term in the polynomial.

In the given polynomial, $$x^{13} - x^{5} y^{2}$$, both terms contain the variable $$x$$. We compare the exponents of $$x$$ in each term:

• In the first term, the exponent of $$x$$ is 13
• In the second term, the exponent of $$x$$ is 5

The smallest exponent is 5, so the GCF is $$x^{5}$$.

The next step is to factor out $$x^5$$ from the polynomial, which simplifies it to:
$$ x^{13} - x^{5} y^{2} = x^{5} (x^{8} - y^{2})$$. This expression is now easier to work with for further factorization.

Difference of Squares

A difference of squares is a specific type of polynomial where two squared terms are subtracted.

The general formula is: here,
$$a^2 - b^2 = (a - b)(a + b).$$

In our polynomial $$x^{8} - y^{2}$$, we can recognize it as a difference of squares:

• $$x^8$$ can be written as $$(x^{4})^{2}$$
• $$y^2$$ is already in the desired form $$(y)^{2}$$

Thus, we have:
\begin{split} a & = x^4 \ b & = y \text\br> With this we use the difference of squares formula:
$$x^8 - y^2 = (x^4)^{2} - (y)^{2} = (x^4 - y)(x^4 + y).$$

Prime Polynomials

A prime polynomial is a polynomial that cannot be factored into the product of two non-constant polynomials.

They are the equivalent of prime numbers in algebra.
(with all simple)
For example, the polynomial $$x^2 + 1$$ is prime because it cannot be factored into real numbers.

In our example, we have further factored our original polynomial: $$x^5 (x^4 - y)(x^4 + y)$$.

After recognizing and factoring out the GCF and using the difference of squares, we've simplified the expression into a product of polynomials.
Each polynomial part, $$x^5$$, $$x^4 - y$$, and $$x^4 + y$$,

cannot be factored further using real numbers, making them prime.

Recognizing prime polynomials helps in knowing when your factorization process is complete. Explanation of exercises, especially those involving polynomials, can seem complex initially.

However, breaking steps down, identifying the greatest common factor, recognizing patterns such as the difference of squares, and understanding prime polynomials, can significantly simplify the process.