Problem 51 - Elementary Algebra

Question

Use a pattern to factor. Check. Identify any prime polynomials. $$ n^{3}+1000 $$

Step-by-Step Solution

Answer

The factorization of n^3 + 1000 is (n+10)(n^2 - 10n + 100). Both factors are prime polynomials.

Step 1: Recognize the Sum of Cubes Pattern

The given expression is a sum of cubes. Recall the sum of cubes formula: The given expression can be written as o

Step 2: Apply the Sum of Cubes Formula

Using the sum of cubes formula, we can factor the expression into: (a + b)(a^2 - ab + b^2)

Step 3: Substitute Values

In this case, a = n and b = 10. Substituting these values into the formula gives: n^{3}+1000 = (n+10)(n^2 - 10n + 100)

Step 4: Check the Factors

Expand the factored form to check: (n + 10)(n^2 - 10n + 100) = n^3 + 1000. Since the original expression and the expanded form are equal, the factorization is correct.

Step 5: Identify Prime Polynomials

Check if any of the factors can be factored further. n+10 is a linear polynomial and is prime. n^2 - 10n + 100 is a quadratic polynomial and irreducible over the real numbers, thus it is also prime.

Key Concepts

sum of cubes

The given polynomial, \(n^3 + 1000\), is categorized as a sum of cubes. A sum of cubes is an equation of the form \(a^3 + b^3\). To factor it, we need to use a specific formula:
The sum of cubes formula is:
\(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\). Let's apply this to our polynomial.
Here, \(n^3\) can be written as \((n)^3\) and \(1000\) can be written as \((10)^3\).
By substituting \(a = n\) and \(b = 10\) into the formula, we factor \(n^3 + 1000\) as:
\( (n + 10)(n^2 - 10n + 100)\)

prime polynomials

A prime polynomial is one that cannot be factored into simpler polynomials over the given number set (such as real numbers).
After factoring the sum of cubes, we get two factors: \(n+10\) and \(n^2 - 10n + 100\).
Let's check their factorability.
The first factor, \(n+10\), is a linear polynomial and is prime. It cannot be factored further.
The second factor, \(n^2 - 10n + 100\), is a quadratic polynomial. Using the quadratic formula or factoring techniques, we see it cannot be reduced to simpler polynomials with real coefficients. Therefore, it is also prime.
We conclude both factors are prime.

quadratic polynomials

Quadratic polynomials are expressions of the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants.
In our example, after factoring the sum of cubes, we obtained: \(n^2 - 10n + 100\), which is a quadratic polynomial.
To determine if it is prime, we apply the quadratic formula:
\[n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
For \(n^2 - 10n + 100\), we have \(a=1\), \(b=-10\), and \(c=100\). Substituting these values into the formula:
\[\frac{10 \pm \sqrt{(-10)^2 - 4(1)(100)}}{2(1)}\]
Simplifies to:
\[\frac{10 \pm \sqrt{100 - 400}}{2} = \frac{10 \pm \sqrt{-300}}{2}\]
Since we get an imaginary number inside the square root (\(\sqrt{-300}\)), there are no real roots, affirming that this quadratic polynomial is prime over the real numbers.