Problem 35 - Elementary Algebra

Question

Factor completely. Identify any prime polynomials. $$ c^{5}+14 c^{3}+48 c $$

Step-by-Step Solution

Answer

The completely factored form is \(c(c^2 + 6)(c^2 + 8)\).

Step 1: Identify Common Factors

First, identify the greatest common factor (GCF) of all the terms in the polynomial. The terms are: \(c^5\), \(14c^3\), and \(48c\). The GCF is \(c\).

Step 2: Factor Out the GCF

Factor out the common factor \(c\) from each term in the polynomial. This yields: \[ c(c^4 + 14c^2 + 48) \]

Step 3: Identify a Quadratic Form

Recognize that the expression inside the parentheses \(c^4 + 14c^2 + 48\) is a quadratic in terms of \(c^2\). Let’s set \(u = c^2\), then the expression becomes: \[ u^2 + 14u + 48 \]

Step 4: Factor the Quadratic Expression

Factor the quadratic expression \(u^2 + 14u + 48\). We need to find two numbers that multiply to 48 and add to 14. These numbers are 6 and 8, so we can write: \[ u^2 + 14u + 48 = (u + 6)(u + 8) \]

Step 5: Substitute Back \(c^2\) for \(u\)

Substitute \(c^2\) back for \(u\) to get: \[ (c^2 + 6)(c^2 + 8) \]

Step 6: Write the Final Factored Form

Combine all the factored parts to give the completely factored form of the original polynomial: \[ c(c^2 + 6)(c^2 + 8) \]

Key Concepts

greatest common factor (GCF)

The greatest common factor, or GCF, is the largest factor that divides two or more numbers. It’s a crucial first step in polynomial factorization because it simplifies the original expression.

For the polynomial \(c^5 + 14c^3 + 48c\), we need to find the GCF of the terms \(c^5\), \(14c^3\), and \(48c\).
Here, each term has at least one \(c\), so the GCF is simply \(c\).

By factoring out the GCF, we make the polynomial easier to handle:
\[ c(c^4 + 14c^2 + 48) \]
This step significantly simplifies the expression and prepares it for further factorization.

quadratic form

Once the GCF is factored out, the simplified polynomial takes on a quadratic form. A quadratic form refers to an expression of the type \(ax^2 + bx + c\).

In our case, we have:
\[ c^4 + 14c^2 + 48 \]
Notice that if we set \(u = c^2\), the expression converts to a more recognizable form for factoring:
\[ u^2 + 14u + 48 \]
Recognizing this quadratic form is essential because it allows the use of standard quadratic factorization techniques.

factoring quadratics

Factoring a quadratic polynomial involves finding two binomials that multiply together to give the original quadratic expression.

For the polynomial \(u^2 + 14u + 48\), we need two numbers that multiply to 48 and add to 14.

The numbers are 6 and 8 because:

• 6 * 8 = 48
• 6 + 8 = 14

This allows us to factor the polynomial as:
\[ (u + 6)(u + 8) \]
Substituting back \(c^2\) for \(u\) gives:
\[ (c^2 + 6)(c^2 + 8) \]
Finally, combining it with the GCF extraction, the fully factored form of the original polynomial is:
\[ c(c^2 + 6)(c^2 + 8) \]

prime polynomial

A prime polynomial cannot be factored any further over the set of integers. After factorization, inspect each part to ensure it can't be simplified any more.

For the polynomial \(c(c^2 + 6)(c^2 + 8)\):

• \(c\) is already in its simplest form
• \(c^2 + 6\) is a sum of squares and cannot be factored further over the integers
• \(c^2 + 8\) is also a sum of squares and is already a prime polynomial

Therefore, the factors \(c\), \(c^2 + 6\), and \(c^2 + 8\) are all in their simplest forms.

Understanding prime polynomials is important to verify the completeness of our factorization.