Problem 72 - Algebra for College Students

Question

Should help you pull together all of the factoring techniques of this chapter. Factor completely each polynomial, and indicate any that are not factorable using integers. $$2 n^{3}+6 n^{2}+10 n$$

Step-by-Step Solution

Answer

The polynomial factors to \(2n(n^2 + 3n + 5)\) and cannot be factored further using integers.

Step 1: Identify Common Factors

Examine each term in the polynomial to see if there is a common factor. The terms are \(2n^3\), \(6n^2\), and \(10n\). All of these terms have a common factor of \(2n\).

Step 2: Factor Out the Common Factor

Factor out \(2n\) from each term of the polynomial: \[2n(n^2 + 3n + 5).\]Now the expression inside the parentheses \(n^2 + 3n + 5\) is what remains.

Step 3: Examine the Quadratic

The quadratic inside, \(n^2 + 3n + 5\), needs to be factored further if possible. The standard method for factoring such a quadratic is to find two numbers that multiply to the constant term (5) and add to the linear term (3). However, in this case, there are no such integer pairs.

Step 4: Conclude Factorization

Since the quadratic \(n^2 + 3n + 5\) cannot be factored any further using integers, the factorization is complete. Thus, the polynomial \(2n^3+6n^2+10n\) factors to \(2n(n^2 + 3n + 5)\).

Key Concepts

Common Factor

In polynomial factorization, the first step is often to identify and extract a common factor from all terms. This simplifies the expression and paves the way for further factorization. A common factor is a number or variable all terms in a polynomial share.

To identify a common factor, check the coefficients and variables of each term in the polynomial. For instance, in the polynomial \(2n^3 + 6n^2 + 10n\), all terms have the common factor \(2n\).

Here's how you can spot and extract it:

• Break down each term to its prime factors and variables.
• In this case, \(2n^3\) is \(2 \times n \times n \times n\), \(6n^2\) is \(2 \times 3 \times n \times n\), and \(10n\) is \(2 \times 5 \times n\).
• Recognize that \(2n\) is present in all terms.

After factoring out \(2n\), the polynomial becomes \(2n(n^2 + 3n + 5)\). This makes the polynomial simpler and easier to work with for further factorization attempts.

Quadratic Equation

A quadratic equation is a polynomial of degree two, typically expressed in the form \(ax^2 + bx + c\). In many polynomial problems, part of factorization involves dealing with such quadratic equations.

For the polynomial \(2n^3 + 6n^2 + 10n\), after factoring out the common term, you're left with the quadratic \(n^2 + 3n + 5\).

To factor a quadratic equation, look for two numbers that multiply to the product of the leading coefficient (\(a\)) and the constant term (\(c\)), while adding to the linear coefficient (\(b\)). Here's how you typically approach it:

• Write down all factor pairs of \(c\) that might add up to \(b\).
• In this case, you need numbers that multiply to \(5\) and add to \(3\).
• However, such a pair does not exist within the set of integers.

Thus, the quadratic \(n^2 + 3n + 5\) is termed as prime over the integers, meaning it can't be factored further using integers.

Polynomial Factorization

Polynomial factorization involves breaking down a polynomial into simpler polynomials that, when multiplied together, give the original polynomial. This technique is essential for simplifying complex expressions.

In the exercise \(2n^3 + 6n^2 + 10n\), factorization begins with identifying a common factor, which simplifies the polynomial to \(2n(n^2 + 3n + 5)\).

Further breaking down requires examining the quadratic \(n^2 + 3n + 5\) for possible factorization. However, as observed, this quadratic is considered irreducible over the integers because it doesn't factor neatly.

The process of polynomial factorization is vital because:

• It simplifies polynomials, making them easier to work with in equations.
• It aids in finding roots and solving equations.
• It reveals crucial properties of the polynomial like divisors.

In conclusion, for some polynomials, like this one, complete factorization using integers might hit a dead-end at the quadratic stage.