Problem 38 - Algebra 1: Concepts and Skills

Question

Factor the expression completely. \(5 s^{3}+30 s^{2}+40 s\)

Step-by-Step Solution

Answer

The factored form of the polynomial is \(5s(s+2)(s+4)\).

Step 1: Identify Common Factors

First, identify the common factors in the polynomial \(5s^{3} + 30s^{2} + 40s\). In this case, the common factor is \(5s\).

Step 2: Factor out the Common Factors

Next, factor out the common factor that was identified. Making \(5s\) the common factor, we can transform the polynomial to \(5s(s^{2} + 6s + 8)\).

Step 3: Factor the Remaining Polynomial

Factor the polynomial in the parenthesis \(s^{2} + 6s + 8\). This polynomial can be factored to \((s+2)(s+4)\).

Step 4: Combine the Factored Terms

Combine the factored common factor and factored polynomial. Therefore, the factored form of the original polynomial is \(5s(s+2)(s+4)\).

Key Concepts

Common Factor

A common factor in a polynomial is a term that divides each component of the polynomial without leaving a remainder. Identifying a common factor is usually the first step in polynomial factoring. In the given example, the polynomial is \(5s^3 + 30s^2 + 40s\).
To find the common factor, look at the coefficients 5, 30, and 40. All these numbers can be divided by 5 evenly, making 5 the numerical common factor. Also, each term contains the variable \(s\). Specifically, the smallest power of \(s\) in each term is \(s^1\). Therefore, \(s\) is also part of the common factor.
When you combine these, the common factor is \(5s\). Factoring the polynomial by the common factor simplifies further steps: \(5s \cdot (s^2 + 6s + 8)\). Recognizing and extracting common factors simplifies the polynomial and prepares it for further factoring steps.

Polynomial Factoring

Polynomial factoring involves breaking down a polynomial into simpler terms or factors that can be combined to reconstruct the original polynomial. After identifying and factoring out the common factor, the next stage focuses on the remaining polynomial.
In the example \((s^2 + 6s + 8)\), after extracting \(5s\) from each term, now only the expression inside the parentheses needs further factoring. This smaller polynomial is still a quadratic trinomial and needs to be simplified if possible.

• Look for two numbers whose product is the last term (8) and sum is the middle term (6).
• These numbers are 2 and 4, as \(2 \cdot 4 = 8\) and \(2 + 4 = 6\).

Thus, the expression \(s^2 + 6s + 8\) factors into \((s + 2)(s + 4)\). By breaking it into two binomial expressions, we've completely factored the polynomial, helping to solve further algebraic equations using these simpler building blocks.

Factoring Quadratics

Factoring quadratics is a fundamental algebraic skill. It involves expressing quadratic equations in their simplest factor form, which is crucial in solving equations and understanding the properties of quadratic functions.
The typical quadratic, \(ax^2 + bx + c\), can usually be factored by identifying two numbers that multiply to \(c\) and add up to \(b\). In our polynomial \(s^2 + 6s + 8\), the numbers are 2 and 4 because they satisfy both conditions: they add to give 6 and multiply to give 8.Factoring quadratics effectively transforms the polynomial into a product of two binomials and reveals the solutions or roots of the quadratic equation, when set equal to zero. Here, by factoring \(s^2 + 6s + 8\) into \((s + 2)(s + 4)\), we demonstrate finding roots of the equation \(s^2 + 6s + 8 = 0\) which are \(s = -2\) and \(s = -4\), solving the equation gracefully.
Understanding the quadratic factoring method not only simplifies expressions but also enhances problem-solving abilities in algebra and beyond.