Problem 54 - Precalculus Mathematics for Calculus

Question

Use the Factor Theorem to show that \(x-c\) is a factor of \(P(x)\) for the given value(s) of \(c\). $$P(x)=x^{3}+2 x^{2}-3 x-10, \quad c=2$$

Step-by-Step Solution

Answer

Since \(P(2) = 0\), \(x-2\) is a factor of \(P(x)\).

Step 1: State the Factor Theorem

The Factor Theorem states that \(x - c\) is a factor of \(P(x)\) if and only if \(P(c) = 0\). This means that if you substitute \(c\) into \(P(x)\) and the result is zero, then \(x-c\) is a factor of \(P(x)\).

Step 2: Substitute \(c = 2\) into \(P(x)\)

Substitute \(c = 2\) into the polynomial to find \(P(2)\).\[P(2) = 2^3 + 2(2)^2 - 3(2) - 10\]

Step 3: Calculate Each Term in \(P(2)\)

Calculate each term separately:- The term \(2^3 = 8\).- The term \(2(2)^2 = 8\).- The term \(-3(2) = -6\).- The constant term is \(-10\).

Step 4: Sum the Values to Find \(P(2)\)

Sum the results:\[8 + 8 - 6 - 10 = 0\]

Step 5: Conclude Based on \(P(2)\)

Since \(P(2) = 0\), according to the Factor Theorem, \(x-2\) is a factor of \(P(x) = x^3 + 2x^2 - 3x - 10\).

Key Concepts

Polynomials

Polynomials are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. They're like a magical toolbox in mathematics. Let's break it down. Imagine a polynomial as a recipe, where you combine different ingredients (terms) in a specific way. Each term consists of a variable (usually represented by \(x\)) raised to a power, multiplied by a constant (coefficient). For example, in the polynomial \(x^3 + 2x^2 - 3x - 10\), you have four terms:

• \(x^3\) : This is the third power of \(x\), representing cubic growth.
• \(2x^2\) : Here, \(x\) is squared and multiplied by 2.
• \(-3x\) : This shows linear growth multiplied by negative 3.
• \(-10\) : A constant term, adding a specific value to the polynomial.

All these terms are combined through addition or subtraction. Polynomials can be classified by degree, which is the highest power of their variable. In this case, it's a cubic polynomial because the highest power is \(x^3\). Understanding the components and degree of polynomials is crucial for solving equations and understanding their behavior.

Roots of Polynomials

Roots of polynomials are the values of \(x\) that make the polynomial equal to zero. They're like the magic key to unlocking solutions. When you find a root, it means you've found an \(x\)-value where the entire expression becomes zero. This concept is essential when you're dealing with polynomial equations.

For example, if you substitute \(x = 2\) into the polynomial \(P(x) = x^3 + 2x^2 - 3x - 10\), and find that the result is zero, you've discovered a root. In our exercise, it was demonstrated that \(P(2) = 0\). This shows that \(x = 2\) is indeed a root of the polynomial. When \(x - c\) is a factor, \(c\) is a root. Thus, it tells us that the entire expression can be divided without a remainder by \(x - 2\).

Finding roots is critical in many math and engineering fields, helping solve practical problems such as optimizing systems, understanding curves, and predicting future behavior.

Synthetic Division

Synthetic division is a simplified way of dividing polynomials, particularly useful when dividing by a linear factor like \(x - c\). Think of it as the fast-track method in polynomial long division, allowing for quick calculations without the mess.

When performing synthetic division to test if \(x - c\) is a factor of a polynomial \(P(x)\), you follow a set process using just the coefficients. If \(P(c) = 0\), then there's no remainder, confirming \(x - c\) as a factor. However, the division process itself, though not explicitly done in our solution, validates the Factor Theorem.

To do synthetic division:

• Write down all coefficients of the polynomial, ignoring any terms with zero coefficients for simplicity.
• Use the value of \(c\) and perform operations sequentially on these coefficients.
• Perform the calculation step-by-step to get the result.

This method follows neat arithmetic steps that are well-suited for handling complex problems with ease. By using synthetic division, you can quickly confirm factorization results or simplify polynomial expressions for other operations. It's a practical and efficient tool in your polynomial-solving arsenal!