Problem 54 - Algebra and Trigonometry

Question

\(53-56\) . 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

\( x-2 \) is a factor of the polynomial.

Step 1: State the Factor Theorem

The Factor Theorem states that \( x-c \) is a factor of a polynomial \( P(x) \) if and only if \( P(c) = 0 \). This means we need to evaluate \( P(x) \) at \( x = 2 \) and see if the result is zero.

Step 2: Substitute c into the Polynomial

To verify if \( x - 2 \) is a factor, substitute \( x = 2 \) into the polynomial. We have:\[P(2) = (2)^3 + 2(2)^2 - 3(2) - 10\]

Step 3: Calculate Each Term in the Polynomial

Calculate each term individually:1. \( (2)^3 = 8 \)2. \( 2(2)^2 = 8 \)3. \( -3(2) = -6 \)4. The constant term is \(-10\).

Step 4: Compute the Value of the Polynomial

Add up the values from the previous step:\[P(2) = 8 + 8 - 6 - 10 \]Simplifying further:\[P(2) = 16 - 16 = 0\]

Step 5: Conclusion Based on Factor Theorem

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

Key Concepts

Polynomial Evaluation

Polynomial evaluation is the process of calculating the value of a polynomial for a given input. In our exercise, this involved substituting the value of \( c = 2 \) into the polynomial \( P(x) = x^3 + 2x^2 - 3x - 10 \). By replacing \( x \) with \( 2 \), we transform the polynomial into a number.To evaluate the polynomial:

• Substitute \( x = 2 \).
• Calculate each individual term separately.
• Add up all these individual values.

When we do this, we end up with \( P(2) = 0 \). This step is crucial as it directly connects to the Factor Theorem, allowing us to check if the polynomial has a factor at \( x = 2 \). Evaluating polynomials thoroughly will ensure understanding and set a solid foundation for factorization steps.

Polynomial Factorization

Polynomial factorization involves breaking down a polynomial into simpler polynomials, known as factors, which when multiplied together give the original polynomial. By the Factor Theorem, we know that if \( x - c \) is a factor, then \( P(c) = 0 \). For our exercise, we found that \( P(2) = 0 \), so \( x - 2 \) is indeed a factor of \( P(x) \).Polynomial factorization is key in solving equations, simplifying expressions, and sketching polynomial graphs. Here are steps you typically need to follow:

• Use the Factor Theorem as a starting point.
• Find other potential factors through techniques like synthetic division or long division.
• Check for possible common factors.
• Keep factoring until the polynomial is expressed as a product of irreducible factors.

Factorizing can initially seem complex but achieving smaller, manageable factors is a rewarding step forward.

Algebraic Verification

Algebraic verification is all about confirming our results using algebraic principles. In the context of this exercise, it's about double-checking that our polynomial evaluation and factorization results hold true. After evaluating it and establishing that \( P(2) = 0 \), it's important to further verify that \( x - 2 \) is a factor. This verification can involve:

• Re-evaluating the polynomial at \( x = 2 \) just to ensure no calculation errors were made.
• Using polynomial division to see if dividing \( P(x) \) by \( x - 2 \) results in no remainder.

Conducting an algebraic verification increases the reliability of your work. It ensures that the solution is correctly derived and supports deeper understanding.