Problem 53 - Algebra and Trigonometry

Question

Use the Factor Theorem to show that \(x-c\) is a factor of \(P(x)\) for the given value(s) of \(c .\) \(P(x)=2 x^{3}+7 x^{2}+6 x-5, \quad c=\frac{1}{2}\)

Step-by-Step Solution

Answer

\( x-\frac{1}{2} \) is a factor of \( P(x) \) because \( P\left(\frac{1}{2}\right) = 0 \).

Step 1: Understand the Factor Theorem

According to the Factor Theorem, a polynomial \( P(x) \) has \( x-c \) as a factor if and only if \( P(c) = 0 \). Our goal is to apply this theorem to verify whether \( x-\frac{1}{2} \) is a factor of \( P(x) = 2x^3 + 7x^2 + 6x - 5 \) by evaluating \( P\left(\frac{1}{2}\right) \).

Step 2: Substitute and Simplify

Substitute \( c = \frac{1}{2} \) into the polynomial \( P(x) \). Calculate \( P\left( \frac{1}{2} \right) = 2\left( \frac{1}{2} \right)^3 + 7\left( \frac{1}{2} \right)^2 + 6\left( \frac{1}{2} \right) - 5 \).

Step 3: Calculate Powers and Multiply

Start computing each term: \( 2\left( \frac{1}{2} \right)^3 = 2\cdot\frac{1}{8} = \frac{1}{4} \); \( 7\left( \frac{1}{2} \right)^2= 7\cdot\frac{1}{4} = \frac{7}{4} \); \( 6\left( \frac{1}{2} \right)= 6\cdot\frac{1}{2} = 3 \).

Step 4: Perform Final Addition

Add all the computed terms: \( \frac{1}{4} + \frac{7}{4} + 3 - 5 \). Convert all numbers to fractions to find a common denominator for easier addition: \( \frac{1}{4} + \frac{7}{4} = \frac{8}{4} = 2 \). Then, \( 2 + 3 - 5 = 0 \).

Step 5: Conclude Using the Factor Theorem

Since \( P\left(\frac{1}{2}\right) = 0 \), according to the Factor Theorem, \( x-\frac{1}{2} \) is a factor of the polynomial \( P(x) \).

Key Concepts

Polynomials

Polynomials are fundamental mathematical expressions that consist of variables and coefficients, arranged in terms of powers in non-negative integers. Each term in a polynomial is a product of a coefficient and a variable raised to an exponent. The general form of a polynomial can be written as \( a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0 \), where each \( a_i \) is a coefficient, \( x \) is the variable, and \( n \) is the degree of the polynomial, indicating the highest power of \( x \) present in the polynomial.

• A polynomial of degree 1 is a linear polynomial, such as \( 3x + 1 \).
• A polynomial of degree 2 is a quadratic polynomial, such as \( x^2 + 5x + 6 \).
• Higher degree polynomials, like cubic polynomials, have higher power terms, such as \( 2x^3 + 7x^2 + 6x - 5 \).

Polynomials are very important in algebra, acting as the building blocks of various mathematical problems. Understanding polynomials is crucial since they can be manipulated in many ways, such as through addition, subtraction, multiplication, division, and factoring.

Factoring Polynomials

Factoring polynomials involves expressing a polynomial as a product of its factors. Factoring is key for solving polynomial equations and helps simplify complex expressions. One of the useful theorems for factoring is the Factor Theorem, which connects roots and factors.

The Factor Theorem states that a polynomial \( P(x) \) has \( x - c \) as a factor if and only if \( P(c) = 0 \). To determine if a given number \( c \) is a root of the polynomial, we substitute \( c \) into the polynomial and check if the result is zero. If it is, then \( x-c \) is indeed a factor of \( P(x) \).

When factoring, especially higher-degree polynomials, other methods may include:

• Finding the greatest common factor (GCF) of the terms.
• Using special formulas, such as difference of squares or sum/product of cubes.
• Employing synthetic division when testing for linear factors.

Factoring is not only about finding factors but also simplifying expressions for ease of solving equations.

Evaluating Polynomials

Evaluating polynomials means finding the value of a polynomial for a specified value of its variable. This process involves substituting a given number for the variable and simplifying the result to find the output of the polynomial at that specific input.

For instance, when evaluating a polynomial \( P(x) = 2x^3 + 7x^2 + 6x - 5 \) at \( x = \frac{1}{2} \), you substitute \( x = \frac{1}{2} \) in place of \( x \) and perform the computations step by step:

• First, calculate each term's value, such as \( 2\left(\frac{1}{2}\right)^3 \), \( 7\left(\frac{1}{2}\right)^2 \), etc.
• Simplify these individual computations.
• Add all the terms together.
• Conclude the evaluation by adding/subtracting to find the final result.

The accuracy in evaluating polynomials depends heavily on proper substitution and performing arithmetic operations correctly. Learning to evaluate polynomials aids in understanding their behavior and in tasks such as solving equations or finding roots.