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Chapter 3: Problem 61

The remainder theorem indicates that when a polynomial \(f(x)\) is divided by \(x-k\) the remainder is equal to \(f(k) .\) For $$f(x)=x^{3}-2 x^{2}-x+2$$ use the remainder theorem to find each of the following. Then determine the coordinates of the corresponding point on the graph of \(f(x)\) $$f\left(-\frac{1}{2}\right)$$

Short Answer

Expert verified
The coordinates are \( \left( -\frac{1}{2}, \frac{15}{8} \right) \).

Step by step solution

01

Understand the Remainder Theorem

The Remainder Theorem states that if a polynomial \( f(x) \) is divided by \( x-k \), the remainder of this division is equal to \( f(k) \). Therefore, to find the remainder when \( f(x) \) is divided by \( x - (-\frac{1}{2}) \) or \( x + \frac{1}{2} \), evaluate \( f \left( -\frac{1}{2} \right) \).
02

Substitute the Value into the Polynomial

Substitute \( x = -\frac{1}{2} \) into the polynomial \( f(x) = x^3 - 2x^2 - x + 2 \). This means we need to find \( f \left( -\frac{1}{2} \right) \).
03

Calculate Each Term

Calculate each term separately for \( x = -\frac{1}{2} \): \[ \left( -\frac{1}{2} \right)^3 = -\frac{1}{8}, \] \[ -2 \left( -\frac{1}{2} \right)^2 = -2 \left( \frac{1}{4} \right) = -\frac{1}{2}, \] \[ -\left( -\frac{1}{2} \right) = \frac{1}{2}, \] \[ 2 \text{ (this remains unchanged)}. \]
04

Combine the Results

Add the results from the previous step together: \[ f \left( -\frac{1}{2} \right) = -\frac{1}{8} - \frac{1}{2} + \frac{1}{2} + 2. \]Simplify this expression to find the value: \[ f \left( -\frac{1}{2} \right) = -\frac{1}{8} + 2 = \frac{-1 + 16}{8} = \frac{15}{8}. \]
05

Determine the Coordinates of the Point

Now that we have \( f \left( -\frac{1}{2} \right) = \frac{15}{8} \), the coordinates of the corresponding point on the graph are \( \left( -\frac{1}{2}, \frac{15}{8} \right) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polynomial Division
Polynomial division helps you break down a complex polynomial into simpler parts. This process is similar to long division with numbers. It is particularly useful when determining how a polynomial behaves when divided by a linear factor like \( x - k \). When dividing, the remainder theorem tells us that the remainder is simply the value of the polynomial evaluated at \( k \). This saves time because you don't have to carry out the full division. Instead, plug in the value and compute.
Function Evaluation
Function evaluation involves finding the value of a polynomial at a specific point. Given a polynomial \( f(x) \), to evaluate it at \( x = k \), replace \( x \) in the polynomial with \( k \).
In our example, \( f(x) = x^3 - 2x^2 - x + 2 \) is evaluated at \( x = -\frac{1}{2} \).
By substituting, we calculate:
  • \( \left( -\frac{1}{2} \right)^3 = -\frac{1}{8} \),
  • \(-2 \left( -\frac{1}{2} \right)^2 = -2 \left( \frac{1}{4} \right) = -\frac{1}{2} \)
  • \(-\left( -\frac{1}{2} \right) = \frac{1}{2} \)
  • \(2 \text{ (this remains unchanged)} \)
Combining these results, we get
\( f \left( -\frac{1}{2} \right) = -\frac{1}{8} - \frac{1}{2} + \frac{1}{2} + 2 = \frac{15}{8} \).
Graph Coordinates
Graph coordinates represent points on a graph and are usually written as \( (x, y) \). In the context of polynomials, \(x\) is the input value and \(f(x)\) (or \(y\)) is the output value. When we evaluated \( f(x) \) at \( x = -\frac{1}{2} \), we found that \( f \left( -\frac{1}{2} \right) = \frac{15}{8} \).
Therefore, the coordinates of the corresponding point on the graph are \( \left( -\frac{1}{2}, \frac{15}{8} \right) \).
This helps us visually understand the behavior of \( f(x) \) by plotting its points on a coordinate plane. Each point gives a glimpse of the polynomial's shape and location.
Algebraic Operations
Algebraic operations like addition, subtraction, multiplication, and division are key in solving polynomial problems. They help in simplifying the polynomial and calculating its values. In our evaluation:
  • Add: Combining individual evaluated terms \( -\frac{1}{8} + 2 \)
  • Subtract: Here, terms cancel out \( -\frac{1}{2} + \frac{1}{2} = 0 \)
  • Multiplication: \( -2 \left( \frac{1}{4} \right) = -\frac{1}{2} \)
Mastering these operations aids in breaking down and understanding polynomials better. Always perform these step-by-step for accuracy.

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