Problem 38
Question
Use synthetic substitution to find \(P(k).\) $$k=3 ; \quad P(x)=x^{2}-x+3$$
Step-by-Step Solution
Verified Answer
The value of \( P(3) \) is 9.
1Step 1: Write down the coefficients
Identify the coefficients of the polynomial \( P(x) = x^2 - x + 3 \). The polynomial is of the form \( ax^2 + bx + c \), so the coefficients are \( a = 1 \), \( b = -1 \), and \( c = 3 \).
2Step 2: Set up the synthetic division
For synthetic substitution, we use a table where the top row consists of the coefficients of the polynomial: \( [1, -1, 3] \). We will substitute \( k = 3 \) along the left side of the table.
3Step 3: Bring down the first coefficient
Immediately bring down the first coefficient (which is 1) in the synthetic division table. This acts as the starting total.
4Step 4: Multiply and add iteratively
Multiply the value we just brought down (1) by \( k = 3 \) and add to the next coefficient (-1): \( 1 \times 3 = 3 \). Adding gives \( -1 + 3 = 2 \). Write 2 as the new second result.
5Step 5: Continue multiplication and addition
Now take the result (2) and multiply by \( k = 3 \). \( 2 \times 3 = 6 \). Add this to the final coefficient (3): \( 3 + 6 = 9 \). Write 9 as the final result of this synthetic division process.
6Step 6: Interpret the final value
The last number from the synthetic division process is the value of \( P(k) \). Therefore, \( P(3) = 9 \).
Key Concepts
Understanding Polynomial EvaluationThe Role of Synthetic Division in Polynomial EvaluationSynthetic Substitution in Precalculus Mathematics
Understanding Polynomial Evaluation
Polynomial evaluation is a core concept in precalculus mathematics. It involves finding the value of a polynomial at a given number, also known as substituting this number into the polynomial. For example, with the polynomial \( P(x) = x^2 - x + 3 \) and given \( k = 3 \), polynomial evaluation requires determining \( P(3) \). This is done by replacing every occurrence of \( x \) with \( 3 \), leading to \( 3^2 - 3 + 3 \).
Calculating step by step:
Calculating step by step:
- Solve \( 3^2 = 9 \)
- Then \( 9 - 3 = 6 \)
- Finally, \( 6 + 3 = 9 \)
The Role of Synthetic Division in Polynomial Evaluation
Synthetic division is a technique closely linked to polynomial evaluation, particularly when using synthetic substitution. It provides a streamlined way to evaluate polynomials at given points, leveraging a tabular format to simplify calculations. The process involves:
In the synthetic substitution with \( P(x) = x^2 - x + 3 \), the coefficients \( [1, -1, 3] \) are used with \( k = 3 \). The calculations inside the synthetic division table directly yield the value of the polynomial, avoiding the traditional multiple substitution and manual arithmetic operations. Ultimately, the last number computed, \( 9 \), represents \( P(3) \).
- Writing the coefficients of the polynomial in a row
- Using the variable \( k \) as the divisor
- Perform interim calculations by multiplying and adding coefficients
In the synthetic substitution with \( P(x) = x^2 - x + 3 \), the coefficients \( [1, -1, 3] \) are used with \( k = 3 \). The calculations inside the synthetic division table directly yield the value of the polynomial, avoiding the traditional multiple substitution and manual arithmetic operations. Ultimately, the last number computed, \( 9 \), represents \( P(3) \).
Synthetic Substitution in Precalculus Mathematics
Precalculus mathematics forms the foundation for advanced algebraic operations, including polynomial evaluation through synthetic substitution. It equips students with efficient methods for complex polynomial calculations which are pivotal in calculus and beyond.
Synthetic substitution, derived from synthetic division, simplifies the evaluation of polynomials at given points. This method reveals relationships and patterns within polynomial functions more clearly.
Synthetic substitution, derived from synthetic division, simplifies the evaluation of polynomials at given points. This method reveals relationships and patterns within polynomial functions more clearly.
- This is particularly useful when dealing with higher degree polynomials.
- It supports a deeper understanding of how coefficients affect a polynomial's graph.
- Offers a practical approach to checking polynomial roots efficiently.
Other exercises in this chapter
Problem 38
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