Problem 42 - A Graphical Approach to Precalculus with Limits

Question

Use synthetic substitution to find \(P(k).\) $$k=\sqrt{3} ; \quad P(x)=x^{4}+2 x^{2}-10$$

Step-by-Step Solution

Answer

\(P(\sqrt{3}) = -5\).

Step 1: Understand Synthetic Substitution

Synthetic substitution is a method to evaluate a polynomial at a given value. It works similarly to synthetic division, but instead of dividing, we use it to find the polynomial's value for a specific input. Here, we will evaluate the polynomial \(P(x) = x^4 + 2x^2 - 10\) at \(k = \sqrt{3}\).

Step 2: Set Up the Synthetic Substitution

Write down the coefficients of the polynomial in descending order of power: 1 (for \(x^4\)), 0 (for \(x^3\), since it is missing), 2 (for \(x^2\)), 0 (for \(x\)), and -10 (the constant term). Then write \(\sqrt{3}\) next to them, which is the value we substitute into the polynomial.

Step 3: Perform Synthetic Substitution

Start with the leading coefficient, which is 1. Multiply it by \(\sqrt{3}\) and write the result under the next coefficient. Continue this process column by column: 1. Write 1 (the first coefficient). 2. Multiply 1 by \(\sqrt{3}\) to get \(\sqrt{3}\). 3. Add 0 (the next coefficient) and \(\sqrt{3}\) to get \(\sqrt{3}\). 4. Multiply \(\sqrt{3}\) by \(\sqrt{3}\) to get 3, write under the next coefficient.5. Add 2 and 3 to get 5.6. Multiply 5 by \(\sqrt{3}\) to get \(5\sqrt{3}\).7. Continue with multiplicative and additive steps until the final constant term.

Step 4: Calculate the Final Result

Following Step 3, the calculations are as follows:1. Starting value is 1.2. Results from multiplying and adding: \(\sqrt{3}, 3, 5, 5\sqrt{3}, -5\).3. The final number at the bottom right of the synthetic division is the value of the polynomial at \(x = \sqrt{3}\), which is -5. Therefore, \(P(\sqrt{3}) = -5\).

Key Concepts

Polynomial Evaluation

When evaluating a polynomial, the goal is to find the value of the polynomial for a specific input. This is like plugging numbers into an equation to simplify it. If you have a polynomial, say \( P(x) = x^4 + 2x^2 - 10 \), substituting a value, such as \( k = \sqrt{3} \) gives us \( P(\sqrt{3}) \).
This means replacing every occurrence of \( x \) in the polynomial with \( \sqrt{3} \).
For example:

• Substitute \( x \) in \( x^4 \) with \( \sqrt{3} \), giving \( (\sqrt{3})^4 \).
• Substitute \( x \) in \( 2x^2 \) with \( 2(\sqrt{3})^2 \).
• Lastly, use the remaining constant term, which is \(-10\) in this polynomial.

After performing these substitutions, simplify the terms step by step and combine them together to get the final result. This method is straightforward but can be time-consuming for high-degree polynomials, which is where synthetic substitution comes in handy.

Synthetic Division

Synthetic division is a shortcut method for dividing a polynomial by a linear factor. Interestingly, it's also a great tool for evaluating polynomials quickly. When using synthetic division for polynomial evaluation, you essentially execute a streamlined process.
The key is to set up the coefficients of the polynomial in a simplified layout, which makes it easier to handle, especially with complex numbers like \( \sqrt{3} \).
Here's how it works:

• Write down the coefficients of the polynomial, omitting any terms that equal zero.
• The number you are substituting, such as \( \sqrt{3} \), is written on the left.
• Using synthetic division, you multiply the result from each step by \( \sqrt{3} \) and then add it to the next coefficient.

By following these steps, you work systematically from the highest order term to the last, ultimately finding the value the polynomial takes on at the given point. This technique saves time and reduces potential for errors, especially when handling more complex expressions.

Precalculus Concepts

In precalculus, understanding functions and their evaluations is fundamental. This exercise of synthetic substitution to evaluate a polynomial is a prime example of precalculus concepts in action. It provides a bridge between algebra and calculus by emphasizing understanding and manipulation of functions.
Familiarizing yourself with synthetic substitution as a technique helps deepen your knowledge of polynomial behavior without diving into complex calculus operations. It involves:

• Recognizing function forms, such as quadratic or quartic.
• Applying algebraic rules to simplify expressions.
• Predicting outcomes using manipulation techniques like synthetic division.

These skills are essential as you progress in math. They help you gain more control over functions and prepare you for more advanced topics encountered in calculus and beyond.