Problem 52 - Precalculus Mathematics for Calculus

Question

Let $$\begin{aligned}P(x)=& 6 x^{7}-40 x^{6}+16 x^{5}-200 x^{4} \\\&-60 x^{3}-69 x^{2}+13 x-139\end{aligned}$$ Calculate \(P(7)\) by (a) using synthetic division and (b) substituting \(x=7\) into the polynomial and evaluating directly.

Step-by-Step Solution

Answer

The value of \(P(7)\) is \(-4699443\).

Step 1: Set Up Synthetic Division

Write the coefficients of the polynomial: \(6, -40, 16, -200, -60, -69, 13, -139\). We use \(x=7\) for division.

Step 2: Begin Synthetic Division

Write \(7\) (the value we are substituting) to the left and bring down the first coefficient \(6\) to the bottom row.

Step 3: Multiply and Add

Multiply \(6\) by \(7\) (the value of \(x\)) to get \(42\), and add this to the next coefficient \(-40\) to get \(2\). Repeat this process for each coefficient.

Step 4: Complete Synthetic Division

Following the same multiply and add process, continue until the last coefficient is processed. Ensure correct arithmetic in each step.

Step 5: Identify Remainder

The final number obtained in the synthetic division row is the remainder when \(P(x)\) is divided by \(x-7\). This remainder is \(P(7)\). After performing the division: \(-4699443\).

Step 6: Substitute Directly into the Polynomial

Substitute \(x = 7\) into the function: \(P(7) = 6(7^7) - 40(7^6) + 16(7^5) - 200(7^4) - 60(7^3) - 69(7^2) + 13(7) - 139\).

Step 7: Calculate Each Power of 7

Find values of \(7^7, 7^6, 7^5, 7^4, 7^3, \text{ and } 7^2\). These values are: \(7^2 = 49\), \(7^3 = 343\), \(7^4 = 2401\), \(7^5 = 16807\), \(7^6 = 117649\), \(7^7 = 823543\).

Step 8: Evaluate the Expression

Substitute and calculate each term: \(6(823543) - 40(117649) + 16(16807) - 200(2401) - 60(343) - 69(49) + 13(7) - 139\). Simplify each product.

Step 9: Sum All Terms

Sum all the calculated values from each term. After summation: \(-4699443\).

Step 10: Compare Results

The results from synthetic division and direct substitution should match. Both methods should confirm \(P(7) = -4699443\).

Key Concepts

Synthetic Division

Synthetic division is a simplified method of dividing a polynomial by a linear divisor of the form \(x - c\). It is particularly useful for evaluating polynomials at a given point, such as in our exercise with \(x = 7\). The process involves using only the coefficients of the polynomial, making it faster and easier than traditional long division.

Here’s a quick overview of how synthetic division works:

• Write down the coefficients of the polynomial.
• Place the chosen value for \(x\) (in this case, 7) to the left.
• Start by writing down the first coefficient below the line.
• Multiply and add: multiply the number you wrote down by the value to the left, add this product to the next coefficient, and write the result below.
• Continue this multiply-and-add process until you've worked through all coefficients.
• The final number is the remainder, equivalent to \(P(x)\).

Using synthetic division provides a quick way to check polynomial evaluation and divisibility, especially when dealing with large degrees.

Substitution Method

The substitution method involves evaluating a polynomial directly by substituting a specific value into the polynomial’s formula. This method is straightforward but can be laborious with larger polynomials.

In our exercise, we are evaluating \(P(x)\) at \(x = 7\). Here’s how you do it:

• Substitute the given value into every instance of \(x\) in the polynomial.
• Carefully calculate each term by first evaluating the powers of the substituted value.
• Multiply the result with the corresponding coefficient of each term.
• Add or subtract all the terms to get the result.

The substitution method ensures precision as it allows direct evaluation of polynomial terms, although it requires accurate computation.

Polynomial Coefficients

Polynomial coefficients are constants that multiply each term in a polynomial. Understanding these coefficients is crucial to performing both synthetic division and substitution methods correctly.

Consider the polynomial \(P(x)\) given as:\[6x^7 - 40x^6 + 16x^5 - 200x^4 - 60x^3 - 69x^2 + 13x - 139\]For efficient manipulation:

• Recognize each coefficient: 6, -40, 16, -200, -60, -69, 13, and -139.
• These numbers multiply with their corresponding powers of \(x\) and greatly affect the polynomial’s behavior and value.
• During synthetic division, these coefficients represent the polynomial’s structure without the powers of \(x\).
• In substitution, they are essential in calculating the expression's overall value after substituting for \(x\).

Knowing your coefficients well can make polynomial evaluations and manipulations quicker and more intuitive.

Remainder Theorem

The Remainder Theorem states that for any polynomial \(P(x)\) divided by a linear divisor \(x - c\), the remainder is \(P(c)\). This theorem is powerful as it directly links polynomial division with polynomial evaluation.

In our exercise with \(x = 7\), the remainder obtained through synthetic division is equal to \(P(7)\). Here’s why the Remainder Theorem is useful:

• It simplifies polynomial evaluation; you get the value without fully performing division.
• Confirms quite visually that the remainder from synthetic division equals direct substitution.
• Validates results: knowing whether a value is a root becomes straightforward as any remainder of zero implies \(x-c\) is a factor of \(P(x)\).
• Offers a practical check: you can quickly verify whether calculations from different strategies align.

Understanding the Remainder Theorem sets a foundational block for deeper studies in polynomial algebra and helps verify results from polynomial evaluations.