Problem 44 - Algebra and Trigonometry

Question

Use synthetic division and the Remainder Theorem to evaluate \(P(c) .\) \(P(x)=6 x^{5}+10 x^{3}+x+1, \quad c=-2\)

Step-by-Step Solution

Answer

Using synthetic division, \(P(-2) = -273\).

Step 1: Write Down the Coefficients

Identify and list down the coefficients of the polynomial function.For the polynomial \(P(x) = 6x^{5} + 10x^{3} + x + 1\), the coefficients are 6, 0 (for \(x^{4}\) term), 10, 0 (for \(x^{2}\) term), 1, and 1.

Step 2: Set Up Synthetic Division

Write the value of \(c\) on the left of the division, followed by the coefficients on the right.The setup looks like this:\[\begin{array}{r|rrrrrr}-2 & 6 & 0 & 10 & 0 & 1 & 1 \\hline & \\end{array}\]

Step 3: Begin the Division Process

Bring down the leading coefficient (6) to the bottom row.This gives us:\[\begin{array}{r|rrrrrr}-2 & 6 & 0 & 10 & 0 & 1 & 1 \\hline & 6 \\end{array}\]

Step 4: Multiply and Add

Multiply the leading entry from the bottom row by \(c\) and add to the next coefficient.\(-2 \times 6 = -12\) and \(0 + (-12) = -12\).Add -12 to the fourth line:\[\begin{array}{r|rrrrrr}-2 & 6 & 0 & 10 & 0 & 1 & 1 \\hline & 6 & -12 \\end{array}\]

Step 5: Continue the Process

Repeat the multiply and add step for the rest of the coefficients.Multiply -12 by -2 to get 24: \(10 + 24 = 34\)Then multiply 34 by -2 to get -68: \(0 + (-68) = -68\)Multiply -68 by -2 to get 136: \(1 + 136 = 137\)Finally multiply 137 by -2 to get -274: \(1 + (-274) = -273\)\[\begin{array}{r|rrrrrr}-2 & 6 & 0 & 10 & 0 & 1 & 1 \\hline & 6 & -12 & 34 & -68 & 137 & -273 \\end{array}\]

Step 6: Interpret the Remainder

The final value in the bottom row represents the remainder.Since the remainder is -273, \(P(-2) = -273\) according to the Remainder Theorem.

Key Concepts

Polynomial evaluation

Polynomial evaluation is a process of determining the value of a polynomial function at a given point. For any polynomial function \( P(x) \), you can substitute the value of \( x \) with a specific number \( c \) to evaluate it. In this exercise, the polynomial given is \( P(x) = 6x^5 + 10x^3 + x + 1 \), and we need to find \( P(-2) \). This means we substitute \( x \) with \(-2\) in the polynomial expression. However, instead of directly calculating each term, we use the synthetic division method to efficiently find the value of the polynomial at this point.

Remainder Theorem

The Remainder Theorem is a useful tool in polynomial algebra. It states that the remainder of the division of a polynomial \( P(x) \) by a linear divisor \( x - c \) is \( P(c) \). This means that if you are dividing \( P(x) \) by \( (x - c) \) and you find a remainder \( R \), then \( P(c) = R \). This theorem simplifies the process of evaluating polynomials because, with synthetic division, we can find the remainder, and hence the value of the polynomial at a specific point, without directly computing larger powers. In our exercise, after performing synthetic division with \( c = -2 \), the remainder found is \(-273\), indicating that \( P(-2) = -273 \).

Polynomial coefficients

Polynomial coefficients are the numbers paired with the variables in a polynomial. Each coefficient represents the multiplier of the corresponding power of \( x \). In the polynomial \( P(x) = 6x^5 + 10x^3 + x + 1 \), the coefficients are extracted as follows:

• 6 for \( x^5 \)
• 0 for \( x^4 \) because there is no \( x^4 \) term present
• 10 for \( x^3 \)
• 0 for \( x^2 \) since there's no \( x^2 \) term
• 1 for \( x \)
• 1 for the constant term

Recognizing these coefficients is crucial when setting up synthetic division, as they provide the data necessary to perform the division process and evaluate the polynomial.

Synthetic division process

The synthetic division process is a streamlined method to divide polynomials, specifically when divided by a linear factor. It reduces computational complexity and allows for quick polynomial evaluation.

Setting Up

First, write down the value of \( c \), the divisor, on the left. Align the polynomial coefficients on the right. Begin by bringing down the first coefficient directly to the bottom row.

Main Process

Multiply the number from the bottom row by \( c \), writing this product underneath the next coefficient. Add vertically to fill the bottom row. Repeat this multiply-and-add step across all coefficients. For example, with initial coefficients 6, 0, 10, 0, 1, and 1, and divisor \(-2\), bring down 6. Multiply it by \(-2\) to get \(-12\). Add \(-12\) to the next coefficient 0 to continue the process.

Finding the Remainder

Once completed, the last number in the bottom row represents the remainder. This final number is the polynomial evaluated at \( c \). In our exercise, the remainder is \(-273\), meaning \( P(-2) = -273 \). The synthetic division concludes with this evaluation, demonstrating its efficiency in polynomial calculations.