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Chapter 3: Problem 83

Given \(p(x)=2 x^{452}-4 x^{92},\) is it easier to evaluate \(p(1)\) by using synthetic division or by direct substitution? Find the value of \(p(1)\).

Short Answer

Expert verified
Direct substitution is easier and the value of \(p(1)\) is \(-2\).

Step by step solution

01

Understanding the Problem

The given polynomial is \(p(x) = 2x^{452} - 4x^{92}\). The task is to evaluate \(p(1)\) and determine the easier method: synthetic division or direct substitution.
02

Evaluate Using Direct Substitution

Substitute \(x = 1\) directly into the polynomial. \[p(1) = 2(1)^{452} - 4(1)^{92} = 2(1) - 4(1) = 2 - 4 = -2\]
03

Evaluate Using Synthetic Division

Synthetic division is typically used to divide polynomials. Here, substituting \(x = 1\) into \(p(x)\) directly will be faster and simpler. Therefore, synthetic division is unnecessary for evaluation.
04

Conclude the Easier Method

Direct substitution is easier and quicker for this problem. The value of \(p(1)\) is \(-2\).

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Key Concepts

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

Direct Substitution
Direct substitution is a straightforward method for evaluating polynomials. This method involves replacing the variable in the polynomial with a given number and simplifying the expression. For the given polynomial, the steps are simple:
  • Identify the value to substitute into the polynomial, which in this case is 1.
  • Replace each instance of the variable x with 1.
  • Calculate the resulting values by following the order of operations.

For example, in the polynomial \(p(x) = 2x^{452} - 4x^{92}\), we substitute 1 for x:
  • \(2(1)^{452} - 4(1)^{92} = 2(1) - 4(1) = 2 - 4 = -2\)

As you can see, direct substitution requires minimal steps, making it efficient especially when the variable's exponent results in simple values, like 1 raised to any power. This straightforward approach prevents unnecessary complexity, which is why it was the preferred method for this problem.
Synthetic Division
Synthetic division is a simplified form of polynomial division, usually used when dividing by a linear factor of the form \((x - c)\). It involves fewer steps and less complex arithmetic compared to long division. However, it is not always the most efficient method for polynomial evaluation.
For synthetic division, you typically follow these steps:
  • Write down the coefficients of the polynomial.
  • Use the given value for x (commonly, x = c) to set up the synthetic division.
  • Perform synthetic division by combining, multiplying, and adding coefficients accordingly.

In practice, this method is more beneficial for finding roots and factors of polynomials rather than simple evaluations. For instance, evaluating \(p(1) = 2x^{452} - 4x^{92}\) directly simplifies immediately, making synthetic division unnecessary in such direct evaluation contexts.
Polynomials in Algebra
Polynomials are expressions involving variables raised to non-negative integer powers, often with coefficients. They play a central role in algebra and are foundational for advanced mathematical concepts.
  • Degree: The degree of a polynomial is the highest power of the variable in the expression. For example, the degree of \(2x^{452} - 4x^{92}\) is 452.
  • Coefficients: These are the numerical factors multiplying the variables. In the polynomial given, the coefficients are 2 and -4.
  • Terms: Polynomials are made up of terms, each consisting of a coefficient, variable, and exponent. In the expression 2x^{452} - 4x^{92}, there are two terms: 2x^{452} and -4x^{92}.

Understanding these components is crucial for simplifying, evaluating, and manipulating polynomial expressions. Polynomials are also used in various real-world applications, such as modeling curves in physics and engineering. In algebra, mastering polynomials opens the door to understanding more complex topics like polynomial equations, factorization, and calculus.

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Most popular questions from this chapter

An engineer for a food manufacturer designs an aluminum container for a hot drink mix. The container is to be a right circular cylinder 5.5 in. in height. The surface area represents the amount of aluminum used and is given by \(S(r)=2 \pi r^{2}+11 \pi r,\) where \(r\) is the radius of the can. a. Graph the function \(y=S(r)\) and the line \(y=90\) on the viewing window [0,3,1] by [0,150,10] . b. Use the Intersect feature to determine point of intersection of \(y=S(r)\) and \(y=90\). c. Determine the restrictions on \(r\) so that the amount of aluminum used is at most \(90 \mathrm{in}^{2}\). Round to 1 decimal place.

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Write an equation of a function that meets the given conditions. Answers may vary. \(x\) -intercepts: (4,0) and (2,0) vertical asymptote: \(x=1\) horizontal asymptote: \(y=1\) \(y\) -intercept: (0,8)
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