Problem 48
Question
Use synthetic division and the Remainder Theorem to evaluate \(P(c)\). $$P(x)=-2 x^{6}+7 x^{5}+40 x^{4}-7 x^{2}+10 x+112, \quad c=-3$$
Step-by-Step Solution
Verified Answer
The value of \(P(-3)\) is 100.
1Step 1: Set Up the Synthetic Division
We will perform synthetic division using the root \(c = -3\). Write \(c = -3\) on the left side and the coefficients of \(P(x)\) in descending order: \(-2, 7, 40, 0, -7, 10, 112\). The zero for the \(x^3\) term is crucial.
2Step 2: Begin the Synthetic Division Process
Bring down the first coefficient, \(-2\), as is. This represents our starting point in the process.
3Step 3: Multiply and Add
Multiply \(-3\) (root \(c\)) by \(-2\) (the value just brought down), resulting in \(6\). Then add this to the next coefficient \(7\) to get \(13\).
4Step 4: Continue Multiplication and Addition
Repeat the process: Multiply \(-3\) by \(13\), giving \(-39\). Add \(-39\) to the next coefficient \(40\) to get \(1\).
5Step 5: Iterate Until Completion
Continue this process using the results obtained:- Multiply \(-3\) by \(1\) to get \(-3\), add to \(0\) for \(-3\).- Multiply \(-3\) by \(-3\) to get \(9\), add to \(-7\) for \(2\).- Multiply \(-3\) by \(2\) to get \(-6\), add to \(10\) for \(4\).- Multiply \(-3\) by \(4\) to get \(-12\), add to \(112\) for \(100\).
6Step 6: Interpret the Results from Synthetic Division
The numbers at the bottom, \(-2, 13, 1, -3, 2, 4\), are the coefficients of the quotient polynomial, and the last number \(100\) is the remainder. So, by the Remainder Theorem, \(P(-3) = 100\).
Key Concepts
Remainder TheoremPolynomial DivisionPolynomial Evaluation
Remainder Theorem
The Remainder Theorem is a powerful tool in algebra that helps us to evaluate polynomials quickly. According to this theorem, if you divide a polynomial \( P(x) \) by a linear divisor of the form \( x - c \), the remainder of this division is simply \( P(c) \). This means you don't have to carry out the entire division; you can simply substitute the value \( c \) into the polynomial to find your remainder. This is especially useful because it allows for a fast and efficient way to evaluate complex polynomials at particular values without full polynomial division.
In the exercise at hand, this concept is used effectively with synthetic division, leading us to find \( P(-3) = 100 \). The Remainder Theorem assures us that since the remainder is \( 100 \), this is exactly \( P(-3) \). This makes polynomial evaluation both quick and straightforward.
In the exercise at hand, this concept is used effectively with synthetic division, leading us to find \( P(-3) = 100 \). The Remainder Theorem assures us that since the remainder is \( 100 \), this is exactly \( P(-3) \). This makes polynomial evaluation both quick and straightforward.
Polynomial Division
Polynomial division can be thought of as a process similar to long division but with variables. When dealing with polynomials, this division helps us break down and make sense of complex expressions by dividing them into simpler parts. There are several methods for polynomial division such as traditional long division and synthetic division.
In this exercise, synthetic division is employed. This method is preferred for its simplicity and efficiency, especially when dividing by a linear binomial \( x-c \). During synthetic division, we only focus on the coefficients of the polynomial, making this method quicker and easier than traditional approaches.
In this exercise, synthetic division is employed. This method is preferred for its simplicity and efficiency, especially when dividing by a linear binomial \( x-c \). During synthetic division, we only focus on the coefficients of the polynomial, making this method quicker and easier than traditional approaches.
- Write down the value of \( c \) (in this case, \(-3\)) and the coefficients of \( P(x) \).
- Follow the synthetic division process, which involves multiplying and adding coefficients step-by-step.
Polynomial Evaluation
Polynomial evaluation refers to finding the value of a polynomial for a specific value of its variable \( x \). This involves substituting the given number into the polynomial and performing the arithmetic computations. This process is a fundamental task in algebra that helps us understand how polynomials behave at specific points.
The exercise demonstrates polynomial evaluation through the combination of synthetic division and the Remainder Theorem. By setting \( x = -3 \), we efficiently use synthetic division to find that the remainder is 100. Therefore, the value of the polynomial \( P(x) \) at \( x = -3 \) is 100.
The exercise demonstrates polynomial evaluation through the combination of synthetic division and the Remainder Theorem. By setting \( x = -3 \), we efficiently use synthetic division to find that the remainder is 100. Therefore, the value of the polynomial \( P(x) \) at \( x = -3 \) is 100.
- Set \( x \) to the desired value (\(-3\) here).
- Conduct synthetic division using \(-3\) as the divisor.
- The remainder from this division process gives \( P(-3) \).
Other exercises in this chapter
Problem 48
Find all the real zeros of the polynomial. Use the quadratic formula if necessary, as in Example \(3(a)\). $$P(x)=x^{3}-5 x^{2}+2 x+12$$
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Evaluate the radical expression and express the result in the form \(a+b i\) $$\sqrt{-3} \sqrt{-12}$$
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Find the intercepts and asymptotes, and then sketch a graph of the rational function and state the domain and range. Use a graphing device to confirm your answe
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