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For Problems 1-10, find \(f(c)\) by (a) evaluating \(f(c)\) directly, and (b) using synthetic division and the remainder theorem. $$ f(n)=6 n^{3}-35 n^{2}+8 n-10 \text { and } c=6 $$

Short Answer

Expert verified
f(6) = 74.

Step by step solution

01

Evaluate f(c) Directly

To find \(f(c)\) using direct evaluation, substitute \(c = 6\) into the polynomial \(f(n) = 6n^3 - 35n^2 + 8n - 10\). 1. Calculate \(6 \times 6^3\) = 1296. 2. Calculate \(-35 \times 6^2\) = -1260.3. Calculate \(+8 \times 6\) = 48.4. Finally, subtract 10.Combine these:1296 - 1260 + 48 - 10 = 74.Thus, \(f(6) = 74\).
02

Set Up Synthetic Division

To use synthetic division, start by setting up the division with the divisor \(n - 6\). In synthetic division, you only use the zero of the divisor, which is \(6\). Write the coefficients of the polynomial \(6, -35, 8, -10\) in a row.
03

Perform Synthetic Division

Carry out synthetic division using these steps:1. Bring down the leading coefficient (6) to the bottom row.2. Multiply it by \(6\) (the value of \(c\)) and write the result under the second coefficient: \(6 \times 6 = 36\). 3. Add -35 and 36 to get 1. 4. Repeat the process: Multiply 1 by 6 and add to 8 to get 14.5. Multiply 14 by 6 and add to -10 to get 74.The remainder of this division is the number at the bottom right: 74.
04

Conclude with the Remainder Theorem

According to the Remainder Theorem, the result from synthetic division (74) is \(f(6)\). This matches the direct evaluation result, confirming that \(f(6) = 74\).

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Key Concepts

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

Direct Evaluation
Direct evaluation is a straightforward method to find the value of a polynomial at a given point. This technique involves directly substituting the given value into the polynomial equation and calculating the result. For instance, if you're given a polynomial function \(f(n) = 6n^3 - 35n^2 + 8n - 10\) and need to find \(f(6)\):
  • Start by substituting \(6\) for every instance of \(n\) in the polynomial.

  • Then compute each term step-by-step:
    - \(6 \times 6^3 = 1296\) - \(-35 \times 6^2 = -1260\) - \(8 \times 6 = 48\) - Lastly, subtract \(10\)
  • Add these results together: \(1296 - 1260 + 48 - 10 = 74\).
This calculated result of \(74\) is \(f(6)\). Direct evaluation is often an easy and quick method when the numbers don’t get too large or complex, as long as you're careful with your arithmetic.
Synthetic Division
Synthetic division is a handy shortcut for dividing a polynomial by a linear term. It’s particularly useful when dealing with polynomials, as it simplifies the division process into basic arithmetic. In this example, it's used to check the evaluation of \(f(n)\) at \(c = 6\).
  • Start by listing the coefficients of the polynomial \(6, -35, 8, -10\).

  • The divisor is the zero of the term \(n - 6\), which is \(6\).

  • Perform the division:
    • Bring the first coefficient (\(6\)) straight down.
    • Multiply this number by \(6\) and write it under the next coefficient.
    • Add the numbers in the column, move to the next coefficient, and continue the process.
    • The last number in the bottom row will be the remainder.
In our example, you carry out this process and end up with \(74\) as the remainder. This remainder is crucial as it helps verify the original problem's evaluation.
Remainder Theorem
The Remainder Theorem is a powerful concept in algebra providing a direct link between polynomial division and evaluating polynomials. This theorem states that if you divide a polynomial \(f(n)\) by a linear divisor \(n - c\), the remainder of that division process will be \(f(c)\).

In simpler terms, when you perform synthetic division and find a remainder, that remainder is the same value you get when directly evaluating the polynomial at \(n = c\). This verified result serves as confirmation of accuracy both ways.
  • For the given polynomial \(f(n) = 6n^3 - 35n^2 + 8n - 10\), when divided by \(n - 6\), the remainder \(74\) coincides with the direct evaluation \(f(6) = 74\).
  • This matching result reassures that calculations are correct and provides confidence in using either method for polynomial evaluations.
Understanding and applying the Remainder Theorem bridges the methodical gap between synthetic division results and direct computation, giving you a quick verification tool for polynomials.

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Most popular questions from this chapter

Graph \(f(x)=\frac{1}{x^{2}}\). How should the graph of \(f(x)=\) \(\frac{1}{(x-4)^{2}}, \quad f(x)=\frac{1+3 x^{2}}{x^{2}}\), and \(f(x)=\frac{1}{x^{2}}\) compare to the graph of \(f(x)=\frac{1}{x^{2}}\) ? Graph the three functions on the same set of axes with the graph of $$ f(x)=\frac{1}{x^{2}} \text {. } $$

For Problems \(35-42\), (a) find the \(y\) intercepts, (b) find the \(x\) intercepts, and (c) find the intervals of \(x\) where \(f(x)>0\) and those where \(f(x)<0\). Do not sketch the graphs. $$ f(x)=(x-4)^{2}(x+3)^{3} $$

A polynomial function with real coefficients is continuous everywhere; that is, its graph has no holes or breaks. This is the basis for the following property: If \(f(x)\) is a polynomial with real coefficients, and if \(f(a)\) and \(f(b)\) are of opposite sign, then there is at least one real zero between \(a\) and \(b\). This property, along with our knowledge of polynomial functions, provides the basis for locating and approximating irrational solutions of a polynomial equation. Consider the equation \(x^{3}+2 x-4=0\). Applying Descartes' rule of signs, we can determine that this equation has one positive real solution and two nonreal complex solutions. (You may want to confirm this!) The rational root theorem indicates that the only possible rational solutions are 1,2 , and 4 . Using a little more compact format for synthetic division, we obtain the following results when testing for 1 and 2 as possible solutions: $$ \begin{array}{r|rrrr} & 1 & 0 & 2 & -4 \\ 1 & 1 & 1 & 3 & -1 \\ 2 & 1 & 2 & 6 & 8 \end{array} $$ Because \(f(1)=-1\) and \(f(2)=8\), there must be an irrational solution between 1 and 2 . Furthermore, \(-1\) is closer to 0 than is 8 , so our guess is that the solution is closer to 1 than to 2 . Let's start looking at \(1.0,1.1,1.2\), and so on, until we can place the solution between two numbers. Because \(f(1.1)=-0.469\) and \(f(1.2)=0.128\), the irrational solution must be between \(1.1\) and 1.2. Furthermore, because \(0.128\) is closer to 0 than is \(-0.469\), our guess is that the solution is closer to \(1.2\) than to \(1.1\). Let's start looking at \(1.15,1.16\), and so on. $$ \begin{array}{l|rrrrr} & 1 & 0 & 2 & -4 \\ \ 1.15 & 1 & 1.15 & 3.3225 & -0.179 \\ 1.16 & 1 & 1.16 & 3.3456 & -0.119 \\ 1.17 & 1 & 1.17 & 3.3689 & -0.058 \\ 1.18 & 1 & 1.18 & 3.3924 & 0.003 \end{array} $$ Because \(f(1.17)=-0.058\) and \(f(1.18)=0.003\), the irrational solution must be between \(1.17\) and \(1.18\). Therefore we can use \(1.2\) as a rational approximation to the nearest tenth. For each of the following equations, (a) verify that the equation has exactly one irrational solution, and (b) find an approximation, to the nearest tenth, of that solution. (a) \(x^{3}+x-6=0 \) (b) \(x^{3}-6 x-6=0 \) (c) \(x^{3}-27 x-60=0 \) (d) \(x^{3}-x^{2}-x-1=0 \) (e) \(x^{3}-2 x-10=0\) (f) \(x^{3}-5 x^{2}-1=0 \)

For Problems \(1-20\), graph each rational function. Check first for symmetry, and identify the asymptotes. $$ f(x)=\frac{2 x^{2}}{x^{2}-2 x-8} $$

For Problems \(35-42\), (a) find the \(y\) intercepts, (b) find the \(x\) intercepts, and (c) find the intervals of \(x\) where \(f(x)>0\) and those where \(f(x)<0\). Do not sketch the graphs. $$ f(x)=x^{2}(2-x)(x+3) $$

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