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Use synthetic division to find the quotients and remainders. Also, in each case, write the result of the division in the form \(p(x)=d(x) \cdot q(x)+R(x),\) as in equation (2) in the text. $$\frac{x^{3}-8}{x-2}$$

Short Answer

Expert verified
The quotient is \(x^2 + 2x + 4\) with a remainder of 0.

Step by step solution

01

Set up the synthetic division

Identify the divisor and the dividend from the expression \(\frac{x^3 - 8}{x - 2}\). Here, the dividend is \(x^3 - 8\) and the divisor is \(x - 2\). We set up the synthetic division by writing the divisor's root, \(2\), on the left and the coefficients of the dividend \(x^3 - 8\) (which are 1, 0, 0, and -8 for \(x^3, x^2, x, \text{and the constant term}\) respectively) in a row.
02

Perform the synthetic division

Bring down the leading coefficient "1" to the bottom row. Multiply the root (2) by the number just written (which is 1) and write the product (2) in the next column. Add this product to the next coefficient (0) and write the result (2) below. Repeat this process: multiply the root (2) by this result (2) to get 4, add it to the next coefficient (0) to get 4. Then multiply again: 2 by 4 to get 8, add it to the last coefficient (-8) to get the remainder 0.
03

Interpret the synthetic division result

The bottom row (except the last number) represents the coefficients of the quotient polynomial. Thus the quotient is \(x^2 + 2x + 4\). The last number in the bottom row represents the remainder, which is 0.
04

Write the result in the required form

Substitute back the quotient and remainder into the equation from the problem statement: \[ p(x) = d(x) \cdot q(x) + R(x) \] Using our results, this becomes: \[ x^3 - 8 = (x - 2)(x^2 + 2x + 4) + 0 \] Thus, the division has no remainder and \(x - 2\) is a factor of \(x^3 - 8\).

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Key Concepts

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

Polynomial Division
Polynomial division is similar to long division you might do with numbers, but instead, it involves polynomials. In such divisions, one polynomial, called the dividend, is divided by another polynomial, called the divisor. This process results in a quotient and sometimes a remainder.

Synthetic division is a simpler, quicker way to divide polynomials, especially when dividing by a linear polynomial like \(x - 2\). This method only works when the divisor is a first-degree polynomial. Instead of dealing with all the terms of the polynomials, it uses only the coefficients. This makes it computationally less intensive and quickens the process.

Here, for example, dividing \(x^3 - 8\) by \(x-2\) with synthetic division only required us to focus on the coefficients \(1, 0, 0, -8\). Hence, we were able to find the quotient \(x^2 + 2x + 4\) and confirm the result.
Quotient and Remainder
In polynomial division, the outcome is expressed through the quotient and the remainder. These results allow us to reformulate the original division in a useful mathematical form. This form is:
  • \(p(x) = d(x) \cdot q(x) + R(x)\)
Here, \(p(x)\) denotes the original polynomial or dividend, \(d(x)\) is the divisor, \(q(x)\) is the quotient, and \(R(x)\) is the remainder.

A noteworthy detail about polynomial division is that, unlike numbers, the remainder will often be a smaller degree polynomial than the divisor. If the remainder is zero, it indicates that the divisor is a factor of the dividend. This simplifies the operation significantly in algebra where understanding the structure of polynomials is essential. For our exercise, dividing \(x^3 - 8\) by \(x-2\), the remainder was zero, confirming that \(x - 2\) is indeed a factor of \(x^3 - 8\).
Factor Theorem
The Factor Theorem is an important concept that connects roots and factors of a polynomial. According to this theorem, if a polynomial \(p(x)\) evaluates to zero at some value \(x = r\), then \(x - r\) is a factor of the polynomial.

Using synthetic division, if we end up with a remainder of zero, it provides evidence for the Factor Theorem. For example, dividing \(x^3 - 8\) by \(x - 2\) gave us a remainder of zero. This confirms \(x - 2\) is a factor, implying \(x = 2\) is a root of the polynomial \(x^3 - 8\).

This theorem is extremely helpful in both solving polynomial equations and factoring polynomials, as it offers a clear condition (zero remainder) to establish when a linear expression is a factor of a polynomial.
Polynomial Functions
Polynomial functions are equations formed from the sums of powers of variables, where each term consists of a coefficient and a power of a variable. These functions are foundational in algebra and are typified by expressions such as \(ax^n + bx^{n-1} + \, \ldots + c\).

They play critical roles in various mathematical analyses because their graphs can depict a wide range of behaviors including variable slopes and curvatures. By using methods like synthetic division, we can simplify and solve polynomials to understand their roots and identify features like intercepts and turning points.

In the example of our exercise, dividing \(x^3 - 8\) by \(x - 2\) confirmed the polynomial \(x^3 - 8\) could be expressed in simpler forms, giving insights into its behavior, roots, and factors efficiently. This is vital in higher mathematics for solving complex equations quickly and accurately.

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Most popular questions from this chapter

You are given an improper rational expression. First, use long division to rewrite the expression in the form (polynomial) \(+\) (proper rational expression) Next, obtain the partial fraction decomposition for the proper rational expression. Finally, rewrite the given improper rational expression in the form (polynomial) \(+\) (partial fractions) \frac{6 x^{3}-16 x^{2}-13 x+25}{x^{2}-4 x+3}

In a note that appeared in The Two-Year College Mathematics Journal [vol. \(12(1981), \text { pp. } 334-336],\) Professors Warren Page and Leo Chosid explain how the process of testing for rational roots can be shortened. In essence, their result is as follows. Suppose that we have a polynomial with integer coefficients and we are testing for a possible root \(p / q .\) Then, if a noninteger is generated at any point in the synthetic division process, \(p / q\) cannot be a root of the polynomial. For example, suppose we want to know whether \(4 / 3\) is a root of \(6 x^{4}-10 x^{3}+2 x^{2}-9 x+8=0\) The first few steps of the synthetic division are as follows. $$ \begin{array}{rrrrrr} 4 / 3 & 6 & -10 & 2 & -9 & 8 \\ \hline & & 8 & -8 / 3 & & \\ \hline & 6 & -2 & & & \\ \hline \end{array} $$ since the noninteger \(-8 / 3\) has been generated in the synthetic division process, the process can be stopped; \(4 / 3\) is not a root of the polynomial. Use this idea to shorten your work in testing to see whether the numbers \(3 / 4,1 / 8\). and \(-3 / 2\) are roots of the equation $$ 8 x^{5}-5 x^{4}+3 x^{2}-2 x-6=0 $$

(a) factor the denominator of the given \(\mathrm{ra}\) tional expression; (b) determine the form of the partial fraction decomposition for the given rational expression; and (c) determine the values of the constants in the partial fraction decomposition that you gave in part (b). To help you in spotting errors, use the fact that in part (c), each of the required constants turns out to be an integer. $$\frac{5 x^{2}+2 x+5}{x^{3}+x^{2}+x}$$

Find a polynomial equation \(f(x)=0\) satisfying the given conditions. If no such equation is possible, state this. Degree \(3 ;\) the coefficient are integers; \(1 / 2,2 / 5,\) and \(-3 / 4\) are roots

Determine the partial fraction decomposition for each of the given rational expressions. Hint: In Exercises \(17,18,\) and \(26,\) use the rational roots theorem to factor the denominator. \frac{1}{x^{3}+x^{2}-10 x+8}

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