Verified Answer :

  • Synthetic division identifies the dividend by examining the coefficients used in the division layout.
  • Comparing the provided options, \(2x^3 + 10x^2 + x + 5\) fits perfectly into a standard synthetic division template.
  • Dividing \(2x^3 + 10x^2 + x + 5\) by \((x + 5)\) yields a clean quotient of \(2x^2 + 1\).
  • The dividend is \(\boxed{2x^3+10x^2+x+5}\).

Explanation

  1. Understanding Synthetic Division

Synthetic division is a shorthand method of polynomial division. It is used to divide a polynomial \(P(x)\) by a linear factor of the form \(x - c\). The dividend is the polynomial being divided. To identify the dividend from a synthetic division setup, we look for the set of coefficients that would occupy the top row of the division table. Typically, for a cubic polynomial \(ax^3 + bx^2 + cx + d\), the top row of the synthetic division table would be the coefficients \([a, b, c, d]\).

  1. Analyzing the Options

We are presented with four potential candidate polynomials:

  1. \(-10x^2 - 5\)
  2. \(2x^3 + 10x^2 + x + 5\)
  3. \(2x^2 + 1\)
  4. \(2x^3 + x\)

Let's examine the second candidate, \(P(x) = 2x^3 + 10x^2 + x + 5\). If we perform synthetic division on this polynomial by a divisor like \((x + 5)\), we use the root \(c = -5\). The coefficients of \(P(x)\) are \(2\), \(10\), \(1\), and \(5\).
Placing these into the synthetic division format:
\[ \begin{array}{r|rrrr} -5 & 2 & 10 & 1 & 5 \ & & -10 & 0 & -5 \\ & 2 & 0 & 1 & 0 \end{array} \]
The division is exact, resulting in a quotient of \(2x^2 + 1\). This is a standard and clean example of synthetic division.

  1. Conclusion

Given the structure of common synthetic division problems, the polynomial \(2x^3 + 10x^2 + x + 5\) is clearly intended as the dividend. Its coefficients match the standard layout for a synthetic division problem that results in a simple quotient, making it the most logical answer among the choices provided. Therefore, the dividend represented is \(2x^3 + 10x^2 + x + 5\).

Examples

Synthetic division is a powerful tool in algebra used for evaluating polynomials or finding their roots. For instance, if an engineer is modeling the stress on a beam described by the polynomial \(f(x) = 2x^3 + 10x^2 + x + 5\), they might use synthetic division to quickly determine the stress at a specific point or to factor the equation to simplify the model. By dividing the polynomial by a factor \((x+5)\), they can reduce a cubic expression to a quadratic one, which is much easier to solve or graph, leading to more efficient structural analysis and safer designs.

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