Some of the potential roots of $f(x)=10 x^3+29 x^2-66 x+27$ according to the rational root theorem are shown. Choose three answers that are actual roots.


  • $-\frac{9}{10}$
  • $\frac{3}{5}$
  • $-\frac{9}{2}$
  • 3
  • 1

Verified Answer :

  • Substitute each candidate root into the polynomial \(f(x) = 10x^3 + 29x^2 - 66x + 27\).
  • Identify that \(f(\frac{3}{5}) = 0\), \(f(-\frac{9}{2}) = 0\), and \(f(1) = 0\).
  • Conclude that the three roots are \(\frac{3}{5}\), \(-\frac{9}{2}\), and \(1\).
  • Final Answer: \(\boxed{\frac{3}{5}, -\frac{9}{2}, 1}\)

Explanation

  1. Analyzing the Problem and Objective

The polynomial given is \(f(x) = 10x^3 + 29x^2 - 66x + 27\). We are given a set of five potential roots: \(\{-\frac{9}{10}, \frac{3}{5}, -\frac{9}{2}, 3, 1\}\). A value \(r\) is a root of the polynomial if and only if \(f(r) = 0\). We will evaluate the function at each given point to determine which ones are indeed roots.

  1. Testing the Candidates

We evaluate the polynomial \(f(x) = 10x^3 + 29x^2 - 66x + 27\) for each candidate root:

  • For \(x = -\frac{9}{10} = -0.9\): \(f(-0.9) = 10(-0.729) + 29(0.81) - 66(-0.9) + 27 = -7.29 + 23.49 + 59.4 + 27 = 102.6 eq 0\).
  • For \(x = \frac{3}{5} = 0.6\): \(f(0.6) = 10(0.216) + 29(0.36) - 66(0.6) + 27 = 2.16 + 10.44 - 39.6 + 27 = 0\). Thus, \(\frac{3}{5}\) is a root.
  • For \(x = -\frac{9}{2} = -4.5\): \(f(-4.5) = 10(-91.125) + 29(20.25) - 66(-4.5) + 27 = -911.25 + 587.25 + 297 + 27 = 0\). Thus, \(-\frac{9}{2}\) is a root.
  • For \(x = 3\): \(f(3) = 10(27) + 29(9) - 66(3) + 27 = 270 + 261 - 198 + 27 = 360 eq 0\).
  • For \(x = 1\): \(f(1) = 10(1)^3 + 29(1)^2 - 66(1) + 27 = 10 + 29 - 66 + 27 = 0\). Thus, \(1\) is a root.
  1. Conclusion

Based on the evaluations performed, the roots of the polynomial \(f(x) = 10x^3 + 29x^2 - 66x + 27\) from the given set are \(\frac{3}{5}\), \(-\frac{9}{2}\), and \(1\). These are the three values that result in \(f(x) = 0\). The final answer consists of these three roots.

Examples

Finding roots of a polynomial is essential in fields like engineering and physics, where the roots of a characteristic equation describe the stability or natural frequencies of a system. For example, if you are designing a bridge or a mechanical suspension system, the roots of the governing polynomials help you determine critical load capacities or resonance points that must be avoided. By solving for these roots, engineers ensure that structures remain stable and safe under various operating conditions.