The function below has at least one rational zero.



Use this fact to find all zeros of the function.

$h(x)=6 x^3-41 x^2-8 x+7$



If there is more than one zero, separate them with commas. Write exact values, not decimal approximations.

Verified Answer :

  • Use the Rational Root Theorem to identify potential rational zeros from the factors of the constant and leading terms.
  • Test the candidates to find the roots \(x = 7\), \(x = -\frac{1}{2}\), and \(x = \frac{1}{3}\).
  • Verify the roots by factoring the polynomial into the form \(h(x) = (x - 7)(2x + 1)(3x - 1)\).
  • Conclude that the zeros of the function are \(\boxed{7, -\frac{1}{2}, \frac{1}{3}}\).

Explanation

  1. Finding Rational Zeros

To find the roots of the cubic function \(h(x) = 6x^3 - 41x^2 - 8x + 7\), we can start by applying the Rational Root Theorem. The potential rational zeros \(\frac{p}{q}\) are determined by the factors of the constant term \(7\) (\(p \in \{\pm 1, \pm 7\}\)) and the factors of the leading coefficient \(6\) (\(q \in \{1, 2, 3, 6\}\)). This gives us a list of possible rational zeros: \(\pm 1, \pm 7, \pm \frac{1}{2}, \pm \frac{7}{2}, \pm \frac{1}{3}, \pm \frac{7}{3}, \pm \frac{1}{6}, \pm \frac{7}{6}\). Testing these values in the function \(h(x)\), we identify that \(x = 7\), \(x = -\frac{1}{2}\), and \(x = \frac{1}{3}\) are roots of the equation because \(h(7) = 0\), \(h(-\frac{1}{2}) = 0\), and \(h(\frac{1}{3}) = 0\).

  1. Verification and Conclusion

To verify that these are all the zeros, recall that a cubic polynomial of the form \(ax^3 + bx^2 + cx + d\) has exactly three roots. We have already found three: \(7\), \(-\frac{1}{2}\), and \(\frac{1}{3}\). We can express the polynomial in factored form as \(h(x) = k(x - r_1)(x - r_2)(x - r_3)\). Here, \(k\) must be the leading coefficient, which is \(6\). So, we test:
\(6(x - 7)(x + \frac{1}{2})(x - \frac{1}{3}) = 6(x - 7)(\frac{2x+1}{2})(\frac{3x-1}{3}) = (x - 7)(2x + 1)(3x - 1)\).
Expanding this product:
\((x - 7)(6x^2 - 2x + 3x - 1) = (x - 7)(6x^2 + x - 1) = 6x^3 + x^2 - x - 42x^2 - 7x + 7 = 6x^3 - 41x^2 - 8x + 7\).
This expansion confirms the original polynomial. Therefore, the zeros are exactly \(7\), \(-\frac{1}{2}\), and \(\frac{1}{3}\).

Examples

In engineering, finding the zeros of a cubic polynomial is essential for determining equilibrium states in dynamic systems, such as a bridge oscillating under wind loads. By solving \(h(x) = 0\), engineers can identify the critical frequencies or displacement values where the system reaches stability or resonance, ensuring the design is safe and robust. The methods used here provide a foundational approach for analyzing such physical models.