Solve the polynomial equation \[ 4 x^{3}+x^{2}-36 x-9=0 \]

To solve the polynomial equation \( 4x^{3}+x^{2}-36x-9=0 \), we can apply the Rational Root Theorem, synthetic division, or the method of finding roots using numerical techniques. However, for this case, let’s first look for rational roots by testing possible factors of the constant term (-9) and the leading coefficient (4). The possible rational roots can be given by ± factors of -9 divided by ± factors of 4: Possible rational roots: \( \pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{4} \). Let's start testing some of these values: 1. Testing \( x = 3 \): \[ 4(3)^3 + (3)^2 - 36(3) - 9 = 4(27) + 9 - 108 - 9 = 108 + 9 - 108 - 9 = 0 \] So, \( x = 3 \) is a root. Now we can perform synthetic division of the polynomial by \( x - 3 \). Performing synthetic division: \[ \begin{array}{r|rrrr} 3 & 4 & 1 & -36 & -9 \\ & & 12 & 39 & 9 \\ \hline & 4 & 13 & 3 & 0 \\ \end{array} \] This gives us a quotient of \( 4x^2 + 13x + 3 \). Next, we can factor or use the quadratic formula to solve \( 4x^2 + 13x + 3 = 0 \). Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-13 \pm \sqrt{13^2 - 4(4)(3)}}{2(4)} = \frac{-13 \pm \sqrt{169 - 48}}{8} = \frac{-13 \pm \sqrt{121}}{8} \] Calculating further: \[ \sqrt{121} = 11 \] Thus, we have: \[ x = \frac{-13 + 11}{8} = \frac{-2}{8} = -\frac{1}{4} \quad \text{and} \quad x = \frac{-13 - 11}{8} = \frac{-24}{8} = -3 \] Hence, the solutions to the polynomial equation \( 4x^3 + x^2 - 36x - 9 = 0 \) are: \[ x = 3, \quad x = -\frac{1}{4}, \quad x = -3 \]