Find all the zeros of \( f(x) \). \( f(x)=2 x^{4}+13 x^{3}+21 x^{2}-8 x-28 \) Arrange your answers from smallest to largest. If there is a double root, list it twice. \( \left.x=-\frac{[?]}{\square}, \square, \square\right] \)

To find the zeros of \( f(x) = 2x^4 + 13x^3 + 21x^2 - 8x - 28 \), we can use synthetic division or the Rational Root Theorem to test for potential rational roots. Testing \( x = -2 \): \[ 2(-2)^4 + 13(-2)^3 + 21(-2)^2 - 8(-2) - 28 = 2(16) - 13(8) + 21(4) + 16 - 28 = 32 - 104 + 84 + 16 - 28 = 0 \] So, \( x = -2 \) is a root. Now we can divide \( f(x) \) by \( x + 2 \) using synthetic division: \[ \begin{array}{r|rrrrr} -2 & 2 & 13 & 21 & -8 & -28 \\ & & -4 & -18 & -6 & 28 \\ \hline & 2 & 9 & 3 & -14 & 0 \\ \end{array} \] This gives us \( f(x) = (x + 2)(2x^3 + 9x^2 + 3x - 14) \). Next, we can apply the Rational Root Theorem to \( 2x^3 + 9x^2 + 3x - 14 \) and test \( x = 1 \): \[ 2(1)^3 + 9(1)^2 + 3(1) - 14 = 2 + 9 + 3 - 14 = 0 \] So, \( x = 1 \) is another root. Dividing \( 2x^3 + 9x^2 + 3x - 14 \) by \( x - 1 \): \[ \begin{array}{r|rrrr} 1 & 2 & 9 & 3 & -14 \\ & & 2 & 11 & 14 \\ \hline & 2 & 11 & 14 & 0 \\ \end{array} \] This results in \( 2x^2 + 11x + 14 \). We can now factor or use the quadratic formula to find the remaining roots: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-11 \pm \sqrt{121 - 112}}{4} = \frac{-11 \pm 3}{4} \] Calculating the roots yields: 1. \( x = \frac{-8}{4} = -2 \) 2. \( x = \frac{-14}{4} = -\frac{7}{2} \) Summarizing all the roots, we have: 1. \( x = -2 \) (double root) 2. \( x = -\frac{7}{2} \) 3. \( x = 1 \) So the zeros of \( f(x) \) arranged from smallest to largest are: \[ \left.x = -2, -\frac{7}{2}, 1\right] \] Thus: First additional information: In polynomials, finding roots helps us understand the function's behavior, such as where it crosses the x-axis, which can give insights into the function's maximums and minimums. Second additional information: These techniques of synthetic division and the Rational Root Theorem are not just valuable for polynomials but also form the foundation for solving higher-degree polynomials and understanding polynomial inequalities in calculus and algebra!