Find all complex zeros of the polynomial function Give exact values. List multiple zeros as necessary. f(x)=x^5-9x^4+30x^3-54x^2+81x-81 All complex zeros are square (Type an exact answer using radicals and i as needed. Use a comma to separate answers as needed.)

To find the complex zeros of the polynomial function \( f(x) = x^5 - 9x^4 + 30x^3 - 54x^2 + 81x - 81 \), we can use the Rational Root Theorem and synthetic division to identify possible rational roots and then factor the polynomial.<br /><br />First, let's use the Rational Root Theorem to identify possible rational roots. The Rational Root Theorem states that any rational root of the polynomial \( f(x) \) is a factor of the constant term (-81) divided by a factor of the leading coefficient (1). Therefore, the possible rational roots are:<br />\[ \pm 1, \pm 3, \pm 9, \pm 27, \pm 81 \]<br /><br />We will test these possible roots by substituting them into the polynomial to see if they yield zero.<br /><br />Let's start with \( x = 1 \):<br />\[ f(1) = 1^5 - 9(1)^4 + 30(1)^3 - 54(1)^2 + 81(1) - 81 = 1 - 9 + 30 - 54 + 81 - 81 = 0 \]<br /><br />Since \( f(1) = 0 \), \( x = 1 \) is a root of the polynomial. We can now perform synthetic division to factor out \( (x - 1) \) from \( f(x) \).<br /><br />Using synthetic division with \( x = 1 \):<br /><br />\[<br />\begin{array}{r|rrrrrr}<br />1 & 1 & -9 & 30 & -54 & 81 & -81 \\<br /> & & 1 & -8 & 22 & -32 & 49 \\<br />\hline<br /> & 1 & -8 & 22 & -32 & 49 & 0 \\<br />\end{array}<br />\]<br /><br />The quotient is \( x^4 - 8x^3 + 22x^2 - 32x + 49 \). So, we have:<br />\[ f(x) = (x - 1)(x^4 - 8x^3 + 22x^2 - 32x + 49) \]<br /><br />Next, we need to find the roots of the quartic polynomial \( x^4 - 8x^3 + 22x^2 - 32x + 49 \). We can use the quadratic formula to solve this quartic polynomial. Let \( y = x^2 \), then the quartic polynomial becomes:<br />\[ y^2 - 8y + 22 = 0 \]<br /><br />Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):<br />\[ y = \frac{8 \pm \sqrt{64 - 88}}{2} = \frac{8 \pm \sqrt{-24}}{2} = \frac{8 \pm 2i\sqrt{6}}{2} = 4 \pm i\sqrt{6} \]<br /><br />So, \( y = 4 \pm i\sqrt{6} \). Since \( y = x^2 \), we have:<br />\[ x^2 = 4 + i\sqrt{6} \quad \text{or} \quad x^2 = 4 - i\sqrt{6} \]<br /><br />Taking the square roots of these:<br />\[ x = \pm \sqrt{4 + i\sqrt{6}} \quad \text{or} \quad x = \pm \sqrt{4 - i\sqrt{6}} \]<br /><br />Thus, the complex zeros of the polynomial \( f(x) = x^5 - 9x^4 + 30x^3 - 54x^2 + 81x - 81 \) are:<br />\[ 1, \sqrt{4 + i\sqrt{6}}, -\sqrt{4 + i\sqrt{6}}, \sqrt{4 - i\sqrt{6}}, -\sqrt{4 - i\sqrt{6}} \]<br /><br />All complex zeros are:<br />\[ 1, \sqrt{4 + i\sqrt{6}}, -\sqrt{4 + i\sqrt{6}}, \sqrt{4 - i\sqrt{6}}, -\sqrt{4 - i\sqrt{6}} \]