Score: $$ 1 / 4 \quad $$ Penalty: 0.5 off Question The function $$ f(x)=x^{3}-x^{2}-6 x+6 $$ has at least one integer root. Use the integer root theorem to find that root, then proceed to find all complex roots. (Note: roots may be integer, rational, irrational, and/or complex.) Answer Attempt 1 out of 10 There is Submit Answer Show Examples Thatch Video

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1. Use the Integer Root Theorem. The possible integer roots are the divisors of the constant term $$6$$ (i.e., $$\pm 1$$, $$\pm 2$$, $$\pm 3$$, $$\pm 6$$).  
2. Test $$x=1$$:  
   $$
   f(1) = 1^3 - 1^2 - 6(1) + 6 = 1 - 1 - 6 + 6 = 0
   $$  
   Since $$f(1)=0$$, $$x=1$$ is a root.

3. Factor the polynomial by dividing $$f(x)$$ by $$(x-1)$$. Using synthetic division with coefficients $$1$$ (for $$x^3$$), $$-1$$ (for $$x^2$$), $$-6$$ (for $$x$$), and $$6$$ (the constant term):

$$
   \begin{array}{c|cccc}
   1 & 1 & -1 & -6 & 6 \
     &   & 1  & 0  & -6 \
   \hline
     & 1 & 0  & -6 & 0 \
   \end{array}
   $$

The quotient is $$x^2 - 6$$, so we have:  
   $$
   f(x) = (x-1)(x^2-6)
   $$

4. Set the quadratic factor equal to zero and solve for $$x$$:  
   $$
   x^2 - 6 = 0 \quad \Rightarrow \quad x^2 = 6 \quad \Rightarrow \quad x = \sqrt{6} \text{ or } x = -\sqrt{6}
   $$

5. Therefore, the roots of the function are:  
   $$
   x = 1,\quad x = \sqrt{6},\quad x = -\sqrt{6}
   $$
The integer root is $$ x = 1 $$. The complex roots are $$ x = \sqrt{6} $$ and $$ x = -\sqrt{6} $$.

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