B; The solutions are 5, -2, 3i, and -3i
Let's look at the original equation:
\(x^4-3x^3-x^2-27x-90=0\)
Our goal is to get the left hand side into parts that are more manageable to work with. I'll try to explain this. We'll use the rational root theorem. This says that if a polynomial equation can be written in the form \(a_nx^n+a_{n-1}x^{n-1}+...+a_0\) and if \(a_0\) and \(a_n\) are integers, then the factor can be found by checking numbers produced by doing \(\pm\frac{dividers\hspace{1mm}of\hspace{1mm}a_0}{dividers\hspace{1mm}of\hspace{1mm}a_n}\).
Dividers of \(a_0:\hspace{1mm}1,2,3,5,6,9,10,15,18,30,45,90\)
Dividers of \(a_n:\hspace{1mm}1\)
Therefore, check for one rational solution by doing: \( \pm\frac{1,2,3,5,6,9,10,15,18,30,45,90}{1}\)
-2 happens to be the first value that satisfies this equation. The opposite of that is 2, so a factor is x+2.
\(\frac{x^4-3x^3-x^2-27x-90}{x+2}=x^3-5x^2+9x-45\), so the monstrosity we had before goes to \((x+2)(x^3-5x+9x-45)=0\).
Thankfully, this is the hardest part. Now, we have to factor \((x^3-5x^2+9x-45)\) further. Find another factor by grouping:
\((x^3-5x^2)+(9x-45)\)
Factor out an x^2 from the first set of parentheses and 9 from the second set and then factor the resulting common term, which happens to be (x-5):
\((x^3-5x^2)+(9x-45)=x^2(x-5)+9(x-5)=(x^2+9)(x-5)\)
After all this effort, we have transformed \(x^4-3x^3-x^2-27x-90\) to \((x+2)(x-5)(x^2+9)\).
Okay, now use the zero factor principle. In other words, set each factor to zero:
\(x+2=0\hspace{1cm}x-5=0\hspace{1cm}x^2+9=0\)
Let's focus on the easy ones first:
\(x=-2 \) and \(x=5\).
Now the harder factor:
\(x^2+9=0\)
\(x^2=-9\)
\(x=\pm\sqrt{-9}\)
\(x=\pm3i\)
Solution set is 5,2,3i, -3i
