Which shows one way to determine the factors of x3 – 9x2 + 5x – 45 by grouping?

Factoring by grouping usually pairs up the first 2 sets of expressions with the second 2 sets.  Ours looks like this, then:  [tex](x^3-9x^2)+(5x-45)=0[/tex].  If we factor out the common x-squared in the first set of parenthesis, along with factoring out the common 5 in the second set, we get this:  [tex]x^2(x-9)+5(x-9)[/tex].  Now the common expression that can be factored out is the (x-9).  When we do that, here's what it looks like:  [tex](x-9)(x^2+5)[/tex].  I'm not sure how far you are going with this.  You could set each of those equal to 0 and solve for x in each case.  The first one is easy.  If x - 9 = 0, then x = 9.  The second one involves the imaginary i since x^2 = -5.  In that case,  [tex]x=i \sqrt{5},-i \sqrt{5} [/tex].  Hopefully, in what I have given you, you can find what you're looking for.