Question

can someone solve +9 for me? i haven't had any luck and I've been on it for 2 days...

Answer 1

i meant #9

Answer 2

i appreciate all the help :)

Answer 3

Let's begin by writing out our roots in the form specified by the question. So,
\[\left( x-3 \right)\left( x-(3+i) \right)\left( x - (3-i) \right) = 0\]

So that should be Part I. For Part 2, we FOIL out the complex roots. Thus, we have
\[\left( x-3 \right)\left( x ^{2} - x(3-i) - x ( 3+i)+(9 - i ^{2})\right) = 0\]

Simplifying, we have
\[(x-3)(x^2 - 3x +ix -3x -ix + 9 + 1) = 0\]
\[(x-3)(x^2 - 6x +10) = 0\]
Here, it's important to recall that
\[i^2 = -1\]
Lastly, we carry out the distribution of the last root giving us
\[(x-3)(x^2 - 6x +10) = 0\]
\[x^3 - 6x^2 + 10x - 3x^2 +18x - 30 = 0\]
\[x^3 - 9x^2 +28x - 30 = 0\]

Answer 4

thank you so much :)

Answer 5

I hope that helps. Try and go through it on your own to make sure that you understand the concepts involved here; they are fundamental and quite important.

Answer 6

i did! i just went through it @RBauer4 but on #8, i am not sure what to do because its quadratic and not cubic..

Answer 7

hey @RBauer4 thank you so much man!

Answer 8

Hmmm. Well, for #8 I would say that you need to make sure you are employing the Rational Root Test properly. The possible roots should be
\[x = \frac{ \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12}{ 1 }\]

Once you got that, you should test each possible root in the polynomial by substituting each of them for x and seeing which three result in zero.