Question

Using the given solutions, find the others:
x^3+x^2+9x-9=0 ;1
Can someone get me started on this problem?

Answer 1

look the image
are you sure that 1 is root ?
or something is wrong ?
for example x3-x2+9x-9=0

Answer 2

1 was a given solution

Answer 3

1 is not a root look the image
it has one root and its not 1
its 0.85

Answer 4

But i don't have graphs on my worksheet. How would you solve the equation. Sorry i don't get the graph

Answer 5

Are you certain the equation isn't \[x^3-x^2+9x-9=0\] which does have a root at x=1?

Answer 6

your problem is wrong !
or you type it wrong
if it was x3-x2+9x-9 =0
on of the roots is 1

Answer 7

Yes it is that equation sorry!

Answer 8

now its ok ?

Answer 9

Well, if x = 1 is a known root, that means we can factor (x-1) out of the equation. Eventually we'll end up with (x-r_1)(x-r_2)(x-r_3) = 0 where r_1, r_2, r_3 are our roots.

\[x^3-x^2+9x-9 = (x-r_1)(x-r_2)(x-r_3) = 0, r_1=1\]

Answer 10

Do you know how to do division of polynomials?

Answer 11

yes i know
but you know?
if ok you can divide by (x-1) then you have degree 2 polynomial

Answer 12

Thank you sorry i didn't get it at first but now i do! :)

Answer 13

The other roots may not be real, but the problem didn't excuse us from finding them!

\[(x^2+9)(x-1)=0\]x=1 is a solution, as we already know. \[(x^2+9) = 0\] can be solved by observing that \[(x+a)(x-a)=x^2+ax-ax-a^2=x^2-a^2 = x^2+i^2a^2\] where \[i = \sqrt{-1} \]and\[i^2=-1\]If we take \[a^2=9\] then we have \[(x^2+9)=(x+3i)(x-3i)=0\] and our other two roots are x = -3i and x = 3i.

If we substitute x=3i into our original equation, let's see what we get:
\[(3i)^3-(3i)^2+9(3i)-9=27i^3-9i^2+27i-9\]Recalling the relationship between i and -1
\[=27i(-1)-9(-1)+27i-9=-27i+9+27i-9=0\]so our root x=3i checks out. Verifying x=-3i is left as an exercise for the reader.