Question 223550
x+1=9x^3 + 9x^2 solve the polynomial equation. 
Been working on this one for awhile and i cant seem to get this one. 4 answers it needs. 
<pre><font size = 4 color = "indigo"><b>
No, it only needs 3 "answers", "solutions" or "roots", whichever
label you wish to give them.  That's because the degree (largest
exponent of x) is 3.

{{{x+1=9x^3 + 9x^2}}}

Get zero on the right:

{{{-9x^3 - 9x^2+x+1=0}}}

Multiply through by -1 so leading coefficient will be
positive:


{{{9x^3 + 9x^2-x-1=0}}}

Possible rational zeros are

±1, ±{{{1/3}}}, and ±{{{1/9}}}

By DesCartes' rule of signs, there is one positive
answer (solution or root), and 2 or 0 negative ones.

Let's try x=1, that is, divide by x-1 to see if
1 is a solution:

1 | 9   9   -1   -1
  |     9  -18  -19
   ----------------
    9 -18  -19  -20

Nope, the remainder is -20, not zero, so x-1 is not a
divisor of {{{9x^3 + 9x^2-x-1}}}, and so x=1 is not
a solution.

Let's try x=-1, that is, divide by x+1 to see if
-1 is a solution:

-1 | 9   9  -1 -1
   |    -9   0  1
   --------------
     9   0  -1  0

Aha! The remainder is zero, so x+1 is a divisor of
of {{{9x^3 + 9x^2-x-1}}}, and therefore x=-1 is a
solution.

Anytime we divide by synthetic division and get a zero 
remainder, we have factored the polynomial.  In this
case we have factored {{{9x^3 + 9x^2-x-1}}} and now
we have

{{{(x+1)(9x^2+0x-1)=0}}}

or eliminating the {{{0x}}} term:

{{{(x+1)(9x^2-1)=0}}}

Now we can factor the expression in the second parentheses
as the difference of two perfect squares and get:

{{{(x+1)(3x-1)(3x+1)=0}}}

Setting each factor = 0,

x+1=0 gives x=-1

3x-1=0 gives x={{{1/3}}}

3x+1=0 gives x={{{-1/3}}}

Edwin</pre>