Question 203745
{{{x^3-4x^2+2x+1=0}}}
To use the quadratic formula you must have a quadratic equation. At the start, this problem is not a quadratic equation. What we must do is factor your polynomial and, by doing so, get a factor which is quadratic.<br>
There are many techniques that can be used to factor expressions (GCF, patterns, grouping, trinomial and trial and error of the rational roots, etc.). I'm not going to try to explain all these to you in this answer to your problem.<br>
A technique that works on this polynomial is the last one: trying possible rational roots. One possible rational root is 1. (1 is a possible rational root of <b>every</b> polynomial.) And 1 works. (Try it and see.)
Since 1 is a root of the polynomial, (x - 1) is a factor of the polynomial. In other "words":
x^3-4x^2+2x+1 = (x - 1)(?)
To find the other factor we can use either log or synthetic division. I'll use synthetic:
<pre>
 1    1  -4  2  1
          1 -3 -1
      -----------
      1  -3 -1  0
</pre>
The "1  -3 -1" in front of the 0 tells me that 1x^2 - 3x - 1 is the other factor. So
{{{x^3-4x^2+2x+1 = (x - 1)(x^2 -3x - 1) = 0}}}
In order for this product ot be zero, one of the factors must be zero (Zero Product Property). So
{{{x - 1 = 0}}} or {{{x^2 -3x - 1 = 0}}}<br>
Now we have the quadratic equation we needed. We can use the quadratic formula on {{{x^2 -3x - 1 = 0}}} to find the irrational roots. (I'll leave it up to you to finish this.) In the end we will have three roots. One rational root (which is 1) and the two irrational roots you will get from the quadratic formula.