Question 825414
The possible rational zeros are <u>+</u>1/1 and <u>+</u>2/1 (which simplify to <u>+</u>1 and <u>+</u>2). Trying these we find that only -2 is actually a zero:
<pre>
-2  |   1   -4   -11   2
-----       -2    12  -2
       ------------------
        1   -6     1   0
</pre>The zero in the lower right corner is the remainder. (It is also f(-2)!) Since it is zero then what we divided by (x-(-2)) [or (x+2)] divides evenly. So (x+2) is a factor of f(x). Not only that but the rest of the bottom row tells us what the other factor is. The "1 -6 1" translates into {{{x^2-6x+1}}}. So now
{{{f(x) = x^3-4x^2-11x+2 = (x+2)(x^2-6x+1)}}}
The other zeros of f(x) will come from the second factor. Since it is a quadratic that won't factor, we can use the quadratic formula:
{{{x = (-(-6)+-sqrt((-6)^2-4(1)(1)))/2(1)}}}
Simplifying...
{{{x = (6+-sqrt(36-4(1)(1)))/2}}}
{{{x = (6+-sqrt(36-4))/2}}}
{{{x = (6+-sqrt(32))/2}}}
{{{x = (6+-sqrt(16*2))/2}}}
{{{x = (6+-sqrt(16)*sqrt(2))/2}}}
{{{x = (6+- 4*sqrt(2))/2}}}
{{{x = (2(3+- 2*sqrt(2)))/2}}}
{{{x = (cross(2)(3+- 2*sqrt(2)))/cross(2)}}}
{{{x = 3+- 2*sqrt(2)}}}
which is short for {{{x = 3+2*sqrt(2)}}} or {{{x = 3-2*sqrt(2)}}}<br>
So the three zeros for f(x) are: -2, {{{3+2*sqrt(2)}}} and {{{3-2*sqrt(2)}}}