Question 216460
Find all real and complex zeros for {{{h(x)=x^3-x^2-7x-15}}}
P: 1,3,5,15
Q:1<br>
Since the function as you given it has no rational roots, I'm going to guess that  you have a typo and {{{h(x)=x^3-x^2-7x+15}}}. If this is wrong, then stop reading  (and you may need to repost your question).<br>
With this h(x) we can find that -3 is a rational root. To illustrate I'll use synthetic division because it will also show how to factor h(x):
<pre>

-3 |    1   -1   -7    15
----        -3   12   -15
       -------------------
        1   -4    5    0
</pre>
The remainder, 0, tells us that h(-3) = 0. So -3 is a root of h(x). And if -3 is a root of h(x), then (x+3) is a factor of h(x). And the other factor is found in the numbers in front of the reaminder above, "1  -4   5", which translates into {{{x^2 -4x + 5}}}.<br>
So {{{h(x) = (x+3)(x^2 -4x + 5)}}}. The other roots for h(x) will come from the roots of {{{x^2-4x+5}}}. This is a quadratic. Since it will not factor we'll use the quadratic formula:
{{{x = (-b +- sqrt(b^2 - 4ac))/(2a)}}}
For our expression, a = 1, b = -4 and c = 5. Substituting these into the formula we get:
{{{x = (-(-4) +- sqrt((-4)^2 - 4(1)(5)))/(2(1))}}}
Simplifying...
{{{x = (4 +- sqrt(16 - 4(1)(5)))/2}}}
{{{x = (4 +- sqrt(16 - 20))/2}}}
{{{x = (4 +- sqrt(-4))/2}}}
With the negative in the square root, we will have complex roots.
{{{x = (4 +- sqrt(-1*4))/2}}}
{{{x = (4 +- sqrt(-1)*sqrt(4))/2}}}
{{{x = (4 +- i*2)/2}}}
{{{x = (2(2 +- i))/2}}}
The 2's cancel leaving:
{{{x = 2 +- i}}}
So our three roots are: -3, 2+i and 2-i