Question 373514
{{{y= x^3+2x^2-5x-6}}}
Since x-1 is a factor, then (x-1) should divide evenly into {{{x^3+2x^2-5x-6}}}. We will use synthetic division to divide:
<pre>
1 |  1  2  -5  -6
---     1   3  -2
    -------------
     1  3  -2  -8
</pre>
The remainder, in the lower right corner, is -8. This means that (x-1) is <b>not</b> a factor of {{{x^3+2x^2-5x-6}}}! (If (x-1) was a factor then the remainder would be zero.) So either the problem was given to you incorrectly or you posted it incorrectly.<br>
If we pretend that we did get a reminder of zero, then the rest of the bottom line represents the other factor. The
1  3  -2
translates into
{{{1x^2 + 3x - 2}}}
Then, to find the other factors, we would factor {{{1x^2 + 3x - 2}}}.