Question 400450
Please post the entire problem. There is no way this is exactly how it was given to you. And use "^" to indicate exponents. For example:
x^4-5x^3+7x^2-5x+6<br>
I'm guessing the problem is something like: "Find the remaining roots (or zeros) of {{{x^4-5x^3+7x^2-5x+6}}} given that one of them is -i."<br>
Roots (or zeros) of a polynomial are values for the variable that make the polynomial equal to zero. And if "r" is a root of a polynomial then (x-r) is a factor of the polynomial.<br>
The fact that -i is a root of the polynomial tells you three things:<ul><li>(x-(-i)), or (x+i), is a factor of the polynomial.</li><li>If there are complex roots, like -i, of a polynomials with real coefficients, like yours, then the complex roots come in conjugate pairs. In general a+bi and a-bi (or a+(-bi)) are a conjugate pair. So if -i, or 0 + (-i), is a root, then its conjugate, 0 + i (or just i) is also a root.</li><li>If i is a root, then (x-i) is a factor.</li></ul>
So we have two roots, i and -i, so far. Your polynomial is a 4th degree polynomial (because the highest exponent is 4) and so it should have 4 roots. To find other two roots we will factor it.<br>
Since (x+i) and (x-i) are factors they will divide into the polynomial evenly. And the quotient will be the other factor. So we will divide by the known factors (using synthetic division):
<pre>
-i |  1   -5     7      -5      6
----      -i    -1+5i    5-6i  -6
     ----------------------------
      1   -5-i   6+5i   -6i     0
</pre>
So {{{x^4-5x^3+7x^2-5x+6 = (x - (-i))(x^3 + (-5-i)x^2 + (6+5i)x -6i) = (x +i)(x^3 + (-5-i)x^2 + (6+5i)x -6i)}}}. Now we will divide the second factor by the other known factor, (x-i):
<pre>
 i |  1   -5-i   6+5i   -6i 
----         i    -5i    6i
     ---------------------------
      1   -5     6       0
</pre>
So now we have:
{{{x^4-5x^3+7x^2-5x+6 = (x +i)(x-i)(x^2-5x+6)}}}
To find the last two roots we will factor the third factor. This is a fairly simple trinomial to factor:
{{{x^4-5x^3+7x^2-5x+6 = (x +i)(x-i)(x-2)(x-3)}}}
Remembering that if "r" is a root then (x-r) is a factor (and vice versa), we can see that the 4 roots of {{{x^4-5x^3+7x^2-5x+6}}} are: -i, i, 2 and 3.