Question 1181501
<br>
In the hopes of making the problem easier, we can, as the other tutor did, assume ("hope") that the polynomial has real coefficients.<br>
In that case, follow the solution given by the other tutor:
if (2+i) is a root then (2-i) is a root;
determine the quadratic factor that comes from those two roots;
divide the given function by that quadratic and observe that the result is another quadratic with no remainder -- confirming that our assumption of real coefficients was okay; and
use the quadratic formula to find the two roots corresponding to that second quadratic factor.<br>
I would like to add a bit to her solution, showing different ways that parts of the work can be done.<br>
She finds the quadratic corresponding to the two root (2+i) and (2-i) by doing the multiplication<br>
{{{(z-(2+i))(z-(2-i))}}}<br>
Note that she makes that multiplication easier by writing it as<br>
{{{((z-2)-i)((z-2)+i)}}}<br>
So that the product can be found as a difference of squares.<br>
Here is another way to form the quadratic corresponding to those two roots.<br>
The linear term of the quadratic is the opposite of the sum of the two roots: -((2+i)+(2-i)) = -4<br>
The constant term of the quadratic is the product of the two roots: (2+i)(2-i)=4+5 = 5<br>
So the quadratic factor having the roots (2+i) and (2-i) is z^2-4x+5.<br>
In the next step, she shows dividing the given polynomial by z^2-4x+5 to get z^2+6x+10 with no remainder.<br>
She doesn't show how she got that.  Synthetic division only works for dividing by linear polynomials, and long division of polynomials is awkward.<br>
So here is the process I like to use to divide a 4th degree polynomial by a quadratic.<br>
We are looking for a quadratic polynomial {{{az^2+bz+c}}} for which<br>
{{{(az^2+bz+c)(z^2-4z+5)=z^4+2z^3-9z^2-10z+50}}}<br>
By looking at the leading coefficient and the constant term of the product, we can immediately determine that a=1 and c=10.  So the multiplication of the two quadratics is<br>
{{{(z^2+bz+10)(z^2-4z+5)=z^4+2z^3-9z^2-10z+50}}}<br>
We can now determine b by looking at where the z^3 term of the product comes from: bz times z^2, plus z^2 times (-4z)<br>
{{{(b)(1)+(1)(-4) = 2}}}
{{{b-4=2}}}
{{{b=6}}}<br>
So our second quadratic factor is {{{z^2+6z+10}}}<br>
That process for finding the second quadratic factor is easier for me than long division of polynomials.<br>