Question 1123541
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Let f(m) = m˄5 - 11m˄3 - 26m˄2 + 48m + 144. Given that  -2 + 2i and  -2 are
roots of f(m), find all the other roots of f(m) and write f(m) as a product of irreducible real quadratic and linear functions.
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Your given complex root AND its conjugate, together will give you a quadratic root or factor of f(m), {{{m^2+4m+8}}}.  Divide f(m) by {{{m^2+4m+8}}}; and then work the rest of the function's factoring...


I'm not showing that process, but {{{m^5+0m^4-11m^3-26m^2+48m+144}}} DIVIDED BY  {{{m^2+4m+8}}} is  {{{m^3-4m^2-8m+18}}}.  Now, use synthetic division to "test for " root {{{-2}}}, and work with the resulting quotient of that.


You're given root of {{{-2}}} for the factor {{{x+2}}}.
<pre>
-2   |   1   -4  -3     18
     |
     |      -2   12    -18
     |______________________
        1   -6    9      0

This meaning, the resulting factor to continue being  {{{m^2-6m+9}}}.
</pre>Recognize that this is a perfect square trinomial,  {{{(m-3)^2}}}.


{{{highlight(f(m)=(x+2)(m^2+4m+8)(m-3)^2)}}}






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Complex roots of polynomial functions occur as conjugate pair.
The given root -2+2i  means that -2-2i is also a root of your function.  You can get the resulting quadratic factor starting as 
{{{(m-(-2+2i))(m-(-2-2i))}}}.
Perform the indicated multiplication of that.  Remember as you go, {{{i^2=-1}}}.