SOLUTION: Let f(m) = m˄5 - 11m˄3 - 26m˄2 + 48m + 144. Given that m = -2 + 2i and m = -2 are roots of f(m), find all the other roots of f(m) and write f(m) as a product of irr

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson  -> Lesson -> SOLUTION: Let f(m) = m˄5 - 11m˄3 - 26m˄2 + 48m + 144. Given that m = -2 + 2i and m = -2 are roots of f(m), find all the other roots of f(m) and write f(m) as a product of irr      Log On


   

Question 1123541: Let f(m) = m˄5 - 11m˄3 - 26m˄2 + 48m + 144. Given that m = -2 + 2i and m = -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.
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
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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%5E2%2B4m%2B8. Divide f(m) by m%5E2%2B4m%2B8; and then work the rest of the function's factoring...

I'm not showing that process, but m%5E5%2B0m%5E4-11m%5E3-26m%5E2%2B48m%2B144 DIVIDED BY m%5E2%2B4m%2B8 is m%5E3-4m%5E2-8m%2B18. 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%2B2.
-2   |   1   -4  -3     18
     |
     |      -2   12    -18
     |______________________
        1   -6    9      0

This meaning, the resulting factor to continue being  m%5E2-6m%2B9.
Recognize that this is a perfect square trinomial, %28m-3%29%5E2.

highlight%28f%28m%29=%28x%2B2%29%28m%5E2%2B4m%2B8%29%28m-3%29%5E2%29





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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
%28m-%28-2%2B2i%29%29%28m-%28-2-2i%29%29.
Perform the indicated multiplication of that. Remember as you go, i%5E2=-1.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


I would go about this in a different order....

Given the root of -2, first use synthetic division to remove that root.

  -2 |  1  0  -11  -26  48  144
     |    -2    4   14  24 -144
     --------------------------
        1 -2   -7  -12  72   0

At this point we know

m%5E5+-+11m%5E3-26m%5E2%2B48m%2B144+=+%28m%2B2%29%28m%5E4-2m%5E3-7m%5E2-12m%2B72%29

Next, given the root -2+2i, we know -2-2i is also a root, because complex roots occur in conjugate pairs.

We can get the quadratic factor corresponding to that pair of roots by using the fact that in the quadratic equation x^2+bx+c=0 the sum of the roots is -b and the product is c.

The sum of these two roots is -4; their product is 4-4i^2 = 4+4 = 8. So the quadratic factor corresponding to these two roots is m^2+4m+8.

So now we know

m%5E5+-+11m%5E3-26m%5E2%2B48m%2B144+=+%28m%2B2%29%28m%5E2%2B4m%2B8%29%28m%5E2%2Bam%2Bb%29

where the coefficients a and b in the second quadratic factor are yet to be determined.

To find those coefficients, we know that

%28m%5E4-2m%5E3-7m%5E2-12m%2B72%29=%28m%5E2%2B4m%2B8%29%28m%5E2%2Bam%2Bb%29}

We can immediately see that b=9 by looking at the constant term: 72 is equal to 8 times b.

And one quick way (with a little practice) to find the coefficient a is to see that the coefficient of the m^3 term, -2, comes from the two partial products (m^2)*(am) and (4m)(m^2). So

-2+=+a%2B4
a+=+-6

So now we know the factorization is

m%5E5+-+11m%5E3-26m%5E2%2B48m%2B144+=+%28m%2B2%29%28m%5E2%2B4m%2B8%29%28m%5E2-6m%2B9%29

And finally we see that the second quadratic factor is reducible, and the final complete factorization is

m%5E5+-+11m%5E3-26m%5E2%2B48m%2B144+=+%28m%2B2%29%28m%5E2%2B4m%2B8%29%28m-3%29%5E2