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

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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)   (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), . Divide f(m) by ; and then work the rest of the function's factoring...

I'm not showing that process, but DIVIDED BY is . Now, use synthetic division to "test for " root , and work with the resulting quotient of that.

You're given root of for the factor .
-2   |   1   -4  -3     18
     |
     |      -2   12    -18
     |______________________
        1   -6    9      0

This meaning, the resulting factor to continue being  .
Recognize that this is a perfect square trinomial, .







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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
.
Perform the indicated multiplication of that. Remember as you go, .
Answer by greenestamps(13200)   (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



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



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

To find those coefficients, we know that

}

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




So now we know the factorization is



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