SOLUTION: Check that (2 + i) is a root of (z^4) + (2(z^3)) - (9(z^2)) - 10z + 50 = 0. What are the remaining 3 roots?

Question 1181501: Check that (2 + i) is a root of (z^4) + (2(z^3)) - (9(z^2)) - 10z + 50 = 0. What are the remaining 3 roots?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52797)   (Show Source): You can put this solution on YOUR website!
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Check that (2 + i) is a root of (z^4) + (2(z^3)) - (9(z^2)) - 10z + 50 = 0. What are the remaining 3 roots?
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            Yes, from the first glance, it is almost impregnable fortress.
            Let apply a military ruse.

Our polynomial is with real ( even with integer (!) ) coefficients. Hence, if (2+i) is a root, then (2-i) is a root, too (!) If so, then our polynomial must be divisible by (z-(2+i))*(z-(2-i)) = ((z-2)-i)*((z-2)-i) = (z-2)^2 + 1 = z^2 - 2z +5. Lets make long division to check if it is TRUE: = with NO REMAINDER (!) Thus we checked that the given polynomial is a multiple of the polynomial z^2 - 2z + 5. It means that (2+i) REALLY is a root of the given polynomial (without direct calculations (!)) +------------------------------------------------------------------------------------------------+ | So, we know now that | | | | (2+i) really is the root of the given polynomial; | | (2-i) is the other root; | | one factor to the given polynomial is z^2 - 2z + 5 with the roots (2+i) and (2-i); | | the other factor is the polynomial z^2 +6z + 10. | +------------------------------------------------------------------------------------------------+ At this point, the last step to do is to find the roots of the quadratic polynomial z^2 + 6z + 10. Apply the quadratic formula to get = = = = -3 +- i. The problem is just solved in full. We checked and proved that 2+i is the root. From it, we concluded that 2-i is the root, too. And finally, we found two remaining roots -3+i and -3-i.

Solved, answered, carefully explained and totally completed.

            T R I U M P H     (!)

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In his post, tutor @greenestamps writes (cited)

In the hopes of making the problem easier, we can, as the other tutor did, assume ("hope") that the polynomial has real coefficients.

I want to HIGHLIGHT that it is not an assumption: the given polynomial REALLY has integer (hence, real) coefficients.

It is not an assumption: it is a FACT.

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

In the hopes of making the problem easier, we can, as the other tutor did, assume ("hope") that the polynomial has real coefficients.

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.

I would like to add a bit to her solution, showing different ways that parts of the work can be done.

She finds the quadratic corresponding to the two root (2+i) and (2-i) by doing the multiplication

Note that she makes that multiplication easier by writing it as

So that the product can be found as a difference of squares.

Here is another way to form the quadratic corresponding to those two roots.

The linear term of the quadratic is the opposite of the sum of the two roots: -((2+i)+(2-i)) = -4

The constant term of the quadratic is the product of the two roots: (2+i)(2-i)=4+5 = 5

So the quadratic factor having the roots (2+i) and (2-i) is z^2-4x+5.

In the next step, she shows dividing the given polynomial by z^2-4x+5 to get z^2+6x+10 with no remainder.

She doesn't show how she got that. Synthetic division only works for dividing by linear polynomials, and long division of polynomials is awkward.

So here is the process I like to use to divide a 4th degree polynomial by a quadratic.

We are looking for a quadratic polynomial for which

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

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)

So our second quadratic factor is

That process for finding the second quadratic factor is easier for me than long division of polynomials.

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