Alexander L.

asked • 05/27/19

Finding the complex roots of a polynomial of the fourth degree.

I was working throught a question in my math book to which I found an answer, but an incorrect one. I was hoping that you guys could help me find the error in my calculations.


Question: The equation x4 - 4x3 + 9x2 - 4x + 8 = 0 has a factor of x2 + 1. Find all roots.


My answer: Because the equation has a root of x2 + 1 i rewrote the equation in the form of (x2 + 1)*q(x) = 0, where q(x) is a polynomial. Because the original equation has no coefficient before 4th-degree term, q(x) = (x + a)(x+b) where a, b are constants.


I then wrote the equation x4 - 4x3 + 9x2 - 4x + 8 = (x2 + 1)(x + a)(x+b) and simplified both sides until I got to the point where x4 - 4x3 + 9x2 - 4x + 8 = x4 + (a+b)x3 + (ab + 1)x2 + (a+b)x + ab. At that point I deduced that the coefficients of the nth-degree term must be equal on both sides, and therefore got the equations:

a + b = -4

ab = 8


From there on I solved the system of equations by isolating a in the first equation, resulting in a = -(b+4). Then I substituted a into the second equation, getting -b(b+4) = 8, which simplifies to b2 + 4x + 8 = 0, which I solved using the quadratic formula, getting the result of b = 2 ± 2i.


Then I solved for a by plugging b in to the first equation, getting a + 2 ± 2i = -4. Solving that equation yields a = -6 ± 2i.


Because the first polynomial could be written in the form of (x2 + 1)(x + a)(x+b) = 0, either:

x2 + 1 = 0, and then x = ±i

x + a = 0, and then x = -a = 6 ± 2i

x + b = 0, and then x = -b = -2 ± 2i


Only x = ±i of my three answers was correct. Thanks in advance!

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