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

Question

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!

Answer 1

Alexander, I did long division by x² + 1 and got

x² - 4x + 8, as opposed to your x² + 4x + 8.

Using the quadratic formula, I got x = 2 +/- 2i

Your analysis appears reasonable, but I think you dropped a sign.

When you simplified to b² + 4b + 8, you used the quadratic formula, which should have given

(1/2)(-4 +/- sqrt(16-32)), resulting in

b = -2 +/- 2i instead of your b = 2 +/- 2i

Then, substituting so as to solve for a, you would have gotten

a - 2 +/- 2i = -4

and therefore

a = -2 +/- 2i, the same as b.

Then (x²+1)(x-2+/-2i)² = 0

so x = +/- i or

x = 2 +/- 2i