Factorizing a polynomial of degree 4 that has complex roots

Question

While working on differential equations with constant coefficients I came across the following auxiliary equation: $r^4 - 4r^3 + 9r^2 - 10r + 6 = 0$.

Initially I tried the hit and trial method for common real values of r. That did not work out. The answer in the book shows that this equation has only complex roots.

How do I go about factorizing this equation?

Answer 1

It factors as a product of two quadratics. If you set $$r^4 - 4r^3 + 9r^2 - 10r + 6 = (r^2 + ar + b)(r^2 + cr + d),$$ equate corresponding coefficient and solve, you will get $$r^4 - 4r^3 + 9r^2 - 10r + 6 = (r^2 - 2r + 2)(r^2 - 2r + 3).$$

Answer 2

In Maxima CAS :

$$factor(r^4−4*r^3+9*r^2−10*r+6);$$

gives :

$$(r^2−2*r+2)*(r^2−2*r+3)$$

Answer 3

Hint: Take a good look at the fourth row of Pascal's triangle: What do you notice ? Spoilers below:

Completing the quadric, we have $(x-1)^4+3x^2-6x+5=(x-1)^4+3(x-1)^2+2$. Now let $t=(x-1)^2$, and then solve or factor the quadratic.