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?
Best Answer
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 $$r^4 - 4r^3 + 9r^2 - 10r + 6 = (r^2 - 2r + 2)(r^2 - 2r + 3).$$