# Factorizing a polynomial of degree 4 that has complex roots

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?

• – lab bhattacharjee Dec 30 '14 at 16:31
• Any real polynomial can be factored as product of polynomials of degree not larger than $2$. Also, if it is monic, it is the product of monic polynomials. Therefore, for some $a,b,c,d\in \mathbb R$, you can write $$r^4 - 4r^3 + 9r^2 - 10r + 6 = \left(r^2 + ar + b\right)\left(r^2 + cr + d\right)$$ and solve a system of equations. – Git Gud Dec 30 '14 at 16:36
• @GitGud Thanks for your help. This comment perfectly answers my query. – Deepabali Roy Dec 31 '14 at 3:33

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

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.

In Maxima CAS :

$$factor(r^4−4*r^3+9*r^2−10*r+6);$$

gives :

$$(r^2−2*r+2)*(r^2−2*r+3)$$

• This doesn't help someone learn any math. – Tim Raczkowski Dec 30 '14 at 16:50
• One can look at the code to see that there are rules to solve such questions. – Adam Jan 1 '15 at 16:05