# Thread: factorising a polynom with it's roots

1. ## factorising a polynom with it's roots

The polynom $A= (a-b)x^4+(c+d)x^3+(e-f)x^2+(g+h)x$ has the roots 0 and the roots of unity, so A is a multiplicity of $x(x^3-1)$. That means $c+d=0, e-f=0$. And one can say 4a-b=1$and$g+h=-1$without loss of generality. My question ist: why it is true only with w.l.o.g ? 2. Originally Posted by ABC3 The polynom$A= (a-b)x^4+(c+d)x^3+(e-f)x^2+(g+h)x$has the roots 0 and the roots of unity, so A is a multiplicity of$x(x^3-1)$. That means$c+d=0, e-f=0$. And one can say 4a-b=1$ and $g+h=-1$ without loss of generality. My question ist: why it is true only with w.l.o.g ?
$\displaystyle (a-b) x^4 +(c+d)x^3+(e-f)x^2 +(g+h)x = x^2 [(a-b)x^2 + (c+d) ] + x [(e-f)x + (g+h) ]$

P.S. what is w.l.o.g. ? not all have native English

3. that means: without loss of generality

4. You are given $\displaystyle (a- b)x^4+ (c+ d)x^3+ (e- f)x^2+ (g+ h)x= x((a-b)x^3+ (c+d)x^2+ (e-f)x+ (g+h))$ and are told that this is equal to $\displaystyle Ax(x^3- 1)= Ax^4- A$ where A= a- b. Then, obviously, c+ d= e- f= 0.

Also, g+ h= -A= -a+ b. I don't understand that "4" in "4a- b" at all. If it had said "we can take a- b= 1 and g+ h= -1" I could understand that- factor out (a- b).

5. Thank You HallsofIvy for your Answer. "a-b=1" ist correct without "4". Now about my question: we can take a-b=1 and g+h=-1, but that is only one of the possible answers ! so it ist not generally correct, am I right?