bergausstein said:
our instructor gave us this problem. i don't know from where did he get this.
i want to learn how to solve this. I've used all the method i know to solve this but no success.
please help me.
paulmdrdo said:
would you help me with this
http://mathhelpboards.com/pre-algebra-algebra-2/challenging-factorization-8267.html
please help me. thanks!
Hello.
Another thing that the brute force I can't think.(Headbang)
If the question, allows you to easily factor with respect to the "x" (6), we can divide, at least in two factors.
1º) We ordered:
x^6(-18y)+x^5(30y^2-6)+x^4(24y^2+10y+24)+x^3(33y^2-32y)+x^2(35y^3-14y)+x(28y^3-72y)+63y^3
2º)
(ax^3+bx^2+cx+d)(ex^3+fx^2+gx+h)=
=x^6(ae)+x^5(af+be)+x^4(ag+bf+ce)+x^3(ah+bg+cf+ed)+x^2(bh+cg+df)+x(ch+dg)+dh
without losing generality, we can assume:
d=7 \ and \ h=9
at the end there is to find the component "y".
x(ch+dg)=x(9c+7g)=x(28y^3-72)
deducting:
c=-8 \ and \ g=4
x^2(bh+cg+df)=x^2(9b-32+7f)=x^2((35y^3-14y)
If \ f=-2 \ then \ b \cancel{\in}{Z}
f=5 \ and \ b=2 \ , \ (9b-32=-14)
And so on. You, see if the system works should try.
Regards.