If you figured out x = 1, -1, 2, -3

Then you solved x

I dont get what you're asking.

 

I got that solution from the back of the book, but it doesnt show how to actually do it. i need the proceedure itself.

 

answer:

show me one example in the real world where i will need to apply this for my chosen career path


i bet the prof likes that one

 

Well, its not that. This is actually not for an algebra class, this is for a Differential Equations class, which is after Calc 1,2, and 3.

If you would be studying to be an Engineer, like me, I supose you would need it. But then egain, you would not have to solve it by hand...so yeah.

 

http://www.algebrahelp.com/

 

You da man! :cheers:

Found what I was looking for.

This is how its done if anyone's interested:

x^4 + x^3 -7x^2 - x + 6 = 0

(x^4 - 7x^2 + 6) + x(x^2 - 1)=0

(x^2 -6)(x^2 - 1) + x(x^2 - 1)=0

(x^2 +x - 6)(x^2 - 1)=0

(x + 3)(x - 2)(x^2 - 1)=0

x = -3, x = 2, x= +/-1

 

Looks like I came just a tad too late h:

 

Dont worry, I got one for ya.

When you have four equations whats the best way to solve for the constants c?

c1 + c2 + c3 + c4 = 1

c1 - c2 + 2c3 - 3c4 = 0

c1 + c2 + 4c3 + 9c4 = -2

c1 - c2 + 8c3 - 27c4 = -1

I cant seem to get a single equation with only one variable.

Thanks in advance.

 

Im stumped lol

 

yeh, i have no idea how to do this shit.

.