# Thread: hard quartic equation problem.

1. ## hard quartic equation problem.

this one is giving me some trouble. I am not sure what method to use.

I figure first expand the equation which gives me quartic form.

but after that I am stuck. I don't think I can factor with the methods I know. I tried a few ways. and substitution seems to fail to get anywhere.

the solution is which does expand to the the same quartic as above. I have NO idea how to do this.

2. Let .

.

Expanding this out and collecting like terms will give you the required form.

3. I tried substitution but not the way you did it in the quartic thanks!

how did you know to use (u+1)?

4. You let u equal whatever you want your function to be a function of, in this case, x - 1.

5. I see, thanks.

6. Hello, skoker!

We have: .(3x + 2)
[(x-1)^3 + 3(x-1)^2 + 2(x-1) + 1]

. . . . . = .
[3(x-1) + 5] [(x-1)^3 + 3(x-1)^2 + 2(x-1) + 1]

. . . . . = . 3(x-1)^4 + 9(x-1)^3 + 6(x-1)^2 + 3(x-1)
. . . . . . . . . . . . . . + 5(x-1)^3 + 15(x-1)^2 + 10(x-1) + 5

. . . . . = . 3(x-1)^4 + 14(x-1)^3 + 21(x-1)^2 + 13(x-1) + 5

7. thank you soroban. that is close to how I was trying to do substitution, before distribution. but how do you get the 5 in [3(x-1)+5]?