Question

Please solve the equations:
x^4+4x^3=x^2+16x+12

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

Answer 1

for: x^4-3x^3+4x^2-6x+4=0
x=1 and x=2 the other are complex

Answer 2

yes, but how i should count that?

Answer 3

do you know how to do synthetic division yet?

Answer 4

rather yes, but in this equation i can't do this :(

Answer 5

solutions to polynomials like this takes a lot of guessing or try to graph them

Answer 6

for this one x^4-3x^3+4x^2-6x+4=0
the roots maybe are: 1,3,4,2,etc you have to try to do synthetic divisions here using these as trials

Answer 7

ok, so the roots depens from " ^ " ? yes?

Answer 8

like for this one,
x^4+4x^3=x^2+16x+12 simplify it first to
x^4+4x^3-x^2-16x-12=0, now the possible roots are 1,-1,2,-2,3,-3,4,-4 etc, now do the synthetic divisions here using these numbers

Answer 9

ok, but how i should to know these numbers?

Answer 10

x^4-3x^3+4x^2-6x+4=0 the coefficients of the polynomials are, 1,-3,4,-6 did you see them? so the possible roots are + and negative of them

Answer 11

positive or negative multiples of them

Answer 12

ok, i see :) thanks for your help :)

Answer 13

ok yw, good luck now

Answer 14

thank you very much :)

Answer 15

by the way, do your synthetic divisions and check your answer,
x^4+4x^3-x^2-16x-12=0 for this the roots are -3.-2,-1,2 ans...

Answer 16

yes, i have the same anwers, thanks once again :)