Log and Polynomial Question

[Darken1]

First Question:
Solve the following system of equations
log{x+1}y=2
log{y+1}x=1/4

Work:
Turned them into equations
(x+1)^2=y (y+1)^(1/4)=x

Substituted second equation into the first equation
((y+1)^(1/4)+1)^2=y

factored out and eventually got
((y+1)^1/4)^2+2((y+1)^1/4)+1=y
Tried to use quadratic formula (a=1,b=2,c=1) and got y=-1. -1=-1 Didn't seem relevant.

Tried to factor it out by substituting x = (y+1)^1/4
(x+1)(x+1)=x^4-1 x^4-1=(x^2+1)(x^2-1)=(x^2+1)(x+1)(x-1)
(x+1)(x+1)=(x^2+1)(x+1)(x-1)
(x+1)=(x^2+1)(x-1)(x+1)=(x^2+1)(x-1)
-(x+1) = -(x+1)
0=(x^2+1)(x-1)-(x+1)
x^3-x^2-2=0
Couldn't factor it out.

Also noticed
(x+1)(x+1)=y
x=-1,y=0
Second Question:
Consider the polynomial p(x)=x^4+ax^3+bx^2+cx+d, where a,b,c,d are real numbers.

Given that 1+i and 1-2i are zeroes of p(x), find the values of a,b,c,d.

Work:
x=1+i and x=1-2i are zeroes
(x-1-i)(x-1+2i)=0
Factored it out
x^2-2x+ix+3
Seems to be missing another factor to get to the original equation.
Perhaps I could divide the original equation by the equation I factored out to get the last factor?
No idea at all

Edit: I need sleep geez...

 

[Opalg]

First Question:
Solve the following system of equations
log{x+1}_{2} log{y+1}_{¼}

I don't understand the problem here, because the "system of equations" has no equals signs and therefore doesn't consist of equations.

Eventually
0=x^3-x^2+x-1
Couldn't factor it out.

However, if you are correct in getting to the equation $0 = x^3 - x^2 + x - 1$ then you can solve it by writing it as $0 = x^2(x-1) + x-1 = (x^2+1)(x-1).$ That has just one real solution $x=1$.

Second Question:
Consider the polynomial p(x)=x^4+ax^3+bx^2+cx+d, where a,b,c,d are real numbers.

Given that 1+i and 1-2i are zeroes of p(x), find the values of a,b,c,d.

Work:
x=1+i and x=1-2i are zeroes
(x-1-i)(x-1+2i)=0
Factored it out
x^2-2x+ix+3
Seems to be missing another factor to get to the original equation.
Perhaps I could divide the original equation by the equation I factored out to get the last factor?
No idea at all

Hint: If a polynomial with real coefficients has a complex zero $u+iv$, then the complex conjugate $u-iv$ is also a zero of the polynomial.

 

[Darken1]

I don't understand the problem here, because the "system of equations" has no equals signs and therefore doesn't consist of equations.

However, if you are correct in getting to the equation $0 = x^3 - x^2 + x - 1$ then you can solve it by writing it as $0 = x^2(x-1) + x-1 = (x^2+1)(x-1).$ That has just one real solution $x=1$.Hint: If a polynomial with real coefficients has a complex zero $u+iv$, then the complex conjugate $u-iv$ is also a zero of the polynomial.

My bad, fixed the problem.My equation was wrong.
So from the previous equation
(x+1)=(x^2+1)(x-1)
I just factored out right hand side and minused (x+1) from both sides to get x^3 - x^2 + x - 1-(x+1)=0.
Supposed to have been
x^3-x^2-2=0
Don't know how to factor this one out.
Side Note: How did you convert x^2(x−1)+x−1 into (x^2+1)(x−1)? Is there an equation for it?

Hm so I guess that means
x=1+i and x=1-2i
x=1-i and x=1+2i
are all zeroes? Think I can go from there.

 

[HallsofIvy]

My bad, fixed the problem.My equation was wrong.
So from the previous equation
(x+1)=(x^2+1)(x-1)
I just factored out right hand side and minused (x+1) from both sides to get x^3 - x^2 + x - 1-(x+1)=0.

Apparently you multiplied rather than "factored". There was no point in that - the standard method for solving such equations is to factor so if the expression is already factored, leave it that way!

Supposed to have been
x^3-x^2-2=0
Don't know how to factor this one out.

It had already been factored for you!

Side Note: How did you convert x^2(x−1)+x−1 into (x^2+1)(x−1)? Is there an equation for it?

The "distributive rule": ab+ ac= a(b+ c). Here, the "a" is x-1: x^2(x- 1)+ 1(x- 1)= (x^2+ 1)(x- 1).

Hm so I guess that means
x=1+i and x=1-2i
x=1-i and x=1+2i
are all zeroes? Think I can go from there.

No, if (x^2+ 1)(x- 1)= 0 then x^2+ 1= 0 or x- 1= 0. The first of those is the same as x^2= -1 which has roots i and -i. x- 1= 0 only for x= 1. The three roots of the cubic equation are i, -i, and 1. I have no idea where you got "1+i", "1- 2i", "1- i" and "1+ 2i".