Question

what are the answers of this equation?
(x+1)^2+(x+2)^3+(x+3)^4=2

Answer 1

HI!!

Answer 2

maybe you are supposed to multiply out and then try to factor
answer is here
http://www.wolframalpha.com/input/?i=%28x%2B1%29%5E2%2B%28x%2B2%29%5E3%2B%28x%2B3%29%5E4%3D2

Answer 3

\[\begin{align*}
2&=(x+1)^2+(x+2)^3+(x+3)^4\\\\
0&=\color{red}{(x+1)^2-9}+\color{green}{(x+2)^3+8}+\color{blue}{(x+3)^4-1}\\\\
0&=\color{red}{\bigg((x+1)-3\bigg)\bigg((x+1)+3\bigg)}+\color{green}{\bigg((x+2)+2\bigg)\bigg((x+2)^2-2(x+2)+4\bigg)}\\
&\quad\quad+\color{blue}{\bigg((x+3)^2-1\bigg)\bigg((x+3)^2+1\bigg)}\\\\
0&=\color{red}{(x-2)(x+4)}+\color{green}{(x+4)\bigg((x+2)^2-2(x+2)+4\bigg)}\\
&\quad\quad+\color{blue}{\Bigg(\bigg((x+3)-1\bigg)\bigg((x+3)+1\bigg)\Bigg)\bigg((x+3)^2+1\bigg)}\\\\
0&=\color{red}{(x-2)(x+4)}+\color{green}{(x+4)\bigg((x+2)^2-2(x+2)+4\bigg)}\\
&\quad\quad+\color{blue}{(x+2)(x+4)\bigg((x+3)^2+1\bigg)}\\\\
0&=(x+4)\Bigg((x-2)+\bigg((x+2)^2-2(x+2)+4\bigg)+(x+2)\bigg((x+3)^2+1\bigg)\Bigg)
\end{align*}\]
From this rearrangement you know that \(x=-4\) must be a root.

Similarly,
\[\begin{align*}
2&=(x+1)^2+(x+2)^3+(x+3)^4\\\\
0&=\color{blue}{(x+1)^2-1}+(x+2)^3+\color{red}{(x+3)^4-1}\\\\
0&=\color{blue}{\bigg((x+1)+1\bigg)\bigg((x+1)-1\bigg)}+(x+2)^3+\color{red}{\bigg((x+3)^2+1\bigg)\bigg((x+3)^2-1\bigg)}\\\\
0&=\color{blue}{x(x+2)}+(x+2)^3+\color{red}{\bigg((x+3)^2+1\bigg)\bigg((x+3)-1\bigg)\bigg((x+3)+1\bigg)}\\\\
0&=\color{blue}{x(x+2)}+(x+2)^3+\color{red}{(x+2)\bigg((x+3)^2+1\bigg)\bigg((x+3)+1\bigg)}\\\\
0&=(x+2)\bigg(x+(x+2)^2+\left((x+3)^2+1\right)(x+4)\bigg)
\end{align*}\]
tells you that \(x=-2\) is also a root.

Expanding the original quartic gives
\[\begin{align*}2&=(x+1)^2+(x+2)^3+(x+3)^4\\\\
0&=(x^2+2x+1)+(x^3+6x^2+12x+8)+(x^4+12x^3+54x^4+108x+81)-2\\\\
0&=x^4+13x^3+61x^2+122x+88
\end{align*}\]
You can proceed with finding the remaining roots via long/synthetic division.