How to solve $x^4-8x^3+24x^2-32x+16=0$ How can we solve this equation?
$x^4-8x^3+24x^2-32x+16=0.$ 
 A: As $x\ne0,$ dividing either sides by $x^2$ 
$$x^2+\left(\frac4x\right)^2-8\left(x+\frac4x\right)+24=0$$
Now as $\displaystyle x^2+\left(\frac4x\right)^2=\left(x+\frac4x\right)^2-2\cdot x\cdot\frac4x$
Setting $x+\dfrac4x=y,$ we get $\displaystyle y^2-8-8y+24=0\implies(y-4)^2=0\iff y=4$
So, we have $\displaystyle x+\frac4x=4\iff(x-2)^2=0$
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
\color{#c00000}{\Large 0}&=x^{4} - 8x^{3} + 24x^{2} - 32x + 16=
16\bracks{\pars{x \over 2}^{4} - 4\pars{x \over 2}^{3} + 6\pars{x \over 2}^{2}
- 4\,{x \over 2} + 1}
\\[3mm]&=16\left\lbrack{4 \choose 0}\pars{x \over 2}^{4}\pars{-1}^{0}
+{4 \choose 1}\pars{x \over 2}^{3}\pars{-1}^{1}
+{4 \choose 2}\pars{x \over 2}^{2}\pars{-1}^{2}
+{4 \choose 3}\,\pars{x \over 2}^{1}\pars{-1}^{3}\right.
\\[3mm]&\left.\phantom{16\bracks{}}\mbox{}
+ {4 \choose 4}\pars{x \over 2}^{0}\pars{-1}^{4}\right\rbrack
=\color{#c00000}{16\bracks{{x \over 2} + \pars{-1}}^{4}}
\quad\imp\quad\color{#00f}{\Large x = 2}
\end{align}
A: You could factorise it, in the manner of $(x-2)^4=0$.  I saw those factors immediately.
One process is to note that $16$ has divisors, and one can try various combinations of this such that the sum gives eight.
Possibilities include $2, 2, 2, 2$ and $4, 4, 1, -1$.  However, one can not produce the second set to give +16, so trying $(x-2)(x-2)(x-2)(x-2)$ is more likely than $(x-4)(x-4)(x-1)(x+1)$.
A: To start, this polynomial is called a biquadratic equation.
First, factor out the polynomial.
$$x^4 - 2x^3 - 6x^3 + 12x^2 + 12x^2 - 24x - 8x + 16 = 0$$
$$x^3(x - 2) - 6x^2(x - 2) + 12x(x - 2) - 8(x - 2) = 0$$ 
$$(x^3 - 6x^2 + 12x - 8)(x - 2) = 0$$
Then after, continue the factorization of the polynomial.
$$(a + b)^2 = a^2 + 2ab + b^2$$ 
$$(a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3$$ 
$$(a + b)^4 = a^4 + 4a^3b + 6a^2 b^2 + 4ab^3 + b^4$$ 
$$x^4 + 4 \times x^2 \times (-2) + 6 \times x^2 \times (-2)^2 + 4 \times x \times (-2)^3 + (-2)^4 = 0$$ 
So $\mathbf{(x - 2)^4 = 0}$. 
