2

How can we solve this equation? $x^4-8x^3+24x^2-32x+16=0.$

hrkrshnn
  • 6,287
user135118
  • 29
  • 1
    $(x-2)^4=0$. $~$ – Oleg567 Mar 30 '14 at 10:35
  • Guess a root $\alpha$, then divide the equation by $x - \alpha$. Repeat. – Dan Shved Mar 30 '14 at 10:36
  • Since there are a lot of powers of two appearing in the polynomial equation, try $x=2$ and find that it is a solution. Repeat after polynomial division to find that $x=2$ still is a solution, arriving at $(x-2)^2(x^2-4x+4)=(x-2)^2(x-2)^2=(x-2)^4=0$ – AlexR Mar 30 '14 at 10:38
  • 1
    There are lots of ways to solve problems, and for most problems there are lots of methods that will work. There are some that are frequently useful, of course: one of the usual things to try for solving polynomial equations of degree higher than $2$ is the rational root theorem; have you tried it yet, and/or did you have any success? –  Mar 30 '14 at 11:02

4 Answers4

4

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 $\displaystyle x+\frac4x=4\iff(x-2)^2=0$

lab bhattacharjee
  • 274,582
4

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \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}

Felix Marin
  • 89,464
  • $$x^4−8x^3+24x^2−32x+1=x^4-\binom41\cdot x^3\cdot2+\binom41\cdot x^2\cdot2^2-\binom43\cdot x\cdot2^2+2^4=(x-2)^4$$ – lab bhattacharjee Mar 30 '14 at 12:36
  • @labbhattacharjee Yes. It's true. It's a custom, in numerical calculations, to set the $x^{0}$-term equal to one. In that way I arrived to the result. I was aware of the binomial after I did that. Thanks. – Felix Marin Mar 30 '14 at 23:13
1

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 $(x-4)(x-4)(x-1)(x+1)$.

wendy.krieger
  • 6,856
0

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 $\mathbf{(x - 2)^4 = 0}$.