Real values of $x$ satisfying the equation $x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0$

Question

Real values of $x$ satisfying the equation $$x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0$$

We can write it as $$512x^9+576x^6+216x^3-512x+219=0$$

I did not understand how can i factorise it.

Help me

Answer 1

$\bf{I\; have \;Solved\; Like \;This \; Way}$

$$x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0\Rightarrow 512x^9+(9\cdot 64)x^6+(27\cdot 8)x^3-512x+219=0$$

So $$\underbrace{(8x^3)^3+3(8x^3)^2\cdot 3+3(3^2)\cdot 8x^3+3^3}-512x+219-3^3=0$$

So $$(8x^3+3)^3=512x-219\Rightarrow (8x^3+3)^3=512\left(x-\frac{192}{512}\right)=8^3\left(x-\frac{3}{8}\right)^{\frac{1}{3}}.$$

So $$8^3\left(x^3+\frac{3}{8}\right) = 8\left(x-\frac{3}{8}\right)^{\frac{1}{3}}\Rightarrow x^3+\frac{3}{8} = \left(x-\frac{3}{8}\right)^{\frac{1}{3}}$$

Now Let $\displaystyle f(x)=x^3+\frac{3}{8},$ Where $f:\mathbb{R}\rightarrow \mathbb{R}\;,$ Then $\displaystyle f^{-1}(x) = \left(x-\frac{3}{8}\right)^{\frac{1}{3}}\;,$ Where $f:\mathbb{R}\rightarrow \mathbb{R}\;$

So We have To solve $$f(x) = f^{-1}(x)$$

Now We now that $f(x)$ and $f^{-1}(x)$ is Symmetrical about $y=x$ line.

So $$f(x) = f^{-1}(x) =x$$

So $$x^3+\frac{3}{8}=x\Rightarrow 8x^3-8x+3=0$$

So $$(2x-1)\left[4x^2+2x-3\right]=0\Rightarrow x=\frac{1}{2}\;\;,x=\frac{-1\pm \sqrt{13}}{4}$$

Answer 2

If this problem can be solved without computation it is reducible; we assume this.

$f(x)=512x^9+576x^+216x^3-512x+219=0$ has two change-sign and $f(-x)$ has three ones so $f(x)$ has at least $9-5=4$ non real roots. We try to find a quadratic factor using the fact that $219=3\cdot73$ and $512=2^9$; this factor could correspond to real or non-real roots.

Trying with $4x^2+ax\pm 3$ we find at once that $a=2$ and the sign minus fits; furthermore $4x^2+2x-3=0$ has two real roots because $\Delta=1+12>0$.

The quotient gives $$128x^7-64x^6+128x^5+32x^4+80x^3-16x^2+122x-73=0$$ and this equation has necessarily a real root because $7$ is odd; assuming this 7-degree polynomial is reducible and noticing that: $$\begin{cases}(2x)^7-(2x)^6+4(2x)5+2(2x)^4+10(2x)^3-4(2x)^2+61(2x)-73=0\\1-1+4+2+10-4+61-73=0\end{cases}$$ it follows at once that $2x=1$ gives a third real root.

Dividing again, now by $(2x)-1$ one gets $$g(x)=(2x)^6+4(2x)^4+6 (2x)^3+16(2x)^2+12(2x)+73=0$$ It is obvious that $g(x)>0$ for $x>0$ and it is easy to show that for $X<0$ $$X^6+4X^4+16X^2+73>-(6X^3+12X)$$ hence $g(x)$ is always positive so $g(x)=0$ has six non-real roots.

Thus $f(x)=0$ has only three real roots, given by $$\color{red}{4x^2+2x-3=0}\space \text {and}\space\space \color{red}{ 2x-1=0}$$

Answer 3

Using factor theorem:

If $x-a$ divides $f(x)$, then $f(a)=0$.

As $(\frac{1}{2})^{9}=\frac{1}{512}$, it's worth to try $x=\pm \frac{1}{2}$.

Substituting $x=\frac{1}{2}$, LHS vanishes, so $\frac{1}{2}$ is one possible value.

Further factorize gets, $$\left( x-\frac{1}{2} \right) \left( x^{2}-\frac{x}{2}-\frac{3}{4} \right) \left( x^{6}+x^{4}+\frac{3x^{3}}{4}+x^{2}+\frac{3x}{8}+\frac{7}{3} \right)=0$$ and the second factor can be solved by quadratic formula.