MHB Solve $(1+x^2)(1+x^3)(1+x^5)=8x^5$: Real Solutions

[anemone]

Find all the real solutions for $(1+x^2)(1+x^3)(1+x^5)=8x^5$.

 

[Albert1]

Find all the real solutions for $f(x)=(1+x^2)(1+x^3)(1+x^5)=8x^5---(1)$.

my solution:

Spoiler

rearrangement of (1) we have :
$f(x)=(1+x^2)(1+x^3)(1+x^5)-8x^5=(x-1)^2\times h(x)=0$
here $h(x)=x^8+2x^7+4x^6+7x^5+10x^4+7x^3+4x^2+2x+1$
it is easy to check there is no real solution for $h(x)=0$
for all $x\in R , h(x)>0$
$\therefore$ the real solutions for (1):$(1+x^2)(1+x^3)(1+x^5)=8x^5$ is $x=1\,\,$ (repeated root)

 

[anemone]

my solution:

Spoiler

rearrangement of (1) we have :
$f(x)=(1+x^2)(1+x^3)(1+x^5)-8x^5=(x-1)^2\times h(x)=0$
here $h(x)=x^8+2x^7+4x^6+7x^5+10x^4+7x^3+4x^2+2x+1$
it is easy to check there is no real solution for $h(x)=0$
for all $x\in R , h(x)>0$
$\therefore$ the real solutions for (1):$(1+x^2)(1+x^3)(1+x^5)=8x^5$ is $x=1\,\,$ (repeated root)

Hi Albert,

Spoiler

Thanks for participating...

The answer is correct, but could you please demonstrate why there is no other real solution(s) for $h(x)=0$, to complete your solution?

 

[lfdahl]

My solution:

Spoiler

Any root in $(1+x^2)(1+x^3)(1+x^5) = 8x^5$ must be positive, because only the RHS < 0 for $x<0$.

$x = 1$ is obviously root.

By polynomial division, I get:

$(x-1)(x^9+x^8+2x^7+3x^6+3x^5-3x^4-3x^3-2x^2-x-1) = 0$

In the reduced polynomial, $x = 1$ is root again. By polynomial division, I get:

$(x-1)^2(x^8+2x^7+4x^6+7x^5+10x^4+7x^3+4x^2+2x+1) = 0$

The second reduced polynomial contains only positive terms for $x > 0$, thus there are no more real roots.

The only real root(s) are: $x = 1$ with multiplicity $2$.

 

[anemone]

My solution:

Spoiler

Any root in $(1+x^2)(1+x^3)(1+x^5) = 8x^5$ must be positive, because only the RHS < 0 for $x<0$.

$x = 1$ is obviously root.

By polynomial division, I get:

$(x-1)(x^9+x^8+2x^7+3x^6+3x^5-3x^4-3x^3-2x^2-x-1) = 0$

In the reduced polynomial, $x = 1$ is root again. By polynomial division, I get:

$(x-1)^2(x^8+2x^7+4x^6+7x^5+10x^4+7x^3+4x^2+2x+1) = 0$

The second reduced polynomial contains only positive terms for $x > 0$, thus there are no more real roots.

The only real root(s) are: $x = 1$ with multiplicity $2$.

Great work, lfdahl(Cool), and thanks for participating!

 

[Albert1]

My solution:

Spoiler

Any root in $(1+x^2)(1+x^3)(1+x^5) = 8x^5$ must be positive, because only the RHS < 0 for $x<0$.

$x = 1$ is obviously root.

By polynomial division, I get:

$(x-1)(x^9+x^8+2x^7+3x^6+3x^5-3x^4-3x^3-2x^2-x-1) = 0$

In the reduced polynomial, $x = 1$ is root again. By polynomial division, I get:

$(x-1)^2(x^8+2x^7+4x^6+7x^5+10x^4+7x^3+4x^2+2x+1) = 0$

The second reduced polynomial contains only positive terms for $x > 0$, thus there are no more real roots.

The only real root(s) are: $x = 1$ with multiplicity $2$.

Spoiler

$(x-1)^2(x^8+2x^7+4x^6+7x^5+10x^4+7x^3+4x^2+2x+1) = 0$
The second reduced polynomial contains only positive terms for $x > 0$
you should also discuss when $x<0$ , the second polynomial will also $\neq 0$
for example if $x<0$, then $x^8,4x^6,10x^4,4x^2,1>0 $, but $2x^7,7x^5,7x^3,2x<0$
all of them cannot be canceled out, do you agree ?

 

[lfdahl]

Spoiler

$(x-1)^2(x^8+2x^7+4x^6+7x^5+10x^4+7x^3+4x^2+2x+1) = 0$
The second reduced polynomial contains only positive terms for $x > 0$
you should also discuss when $x<0$ , the second polynomial will also $\neq 0$
for example if $x<0$, then $x^8,4x^6,10x^4,4x^2,1>0 $, but $2x^7,7x^5,7x^3,2x<0$
all of them cannot be canceled out, do you agree ?

Hi, Albert

Spoiler

As far as I can see, the case $x < 0$ need not be taken into account at all ... or?

 

[Albert1]

Hi, Albert

Spoiler

As far as I can see, the case $x < 0$ need not be taken into account at all ... or?

Spoiler

Why the case $x<0$ needs not be taken into account ?
for example $x^8+2x^7=0, x=-2$ is one of its roots

 

[lfdahl]

Spoiler

Why the case $x<0$ needs not be taken into account ?
for example $x^8+2x^7=0, x=-2$ is one of its roots

Hi, Albert

Spoiler

If you take a look at the starting equation: $(1+x^2)(1+x^3)(1+x^5) = 8x^5$

- you´ll see, that the $LHS \geq 0, \: \: \: x \in \mathbb{R}$.

On the other hand, the $RHS < 0, \:\:\: x <0$ and $\ge 0$ for $x \ge 0$

Therefore, the case, when $x < 0$ is no option to be further analyzed.

 

[Albert1]

Hi, Albert

Spoiler

If you take a look at the starting equation: $(1+x^2)(1+x^3)(1+x^5) = 8x^5$

- you´ll see, that the $LHS \geq 0, \: \: \: x \in \mathbb{R}$.

On the other hand, the $RHS < 0, \:\:\: x <0$ and $\ge 0$ for $x \ge 0$

Therefore, the case, when $x < 0$ is no option to be further analyzed.

yes,you are right
very smart and sharp observation

 

[anemone]

yes,you are right
very smart and sharp observation

Actually lfdahl has clearly made that point in his first solution post...(Nod)