MHB What are the 6 Complex Roots of this Polynomial?

[mente oscura]

Hello.

Find the 6 complex roots:

x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792

Regards.

 

[anemone]

Hello.

Find the 6 complex roots:

x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792

Regards.

My solution:

Spoiler

Let $f(x)=x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792=(x^2+ax+b)(x^2+px+q)(x^2+mx+n)$ where each of the quadratic factor is greater than zero (since we're told $f(x)$ has all 6 complex roots) and $a,\,b,\,p,\,q,\,m,\,n \in N$.

[TABLE="class: grid, width: 850"]
[TR]
[TD]When $x=-1$, we get:[/TD]
[TD]When $x=1$, we have:[/TD]
[/TR]
[TR]
[TD]$845=(b-a+1)(q-p+1)(n-m+1)$

$5\cdot 13 \cdot 13 =(b-a+1)(q-p+1)(n-m+1)$[/TD]
[TD]$4641=(b+a+1)(p+q+1)(m+n+1)$

$3\cdot 7 \cdot 13 \cdot 17 =(b+a+1)(q+p+1)(n+m+1)$[/TD]
[/TR]
[/TABLE]

If we let $b-a+1=5$, $q-p+1=13$ and $n-m+1=13$, we obtain:

$3\cdot 7 \cdot 13 \cdot 17 =(5+2a)(13+2p)(13+2m)$

Now, if we consider for one more case that is when $x=-2$, that gives

$672=(b-2a+4)(q-2p+4)(n-2m+4)$

$672=(8-a)(16-p)(16-m)$

$2^5\cdot 3 \cdot 7=(8-a)(16-p)(16-m)$

Now, if we focus solely on the conditions

[math]\color{yellow}\bbox[5px,green]{3\cdot 7 \cdot 13 \cdot 17 =(5+2a)(13+2p)(13+2m)}[/math] and [math]\color{green}\bbox[5px,yellow]{2^5\cdot 3 \cdot 7=(8-a)(16-p)(16-m)}[/math],

it's easy to check that $a=4,\,p=2,\,m=4$ satisfy the condition and that yields $b=8,\,p=14,\,m=16$ and hence

$x^2+ax+b=x^2+4x+8=0$ gives the complex roots of $-2 \pm 2i$.

$x^2+px+q=x^2+2x+14=0$ gives the complex roots of $-1 \pm \sqrt{13}i$.

$x^2+mx+n=x^2+4x+16=0$ gives the complex roots of $-2\pm2\sqrt{3}i$.

 

[mente oscura]

My solution:

Spoiler

Let $f(x)=x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792=(x^2+ax+b)(x^2+px+q)(x^2+mx+n)$ where each of the quadratic factor is greater than zero (since we're told $f(x)$ has all 6 complex roots) and $a,\,b,\,p,\,q,\,m,\,n \in N$.

[TABLE="class: grid, width: 850"]
[TR]
[TD]When $x=-1$, we get:[/TD]
[TD]When $x=1$, we have:[/TD]
[/TR]
[TR]
[TD]$845=(b-a+1)(q-p+1)(n-m+1)$

$5\cdot 13 \cdot 13 =(b-a+1)(q-p+1)(n-m+1)$[/TD]
[TD]$4641=(b+a+1)(p+q+1)(m+n+1)$

$3\cdot 7 \cdot 13 \cdot 17 =(b+a+1)(q+p+1)(n+m+1)$[/TD]
[/TR]
[/TABLE]

If we let $b-a+1=5$, $q-p+1=13$ and $n-m+1=13$, we obtain:

$3\cdot 7 \cdot 13 \cdot 17 =(5+2a)(13+2p)(13+2m)$

Now, if we consider for one more case that is when $x=-2$, that gives

$672=(b-2a+4)(q-2p+4)(n-2m+4)$

$672=(8-a)(16-p)(16-m)$

$2^5\cdot 3 \cdot 7=(8-a)(16-p)(16-m)$

Now, if we focus solely on the conditions

[math]\color{yellow}\bbox[5px,green]{3\cdot 7 \cdot 13 \cdot 17 =(5+2a)(13+2p)(13+2m)}[/math] and [math]\color{green}\bbox[5px,yellow]{2^5\cdot 3 \cdot 7=(8-a)(16-p)(16-m)}[/math],

it's easy to check that $a=4,\,p=2,\,m=4$ satisfy the condition and that yields $b=8,\,p=14,\,m=16$ and hence

$x^2+ax+b=x^2+4x+8=0$ gives the complex roots of $-2 \pm 2i$.

$x^2+px+q=x^2+2x+14=0$ gives the complex roots of $-1 \pm \sqrt{13}i$.

$x^2+mx+n=x^2+4x+16=0$ gives the complex roots of $-2\pm2\sqrt{3}i$.

Spoiler

Well done!, anemone, the solutions are correct. I wait a little to my solution, if anyone else dares.

Regards.

 

[mente oscura]

Hello.

My solution:

Spoiler

For my "system":

http://mathhelpboards.com/number-theory-27/polynomials-roots-10020.html

P(x)=x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792

Roots: x_1, \ x_2, \ x_3, \ x_4, \ x_5, \ x_6

1792=2^8*7
I will look for a polynomial, which has the same roots, less a unit.

P_{-1}(x)=x^6+16x^5+135x^4+688x^3+2279x^2+4560x+4641

4641=3*7*13*17P_{-2}(x)=x^6+22x^5+230x^4+1408x^3+5328x^2+11808x+12320

12320=2^5*7*11P_{-3}(x)=x^6+28x^5+355x^4+2568x^3+11167x^2+27724x+31117

31117=37*29^2

This one, it seems to be interesting.

I will look for a polynomial, which has the same roots, more a unit.

P_{+1}(x)=x^6+4x^5+35x^4+88x^3+351x^2+468x+845

845=5*13^2

This one, it seems to be interesting.

On the other hand, the complex roots, they are of the form:

x_{i_1}=p+qi, y x_{i_2}=p-qi.

They will generate the quadratic polynomial:

x^2+ax+b

Such that:

a=-2p, y b=p^2+q^2

Using the independent terms of:

P_{-3}(x) \ y \ P_{+1}(x):

A)

(p-3)^2+q^2=p^2+9-6p+q^2=29.(1)

(p+1)^2+q^2=p^2+1+2p+q^2=13.(2)

The difference is:

8-8p=16 \rightarrow{}p=-1

Substituting in (1) ó (2): q= \pm \sqrt{13}

Then, the polynomial "candidate" would be:

x^2+2x+14(*)

We divide, and observe that it turns out to be "exact". We can say that it has as roots:

x_1=-1+\sqrt{13}i \ y \ x_2=-1-\sqrt{13}i

We try to divide, for the second time for (*), and, We see that it does not turn out to be exact.

B)

(p-3)^2+q^2=p^2+9-6p+q^2=37

(p+1)^2+q^2=p^2+1+2p+q^2=13

The difference is:

8-8p=16 \rightarrow{}p=-2

Therefore, q=2 \sqrt{3}

Then, the polynomial "candidate" would be:

x^2+4x+16. Correct. Roots:

x_3=-2+2 \sqrt{3}i \ y \ x_4=-2-2 \sqrt{3}i

C)

(p-3)^2+q^2=p^2+9-6p+q^2=29

(p+1)^2+q^2=p^2+1+2p+q^2=5

The difference is:

8-8p=16 \rightarrow{}p=-2

Therefore, q=2

Then, the polynomial "candidate" would be:

x^2+4x+8. Correct. Roots:

r_5=-2+2 i \ y \ r_6=-2-2 i

Conclusion:

P(x)=x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792=

=(x^2+2x+14)(x^2+4x+16)(x^2+4x+8)

Regards.