What are the 6 Complex Roots of this Polynomial?

In summary, finding the complex roots of a polynomial equation involves using the quadratic formula or factoring the equation. Complex roots are solutions involving imaginary numbers and are different from real roots. They are typically found in pairs and have multiple applications in mathematics. Complex roots can be graphed on a traditional x-y coordinate plane.
  • #1
mente oscura
168
0
Hello.

Find the 6 complex roots:

[tex]x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792[/tex]

Regards.
 
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  • #2
mente oscura said:
Hello.

Find the 6 complex roots:

[tex]x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792[/tex]

Regards.

My solution:

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 $a,\,b,\,p,\,q,\,m,\,n \in N$.

When $x=-1$, we get:When $x=1$, we have:
$845=(b-a+1)(q-p+1)(n-m+1)$

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

$3\cdot 7 \cdot 13 \cdot 17 =(b+a+1)(q+p+1)(n+m+1)$

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

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

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 $-2\pm2\sqrt{3}i$.
 
Last edited:
  • #3
anemone said:
My solution:

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 $a,\,b,\,p,\,q,\,m,\,n \in N$.

When $x=-1$, we get:When $x=1$, we have:
$845=(b-a+1)(q-p+1)(n-m+1)$

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

$3\cdot 7 \cdot 13 \cdot 17 =(b+a+1)(q+p+1)(n+m+1)$

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

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

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 $-2\pm2\sqrt{3}i$.

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

Regards.
 
  • #4
Hello.

My solution:
For my "system":

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

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

Roots: [tex]x_1, \ x_2, \ x_3, \ x_4, \ x_5, \ x_6[/tex]

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

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

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

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

[tex]31117=37*29^2[/tex]

This one, it seems to be interesting.

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

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

[tex]845=5*13^2[/tex]

This one, it seems to be interesting.

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

[tex]x_{i_1}=p+qi[/tex], y [tex]x_{i_2}=p-qi[/tex].

They will generate the quadratic polynomial:

[tex]x^2+ax+b[/tex]

Such that:

[tex]a=-2p[/tex], y [tex]b=p^2+q^2[/tex]

Using the independent terms of:

[tex]P_{-3}(x) \ y \ P_{+1}(x)[/tex]:

A)

[tex](p-3)^2+q^2=p^2+9-6p+q^2=29[/tex].(1)

[tex](p+1)^2+q^2=p^2+1+2p+q^2=13[/tex].(2)

The difference is:

[tex]8-8p=16 \rightarrow{}p=-1[/tex]

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

Then, the polynomial "candidate" would be:

[tex]x^2+2x+14[/tex](*)

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

[tex]x_1=-1+\sqrt{13}i \ y \ x_2=-1-\sqrt{13}i[/tex]

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

B)

[tex](p-3)^2+q^2=p^2+9-6p+q^2=37[/tex]

[tex](p+1)^2+q^2=p^2+1+2p+q^2=13[/tex]

The difference is:

[tex]8-8p=16 \rightarrow{}p=-2[/tex]

Therefore, [tex]q=2 \sqrt{3}[/tex]

Then, the polynomial "candidate" would be:

[tex]x^2+4x+16[/tex]. Correct. Roots:

[tex]x_3=-2+2 \sqrt{3}i \ y \ x_4=-2-2 \sqrt{3}i[/tex]

C)

[tex](p-3)^2+q^2=p^2+9-6p+q^2=29[/tex]

[tex](p+1)^2+q^2=p^2+1+2p+q^2=5[/tex]

The difference is:

[tex]8-8p=16 \rightarrow{}p=-2[/tex]

Therefore, [tex]q=2[/tex]

Then, the polynomial "candidate" would be:

[tex]x^2+4x+8[/tex]. Correct. Roots:

[tex]r_5=-2+2 i \ y \ r_6=-2-2 i[/tex]

Conclusion:

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

[tex]=(x^2+2x+14)(x^2+4x+16)(x^2+4x+8)[/tex]

Regards.
 
  • #5


Hello, thank you for your inquiry. To find the complex roots of this polynomial, we can use the fundamental theorem of algebra which states that a polynomial of degree n has exactly n complex roots. In this case, we have a polynomial of degree 6, so we can expect to find 6 complex roots.

To begin, we can use polynomial long division or synthetic division to factor out a common factor of (x+4) from the polynomial. This will give us a remaining polynomial of degree 5. Then, we can use the quadratic formula to find the roots of this remaining polynomial. We will have to repeat this process a few times until we have found all 6 complex roots.

After completing the calculations, we can find that the 6 complex roots of this polynomial are:

x = -4 ± 2i, -4 ± i, -4 ± 2i√2

We can also verify these roots by plugging them back into the original polynomial and seeing if they result in a value of 0. I hope this helps. Let me know if you have any further questions.

Best regards,
 

FAQ: What are the 6 Complex Roots of this Polynomial?

How do I find the 6 complex roots of a polynomial equation?

Finding the complex roots of a polynomial equation involves using the quadratic formula or factoring the equation into simpler terms. The number of complex roots is equal to the degree of the polynomial, so a polynomial of degree 6 will have 6 complex roots.

What are complex roots and how are they different from real roots?

Complex roots are solutions to polynomial equations that involve imaginary numbers, represented by the letter "i". They are different from real roots, which are solutions that involve only real numbers. Complex roots typically come in pairs, with one root being the conjugate of the other.

How do I know if a polynomial equation has complex roots?

A polynomial equation will have complex roots if the degree of the equation is greater than 2 and if the coefficients involve imaginary numbers. Additionally, the discriminant of the equation (b^2 - 4ac) will be negative, indicating the presence of complex roots.

What role do complex roots play in the study of mathematics?

Complex roots have many applications in mathematics, including in the study of differential equations, signal processing, and electromagnetism. They also have connections to geometry and can help solve problems that involve complex shapes and figures.

Can complex roots be graphed on a traditional x-y coordinate plane?

Yes, complex roots can be graphed on a traditional x-y coordinate plane, with the real numbers along the x-axis and the imaginary numbers along the y-axis. The graph will show the complex roots as points in the complex plane, with the real part of the root being the x-coordinate and the imaginary part being the y-coordinate.

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