Question

find all the roots if x^4-6x^3+7x^2+6x+1=0

Answer 1

Mind starting off by sharing your thoughts about approaching this problem? Which methods have you used in the past to find the roots of given polynomial equations?

Answer 2

Have you seen and used either synthetic division or long division in the past?

Answer 3

@tom123cruise: your thoughts?

Answer 4

yeah i tried with the synthetic division... with nos 1 to 4 both -&+ but the remainder is not 0.

Answer 5

I have to agree that neither -1 nor 1 appears to be a root, as the remainder after synthetic division is not zero in either case. Have you learned any other methods of solving polynomial equations?

Answer 6

Have you ever heard of Newton's Method?

Answer 7

As a last resort: Please double check to ensure that you've copied this problem down correctly. Using Newton's Method, I'm getting closer and closer to a root, but that root is a negative fraction.

Answer 8

sry i dont know newtons method.

Answer 9

I'd suggest you 1) double check to ensure that you have copied down this problem correctly, and 2) if you don't know Newton's Method, and synthetic division does not succeed, move on to the next problem you have to solve.

Answer 10

yeah the problem is correctly copied. anyway thanks.

Answer 11

By the rational root test, the only possible rational solutions are x = 1 or -1, but none of those work.

Answer 12

I've got it! I think. Using substitution.

Answer 13

can u please post it here

Answer 14

Yes.

Answer 15

I noticed that the coefficients are almost symmetric if written sequentially: 1, -6, 7, 6, 1. That was the clue that substitution could be used.

\(x^4-6x^3+7x^2+6x+1=0\)
Let \(x = iy\).
\(y^4+6iy^3-7y^2+6iy+1=0\)
\(y^2+6iy-7+\frac{6i}y+\frac1{y^2}=0\)
Let \(z = y + \frac1y\) and find \(a, b, c\) such that \(az^2+bz+c = y^2+6iy-7+\frac{6i}y+\frac1{y^2}\).
\(a = 1,\ b = 6i,\ c = -3\)
\(z^2 + 6iz -3 = 0\)
Solve for \(z\), substitute back into the relation between \(z\) and \(y\), solve for \(y\), substitute back into the relation between \(y\) and \(x\), solve for \(x\).

Really tough. The algebra will take long. Was my explanation clear enough?

Answer 16

I have to go to bed soon.

Answer 17

it was gud enough. thanks. im not being rude do u know completing the square method?

Answer 18

Yes, why?

Answer 19

can that method can be applied here in this eqn?

Answer 20

Do you mean the one involving \(z\)?

Answer 21

wat time is at ur location?? here its pretty late so gotta go. bye and thanks

Answer 22

Midnight. Goodbye, you're welcome :)

Answer 23

From Mathematica 9:\[x^4-6 x^3+7 x^2+6 x+1=\left(x^2-3 x-1\right)^2 \]

Answer 24

5 minutes ago! I just found out too.

Answer 25

I got the clue that the roots were double roots by looking at the graph and seeing that it was tangential to the x axis at both roots. This meant that we should (maybe) be able to write the polynomial as
\(\begin{split}
p(x)
&= x^4-6x^3+7x^2+6x+1 \\
&= y^4+6y^3-7y^2+6y+1 \\
&= (y^2+ay+1)(y^2+by+1))
\end{split}\)
By expanding and comparing coefficients we find that \(a = b = 3i\). Substituting \(x=yi\) we get robtobey's expression.

Answer 26

Maybe that's a bit too much guesswork though.

Answer 27

Sorry if you didn't understand my last post. I don't understand my own words. I'll try again.

Observe that the polynomial \(p(x) = x^4-6x^3+7x^2+6x+1\) seems to have a local minimum at each of its roots by graphing it. Assume that that's true. Then the multiplicity of each root is even. Because the sum of the multiplicities is the degree of the polynomial (four), the multiplicity of each root is two and we should be able to write \(p(x) = k(x-x_1)^2(x-x_2)^2 = k((x-x_1)(x-x_2))^2 = k(x^2+ax+b)^2\) for some constants \(k, a, b\). Expand \(p(x) = kx^4+2kax^3 + k(a^2+2b)x^2+2kabx+kb^2\). Compare the coefficients of the original expression with the new expression to deduce \(k=1\), \(a=-3\), \(b=-1\).

Answer 28

You can write the given polynomial as a "polynomial of polynomials":
\[\begin{align*}x^4-6x^3+7x^2+6x-1&=(x^4-6x^3+9x^2)+(-2x^2+6x)-1\\
&=(x^2-3x)^2-2(x^2-3x)-1\\
&=((x^2-3x)^2-1)^2
\end{align*}\]
like @robtobey said.

Answer 29

@SithsAndGiggles how did you know to go from the first expression to the second?

Answer 30

The first three terms kind of looked like a squared binomial, so I rewrote \(7x^2\) as \((9-2)x^2\). The extra \(-2x^2\) happens to let you factor again, and quite nicely.

Answer 31

OK

Answer 32

(but the first three -1's should be +1 and the second ^2 on the last line should not be at all)

Answer 33

Basically it's a matter of completing the square.

And yes, sorry, minor typo :)

Answer 34

So the two highest degree terms determined the "inner polynomial" alone. It doesn't seem very general.

Answer 35

But efficient, yes.

Answer 36

Yeah I don't expect this method to work for all \(n\)th order polynomials :/ Life would a lot easier.

Answer 37

Thanks for the discussion!