Roots and linear combinations

**I-Think** · January 28th 2013, 11:15 AM

Let $f(x)=x^3+x-1$ and $\theta$ be its real root

Express $\frac{\theta^7+4}{\theta+1}$ as a linear combination of $1,\theta$ and $\theta^2$

**ILikeSerena** · January 28th 2013, 12:10 PM

Originally Posted by I-Think

Let $f(x)=x^3+x-1$ and $\theta$ be its real root

Express $\frac{\theta^7+4}{\theta+1}$ as a linear combination of $1,\theta$ and $\theta^2$

Hi I-Think!

Since $\theta$ is a root, it follows that $\theta^3=1-\theta$ .
Substitute and divide?

**Deveno** · January 28th 2013, 12:30 PM

note that:

$\theta^3 = 1 - \theta$ .

hence:

$(\theta + 1)^3 = \theta^3 + 3\theta^2 + 3\theta + 1 = 1 - \theta + 3\theta^2 + 3\theta + 2 = 3\theta^2 + 2\theta + 2$

$= 3\theta^2 + 6\theta + 3 - 4\theta - 1 = 3(\theta + 1)^2 - 4(\theta + 1) + 3$

that is:

$1 = \frac{1}{3}(4(\theta + 1) - 3(\theta + 1)^2 + (\theta + 1)^3)$

so that evidently:

$\frac{1}{\theta + 1} = \frac{1}{3}(4 - 3(\theta + 1) + (\theta + 1)^2) = \frac{1}{3}(2 - \theta + \theta^2)$

this is probably the hard part.

now:

$\theta^7 + 4 = (\theta^3)^2\theta + 4 = (1 - \theta)^2\theta + 4 = \theta - 2\theta^2 + \theta^3 + 4$

$=\theta - 2\theta^2 + 1 - \theta + 4 = 5 - 2\theta^2$

thus:

$\frac{\theta^7 +4}{\theta + 1} = \frac{1}{3}(5 - 2\theta^2)(2 - \theta + \theta^2)$

$= \frac{1}{3}(10 - 5\theta + \theta^2 + 2\theta^3 - 2\theta^4) = \frac{1}{3}(10 - 5\theta + \theta^2 + 2 - 2\theta - 2(1 - \theta)\theta)$

$= \frac{1}{3}(12 - 7\theta + \theta^2 - 2\theta + 2\theta^2) = \frac{1}{3}(12 - 9\theta + 3\theta^2) = 4 - 3\theta + \theta^2$

...if...my arithmetic is ok....

Originally Posted by ILikeSerena

Hi I-Think!

Since $\theta$ is a root, it follows that $\theta^3=1-\theta$ .
Substitute and divide?

the trouble with this is you still have some remainder $\frac{r}{\theta + 1}$ so finding the inverse of $\frac{1}{\theta + 1}$ appears to be unavoidable.

**ILikeSerena** · January 28th 2013, 12:36 PM

Yep. Your arithmetic is ok.

Alternatively:

$\theta^7 +4 = {(\theta^3)^2\theta +4} = {(1 - \theta)^2\theta +4} = {\theta^3-2\theta^2+\theta +4}$

Now do a long division by $\theta + 1$ to find (remainder turns out to be zero):

${\theta^7 +4 \over \theta + 1} = {\theta^3-2\theta^2+\theta +4 \over \theta + 1} = \theta^2-3\theta+4. \quad \Box$

**Deveno** · January 28th 2013, 12:41 PM

i'm curious, how did you know to stop at a cubic?

**ILikeSerena** · January 28th 2013, 12:52 PM

Originally Posted by Deveno

i'm curious, how did you know to stop at a cubic?

The power of the numerator had to go down.
I could easily get it down to a cubic.
Division by a first order polynomial would yield a second order as requested... I just tried - it seemed to be the type of problem that would have a neat solution.

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