If P(x) = 25x^6 - 25x^5 + 74x^4 - 24x^3 + 47x^2 + x - 2 is the complex number -i , a root of P(x)?

it would really help me if you can post an explantion.

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- Oct 25th 2012, 11:53 PMmisqa88Polynomial equation
If P(x) = 25x^6 - 25x^5 + 74x^4 - 24x^3 + 47x^2 + x - 2 is the complex number -i , a root of P(x)?

it would really help me if you can post an explantion. - Oct 25th 2012, 11:58 PMMarkFLRe: Polynomial equation
Check to see if $\displaystyle P(i)=0$. To make it easier, use the fact that $\displaystyle i^n=i^{n-4}$.

- Oct 26th 2012, 12:01 AMmisqa88Re: Polynomial equation
- Oct 26th 2012, 12:05 AMMarkFLRe: Polynomial equation
$\displaystyle P(i)=25i^6-25i^5+74i^4-24i^3+47i^2+i-2$

$\displaystyle i^6=i^2=-1$

$\displaystyle i^5=i^1=i$

$\displaystyle i^4=i^0=1$

$\displaystyle i^3=i\cdot i^2=-i$

$\displaystyle i^2=-1$

Now, using this, do you find $\displaystyle P(i)=0$? - Oct 26th 2012, 12:09 AMmisqa88Re: Polynomial equation
yes P(i) = 0

- Oct 26th 2012, 12:13 AMMarkFLRe: Polynomial equation
Yes, good work! :)

- Oct 26th 2012, 12:16 AMmisqa88Re: Polynomial equation
so is that it? we proved that the complex number -i is a root of P(x)?

Amazing :) Thank you so much. - Oct 26th 2012, 12:19 AMMarkFLRe: Polynomial equation
Yes, that's all you have to do to see if a number is a root of a function. By the conjugate root theorem, we also know then that -i has to be a root as well, and so $\displaystyle x^2+1$ must be a factor of $\displaystyle P(x)$.

- Oct 26th 2012, 12:26 AMmisqa88Re: Polynomial equation
Thank you so much for helping MarkFL2! You are the most helpful. I also have 2 other examples I don't know how to do but they are about surjectivity/injectivity of a function, would you be able to help me?

- Oct 26th 2012, 12:39 AMMarkFLRe: Polynomial equation
I don't know much about that, I never studied it, but there are many helpful people here and I'm sure some of them have studied this.

- Oct 26th 2012, 06:42 AMSalahuddin559Re: Polynomial equation
In general, if you want to find "rational" roots of an polynomial, try to do the following:

For a continuous function, in the neighborhood of a root, both negative and positive values exist. Since polynomials satisfy these conditions, you can look for sign-changes and then do bi-section to narrow on roots. Typical problems have "nice" rational numbers, so, this is more like an exam tip. For example,

(2x - 3)^3 = (4x^2 -12x + 9) * (2x - 3) = 8x^3 - 24x^2 + 18x - 12x^2 + 36x - 27 = 8x^3 - 36x^2 + 54x - 27.

Consider its value at 0, which is -27, negative. Now consider it at 2, which is 1 that is positive. So, the zero must be somewhere in the middle. Now try 1, which is (0 + 2)/2, and so on... Remember, this can work if your roots are "nice" rational numbers, like 3/2, 5/4 etc... And, though it does not fail, you do not have infinite time to pursue this in exams, so, use it only to some extent, to get answer, otherwise, just switch to something else.

In order to verify for higher x, pair successive terms, you end up with small and manageable values. Like, 8x^3 - 36x^2 + 54x - 27 = 4x^2(2x - 9) + 27(3x - 1).

Salahuddin

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