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.
$\displaystyle P(i)=25i^6-25i^5+74i^4-24i^3+47i^2+i-2$
$\displaystyle i^6=i^2=-1$
$\displaystyle i^5=i^1=i$
$\displaystyle i^4=i^0=1$
$\displaystyle i^3=i\cdot i^2=-i$
$\displaystyle i^2=-1$
Now, using this, do you find $\displaystyle P(i)=0$?
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
Maths online