Thread: roots with multiplicity > 1 polynomials

I was solving the problem listed here..

Polynomial x

I understood the logic behind the first line which says that "A root is of multiplicity 1 ==> the polynomial has a root in common with it's own derivative.

But I wasn't able to follow the next where the reply is like this.
Thus the given information tells us that x^4 + x + 6 and 4x^3 + 1 have a common root r. It follows that
0 = 4(r^4 + r + 6) - r(4r^3 + 1) = 3r + 24.

I could not follow the logic behind the line 0 = 4(r^4 + r + 6) - r(4r^3 + 1) = 3r + 24.

I have been searching for the justification behind this step, but have been failing again and again. Can someone throw some light on this?

Thanks

if r is a root of x⁴ + x + 6 and 4x³ + 1, then:

r⁴ + r + 6 = 0, and 4r³ + 1 = 0, hence:

4(r⁴ + r + 6) - r(4r³ + 1) = 4*0 - r*0 = 0.

expanding, we have:

4r⁴ + 4r + 24 - 4r⁴ - r = 0, and the r⁴ terms cancel out leaving:

3r + 24 = 0

@MaxJasper:

the original question deals with a field of characteristic p, so a real-valued plot is not especially helpful, here (since char(R) = 0).

the graph helped with a visual perspective of the problem.

Thanks, max Jasper

Thanks for your help DEVENO
i presume the logic to be this.
since r is a root, r^4 + r + 6 = 0, and 4r^3 + 1 = 0 so, multiplying any constant/variable other than 4 and r will also return a zero. but choosing 'r' and '4' helps us because, we are able to get rid of r^4 helping in a simpler form of a polynomial.

i was able to follow after that...
That was of great help.

Thanks again,