Polynomials N zeros?
Hi,,
I wanted to ask why polynomials can have a maximum of N zeros.
Hi,,
I wanted to ask why polynomials can have a maximum of N zeros.
The number 5. For me it came out: 8a: 25% 8b: 25% 8c: 3/8 = 37.5%
I need to find the antiderivative of integral( sin(x)* cos(x)). How do I do this through SUBSTITUTION?
And if so: Why is this currently the case, or are there tendencies to (again) fill the global metropolis shell with economic global metropolis functions, or will Berlin always be/remain/have an empty shell?
Can someone please help me with task 11? How should I approach it?
I can't do tasks 2, 3, 4, and 5 because I don't understand them. 📚 Can someone please help me? 🆘 The solutions are on the sheet, but I need the calculation method. 🧮
Good day, I'm stuck on this problem. How do you get pi/6 for b and 3 for c? Greetings
A polynomial over a body disintegrates into linear factors in its algebraic conclusion. If the basic ring is zero divider-free, it follows that there are no more than # linear factors= Deg polynomial zeros. In general, a polynomial can have more zerotels, for example, polynomials via Mat(n) or C(R), the quadratic dies via C/continuous function via R.
If a zero x_0 is present, the linear factor (x−x_0) can be split off with a polynomial division, whereby the degree of the polynomial is reduced by one. This can only be done as often as the degree of the original polynomial was.
Suppose a polynomial has n reelle zeros a1 . ., wherein individual zeros ak can be identical (multiple zero). The polynomial can then be represented as follows:
f(x) = b*(x – a1)*(x – a2)* … * (x – an)
When multiplying the n clamps, the highest potency is obtained to b*x^n, i.e. the polynomial is then also of degree n.
Only true about algebraic sealed bodies. For R, b is a product (x^2+|am^^2) and c .