Polynomial Functions w/ zeros.

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To find a polynomial function with the zeros 2, 4+sqrt(5), and 4-sqrt(5), the expression can be simplified. The factors can be combined: (x-(4+sqrt(5)))(x-(4-sqrt(5))) simplifies to (x-4)² - 5, resulting in the polynomial x² - 8x + 11. This represents the monic polynomial of the lowest degree with the given roots. There are infinitely many polynomial functions that can have these zeros, but this is one of the simplest forms.
AznBoi
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Ok I have a probelm with find the polynoimal function which has these zeros:

Zeros: 2, 4+sqrt.(5), 4-sqrt.(5)

Find the polynomial equation with the given zeros.

So far I know:

y=(x-2)(x-(4+sqrt.(5))(x-(4-sqrt.(5))

but is there any way I could make the factor (x-(4+sqrt.(5)) into a better one? For example:

Zeros: -2,-1,0,1,2

I did:
y=x(x^2-1)(x^2-4)

instead of y=x(x+2)(x+1)(x-1)(x-2)

Thanks! :biggrin:
 
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AznBoi said:
Ok I have a probelm with find the polynoimal function which has these zeros:
Zeros: 2, 4+sqrt.(5), 4-sqrt.(5)

Find the polynomial equation with the given zeros.
Not "the" polynomial function- "a" polynomial function. There are an infinite number of polynomial functions having these zeros.

So far I know:

y=(x-2)(x-(4+sqrt.(5))(x-(4-sqrt.(5))

but is there any way I could make the factor (x-(4+sqrt.(5)) into a better one?
Yes, much as you did with -2, 2, and -1, 1 below: (x-(4+\sqrt{5}))(x- (4-\sqrt{5}))= ((x-4)+\sqrt{5})((x-4)-\sqrt{5})= (x-4)^2- 5
= x^2- 8x+ 16- 5= x^2- 8x+ 11
That is the monic polynomial of lowest degree having those roots.
For example:

Zeros: -2,-1,0,1,2

I did:
y=x(x^2-1)(x^2-4)

instead of y=x(x+2)(x+1)(x-1)(x-2)

Thanks! :biggrin:
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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