Creating a polynomial from zeros that passes through point x

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
GreenLantern
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Joined: Sat Mar 07, 2009 10:47 pm
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Creating a polynomial from zeros that passes through point x

Postby GreenLantern » Sat Mar 07, 2009 11:07 pm

Book Question: Find a polynomial f(x) of degree 6 such that 0 and 3 are both zeros of multiplicity 3 and f(2)=-24. Sketch the graph of f.

My Question: I can create the polynomial well enough, but I don't under stand how to make it match up to f(2)=-24.

My work so far:
Zeros: 0, 3
Factors: (x)^3 (x-3)^3
Distributed: f(x)= [x^6] - [9x^5] + [27x^4] - [27x^3]
(I hope I typed this up correctly.)

Now I need to make that polynomial pass through point (2,-24). How do I do that?

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stapel_eliz
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Re: Creating a polynomial from zeros that passes through point x

Postby stapel_eliz » Sat Mar 07, 2009 11:20 pm

GreenLantern wrote:Book Question: Find a polynomial f(x) of degree 6 such that 0 and 3 are both zeros of multiplicity 3 and f(2)=-24. Sketch the graph of f.

My Question: I can create the polynomial well enough, but I don't under stand how to make it match up to f(2)=-24.

My work so far:
Zeros: 0, 3
Factors: (x)^3 (x-3)^3

This gives you:

. . . . .

Why the "a"? Because you don't know what they divided through by, if anything, when they went to solve the original polynomial equation:

. . . . .

GreenLantern wrote:Now I need to make that polynomial pass through point (2,-24). How do I do that?

They've given you values for x and y; namely, x = 2 and y = -24. Plug these in:

. . . . .

. . . . .

. . . . .

. . . . .

Solve to find the value of "a", and then rewrite your polynomial equation with the factors you've already found, and now this numerical multiplier.

You probably don't have to multiply the thing out, by the way.... :wink:

GreenLantern
Posts: 23
Joined: Sat Mar 07, 2009 10:47 pm
Contact:

Re: Creating a polynomial from zeros that passes through point x

Postby GreenLantern » Sat Mar 07, 2009 11:29 pm

Thank you very much! That clears many things up, and the link was very helpful as well.

I was going to far in distributing my factors, and I left out the 'a' coefficient.


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