I have to make a polynomial function that uses these:

Third-degree, with zeros of -3, -1, and 2, and passes through the point (4, 7)

(Angry)

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- Jun 19th 2008, 02:50 PMmankvillContructing Polynomial Functions
I have to make a polynomial function that uses these:

Third-degree, with zeros of -3, -1, and 2, and passes through the point (4, 7)

(Angry) - Jun 19th 2008, 02:52 PMMoo
Hello !

If -3, -1 and 2 are zeros of the polynomial P(x), then x-(-3)=x+3, x+1 and x-2 divide P(x).

But (x+3)(x+1)(x-2) is already a polynomial of degree 3. So P(x) is just proportional to (x+3)(x+1)(x-2).

$\displaystyle P(x)=a(x+3)(x+1)(x-2)$

But we know that $\displaystyle P(4)=7$. Substituting, you can get a (Wink) - Jun 19th 2008, 02:55 PMmankvill
...I'm confused. What is the answer, then?

(Doh) - Jun 19th 2008, 03:03 PMMoo
- Jun 19th 2008, 03:07 PMmankvill
$\displaystyle x^3/10 + x^2/5 - x/2 - 3/5$

Is this correct?

edit: i don't know how to make a nominator and denominator, obviously - Jun 19th 2008, 10:16 PMMoo