Suppose p is a polynomial of degree 3 with three zeros at x=1, x=2, and x=3. Find p if p(x) evaluated at 5 is 48.

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- Nov 6th 2012, 09:35 AMskweres1Polynomials
Suppose p is a polynomial of degree 3 with three zeros at x=1, x=2, and x=3. Find p if p(x) evaluated at 5 is 48.

- Nov 6th 2012, 09:50 AMPaulAdrienMauriceDiracRe: Polynomials
Since we know that 1,2,3 are roots, we can write that:

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

$\displaystyle P(5)=a(5-1)(5-2)(5-3)=24a=48$ -> $\displaystyle a=2$.

Now distribute 2 in $\displaystyle P(x)=a(x-1)(x-2)(x-3)$:

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

$\displaystyle =(2x^2-6x+4)(x-3)=2x^3-12x^2+22x-12$.

So $\displaystyle P(x)=2x^3-12x^2+22x-12$.

I hope you understand now. - Nov 6th 2012, 12:17 PMskweres1Re: Polynomials
Thank you so much! Huge help!!