# Polynomials

• Nov 6th 2012, 10:35 AM
skweres1
Polynomials
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, 10:50 AM
Re: Polynomials
Since we know that 1,2,3 are roots, we can write that:

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

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

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

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

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

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

I hope you understand now.
• Nov 6th 2012, 01:17 PM
skweres1
Re: Polynomials
Thank you so much! Huge help!!