Find a polynomial p of degree 3 such that -3, -1 and 7 are zeros of p and p(1)=2

Question

Find a polynomial p of degree 3 such that -3, -1 and 7 are zeros of p  and p(1)=2

anonymous · Answer

p(1)-2=0

anonymous · Answer

zoom

amistre64 · Answer

(-3+3)(-1+1)(7-7)
  r         r        r


(x+3)(x+1)(x-7)
1       1       1
---------------
4     2        6  = 48

so divide by 24

amistre64 · Answer

1/24 (x+3)(x+1)(x-7)

amistre64 · Answer

not subtractoin; scaled

anonymous · Answer

$$a(x+3)(x+1)(x-7)$$ solve 
$$p(1)=2$$ for a 
$$p(1)=a\times 4\times 2\times -6=2$$ etecs

anonymous · Answer

actually i get 
$$a=-\frac{1}{24}$$

amistre64 · Answer

meh ... lol

anonymous · Answer

on account of 1 - 7 = -6
 but what is a minus sign between friends?

amistre64 · Answer

thats what we got lawyers for :)

anonymous · Answer

i knew they were there for some reason...

anonymous · Answer

so what do I do after this step? 1/24 (x+3)(x+1)(x-7)

anonymous · Answer

(x+1) (x+3)(x-7) 
= (x+1) ( x^2 - 4x - 21)
= x^3 - 4x^2 - 21x + ( x^2 - 4x - 21)
= x^3 - 3x^2 - 25x - 21