Question

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

Answer 1

p(1)-2=0

Answer 2

zoom

Answer 3

(-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

Answer 4

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

Answer 5

not subtractoin; scaled

Answer 6

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

Answer 7

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

Answer 8

meh ... lol

Answer 9

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

Answer 10

thats what we got lawyers for :)

Answer 11

i knew they were there for some reason...

Answer 12

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

Answer 13

(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