Question

What is the third–degree polynomial function such that f(0) = –12 and whose zeros are 1, 2, and 3?


f(x) = 2x3 + 12x2 – 22x – 12

f(x) = 2x3 – 12x2 + 22x – 12

f(x) = 4x3 – 24x2 + 22x – 24

f(x) = 4x3 + 24x2 – 44x – 24 What is the third–degree polynomial function such that f(0) = –12 and whose zeros are 1, 2, and 3?


f(x) = 2x3 + 12x2 – 22x – 12

f(x) = 2x3 – 12x2 + 22x – 12

f(x) = 4x3 – 24x2 + 22x – 24

f(x) = 4x3 + 24x2 – 44x – 24 @Mathematics

Answer 1

Some help?

Answer 2

the factored form of a third degree polynomial looks like
f(x) = k(x-a)(x-b)(x-c)

f(0)=-12 that means when x =0 f(x)=-12
and a=1, b=2, c=3
substituting in equation 1
f(0)= k(0-1)(0-2)(0-3)
f(0) = -6k
therefore -12=-6k
k=2

so the polynomial is
f(x)=2(x-1)(x-2)(x-3)

Answer 3

expand it lol

Answer 4

The second choice is the right one btw:)

Answer 5

haha k, how did you get -6?

Answer 6

u know that f(0) = -12
right?

Answer 7

yes

Answer 8

and f(x)=k(x-a)(x-b)(x-c)
so if u substitute x=0 here
f(0)=k(0-a)(0-b)(0-c)
f(0) = k(-a)(-b)(-c)
a, b and c are the zeros of the polynomial
so a=1 and b=2 and c=3
substituting these values in f(0)
u get
f(0) = k(-1)(-2)(-3)
f(0)= k * -6

and we know that f(0)=-12
so -6k=-12

Answer 9

Oh is this a theorem? or no?

Answer 10

all third degree polynomials look like
f(x)=k(x-a)(x-b)(x-c)
where k is any constant and a,b,c are zeros of polynomial

Answer 11

Oh ok, man I have to find that in my lessons, I dont even know if it mentioned it haha.

Answer 12

hehe, it can be found using other methods, this is the easiest

Answer 13

Oh would you mind telling me the other methods? sorry for all the questions haha I just want to have it down packed