The polynomial f(x) has degree 3. If f(-1) = 15, f(0)= 0, f(1) = -5, and f(2) = 12, then what are the x-intercepts of the graph

I have not worked through this qustion but I'd start with the generic polynomial

\(f(x)=ax^3+bx^2+cx+d\)

Then start substituting so see if you can get any of those coefficients.

Starrt with  f(0)=0

You will probably end up with a couple of equations that you will need to solve simultaneously.

See how you go and post what you find for yourself.

If you then need more help then continue this thread.

Please no one else answer until asker guest has returned with some input of his/her own.

It has no constant because f(0)=0

c_3x^3+c_2x^2+c_1x

-c_3+c_2-c_1=15

c_3+c_2+c_1=-5

8c_3+4c_2+2c_1=12

4c_3+2c_2+c_1=6

-2c_3+2c_2-2c_1=30

*2

6c_3+6c_2+6c_1=-30

*6

add together

4c_3+8c_2+4c_1=0

divide by 4

c_3+2c_2+c_1=0

compare for x intercepts

c_3x^3+c_2x^2+c_1x=c_3+2c_2+c_1

I've figured out the answer to the question. I just had to do some substitution with the equations. The x-intercepts are (-4,0), (0,0) and (1.5,0). =)

Yep. Here is the link to the graph: https://www.desmos.com/calculator/bvpzoqrgnu