Not sure how to do this.. at all

Let p(x) be 3rd degree polynomial w/ real coefficients and leading coefficient one such that p(-1) = 2 and p(3)=2. Find a possible polynomial p(x)


thank youu


P.S. also for another problem, x^2 + 2x - 1, completely factor into a product of linear factors. is the answer (x-(-1-√2)), (x-(-1+√2)) correct?

Your PS problem is correct.

For the first one, let p(x) = x^3 + b x^2 + c x + d.
Now p(-1)=2 so (-1)^3 + b (-1)^2 + c(-1) +d = 2 ie -1+b-c+d=2
Do the same for p(3)=2.
You'll end up with 2 equations involving 3 unknowns (b, c and d).
You won't be able to solve these (there is not an unique solution), but you should be able to get (for example) b in terms of d and also c in terms of d.
Then choose any value you like for d, and calculate b and c accordingly.

This will the give you a possible polynomial. (There are infinitely many possible answers.)

"You'll end up with 2 equations involving 3 unknowns (b, c and d).
You won't be able to solve these (there is not an unique solution), but you should be able to get (for example) b in terms of d and also c in terms of d."

still a bit confused how to do this.. how do i set up an equation to solve for? i'm sorry, i'm just so bad in math