... of a polynomial function with zeros at 2,-2, 1 and a y-intercept of 24
I put (x-2)(x-1)(x+2) +24 and factored out but it didn't work correctly
I got x^{3}-1x^{2}-4x+24
What am I doing wrong here?
you can't just add 24 and expect the zeros to remain the same.
What you can do is scale the entire thing
$A(x-2)(x-1)(x+2)$
Now find A so that the above expression is 24 when x=0.
$A(x-2)(x-1)(x+2)|_{x=0}=24$
$A(-2)(-1)(2)=24$
$4A=24$
$A=6$
so
$6(x-2)(x-1)(x+2)$ has the properties you want.
If the polynomial has zeroes at 2, - 2, and 1, then it is of the form
$f(x) = a(x - 2)^b(x + 2)^c(x - 1)^d = 0,\ where\ b,\ c,\ d \in \mathbb Z^+\ and\ a \ne 0.$
An infinite number of polynomials in that form have a y-intercept of 24.
But there is only one cubic of that form with y-intercept of 24.
$f(x) = a(x - 2)(x + 2)(x - 1) = 0\ and\ f(0) = 24 \implies a(x^2 - 4)(x - 1) = 0 \implies a(x^3 - x^2 - 4x + 4) = 0 \implies$
$4a = 24 \implies a = 6 \implies f(x) = 6x^3 - 6x^2 - 24x + 24.$
$f(-2) = 6(-8) - 6(4) - 24(-2) + 24 = -48 - 24 + 48 + 24 = 0.$ Check
$f(2) = 6(8) - 6(4) - 24(2) + 24 = 48 - 24 - 48 + 24 = 0.$ Check
$f(1) = 6(1) - 6(1) - 24(1) + 24 = 6 - 6 - 24 + 24 = 0.$ Check
$f(0) = 6(0) - 6(0) - 24(0) + 24 = 0 + 0 + 0 + 24 = 24.$ Check