Write a polynomial equation with the following roots. A quartic funciton with roots of $-3$,$-1$ and $4$? Write a polynomial equation with the following roots. A quartic funciton with roots of $-3,-1$ and $4$?
use each root as a factor
$$f(x) = a(x + 3)(x + 1)(x - 4)^2$$
$$f(x) = a(x^2 + 4x + 3)(x^2 - 8x + 16)$$
$$f(x) = a(x^4 - 4x^3 - 13x^2 + 40x + 48)$$
$$f(5) = 16$$
$$16 = a(5^4 - 4(5^3) - 13(5^2) + 40(5) + 48 = 48a \rightarrow a = 3$$
$$f(x) = 3x^4 - 12x^3 - 39x^2 + 120x + 144$$
Is that correct? Please show another way of doing it please with your steps.
 A: You did just fine: you equation:
$$f(x) = a(x + 3)(x + 1)(x - 4)^2;\quad a\neq 0\tag{1}$$ is entirely fine: it is a quartic function with three roots: $x = -3, x = -1, \;x = 4$, where the root $x = 4$ has multiplicity of two.  Your expansion of the factors is spot on, but not required (and doing the extra work leaves room for a silly mistake, though not in your expansion).
One observation: When evaluating $f(5) = 16$ to solve for $a$, note that $16 = 48a \implies a = \dfrac 13$, which will alter the final equation at the very end. 
Note: you could easily have used equation $(1)$ to solve for $a$ as well:
$$f(x) = a(x + 3)(x + 1)(x - 4)^2$$ $$\implies f(5) = 16 = a(5+3)(5+1)(5-4)^2 = 48a $$ $$\iff a = \frac{16}{48} = \frac 13$$
The following functions are also valid quartic functions with the same required roots, as well:
$$f(x) = a(x + 3)^2(x + 1)(x - 4); \quad a\neq 0$$
$$f(x) = a(x + 3)(x + 1)^2(x - 4);\quad a\neq 0$$
A: That's correct.  You made $4$ a double root by including two $(x-4)$ factors.  You could have equally well made $-3$ or $-1$ double roots.
