A 3rd degree polynomial has roots at x=-2i and x=5. The y-intercept is (0,25). Write an equation for this function in factored form with real coefficients.
My first hack is to say that in factored form this is (x-5)(X2+4) as that gets me the roots, but that has y-intercept (0,-20). What am I missing?
Thanks!
Best Answer
To elaborate on SmileyCraft's answer (they give the answer but not the underlying reason why, which I feel is more important)...
The Roots:
On first contending with the roots issue, which the comments hint you had trouble with: we know $-2i$ and $5$ are roots. We know that since $-2i$ is a root, $+2i$ is also a root because it's the conjugate of $-2i$. Then, since $5, -2i, +2i$ are all roots, we know that the polynomial can be factored as
$$f(x) = a(x - 5)(x - 2i)(x + 2i)$$
up to a leading coefficient $a$ that I'll talk about later. (You assumed $a=1$, and I'll explain why that's wrong in the next section.) It should be clear that this form is correct otherwise, since plugging in any of our roots gives $f(x) = 0$.
Since you wanted real coefficients and such, we note that $(x-2i)(x+2i)=x^2 + 4$, thus giving us the form you posed in the question (again, up to that constant $a$).
Finding $a$:
So now you know that the polynomial has the form
$$f(x) = a(x-5)(x^2 + 4)$$
The reason we include this $a$ is because the leading coefficient of the polynomial doesn't actually affect the roots: it does affect certain aspects of its behavior, but not that one.
(Note of import: that is only true when the polynomial is factored in this form. For example, in the other standard form of a cubic, $f(x) = ax^3 + bx^2 + cx + d$, the coefficient $a$ does affect the zeroes. It's only if in this factored form that the leading coefficient does not affect the roots. Play around on a graphing calculator, shouldn't be hard to convince yourself of these facts.)
We also know that the $y$-intercept of the graph is $(0,25)$. Thus, plug in $x = 0$ and $f(x) = 25$ in the equation above, and then you'll be able to figure out what $a$ actually is.
Plugging in these values, then, we see
$$25 = a(-5)(4) \;\;\; \Rightarrow \;\;\; 25 = a(-20) \;\;\; \Rightarrow \;\;\; a = \frac{25}{-20} = -\frac{5}{4}$$
Thus, the polynomial in its factored form is
$$f(x) = \left( -\frac{5}{4} \right)(x-5)(x^2 + 4)$$