Construct a polynomial function with the following properties: third degree, only real coefficients, -5 and 3+I are two of the zeros, y intercept is -50

Answer 1

Recall the Factor Theorem:

Let $F$ be a field, $f(x)\in F[x]$, and $a\in F$. Then $a$ is a root of the polynomial $f(x)$ if and only if $x-a$ is a factor of $f(x)$ in $F[x]$.

Further, recall the following lemma:

If $f(x)$ is a polynomial in $\mathbb{R}[x]$ and $a+bi$ is a root of $f(x)$ in $\mathbb{C}$, then $a-bi$ is also a root of $f(x)$.

Suppose $f(x)\in\mathbb{R}$ with roots $-5$ and $3+i$. Since $-5$ is a root, then by the factor theorem $x-(-5) = x+5$ is a factor of $f(x)$. Note, $\mathbb{C}$ is an extension field of $\mathbb{R}$ $\Rightarrow$ $\mathbb{C}[x]$ is an extension field of $\mathbb{R}[x]$. In $\mathbb{C}[x]$, since $3+i$ is a root, then by the factor theorem $x-(3+i) = x-3-i$ is another factor of $f(x)$. Also, by the lemma, since $3+i$ is a root, $3-i$ is a root, and $x-(3-i) = x-3+i$ is another factor of $f(x)$. Thus,

\begin{align*}

f(x) &= f_0(x+5)(x-3-i)(x-3+i), \\

&= f_0(x+5)(x^2-3x+ix-3x+9-3i-ix+3i-i^2), \\

&= f_0(x+5)(x^2-6x+9-i^2), \\

&= f_0(x+5)(x^2-6x+9-(-1)), \\

&= f_0(x+5)(x^2-6x+10), \\

&= f_0(x^3-6x^2+10x+5x^2-30x+50), \\

&= f_0(x^3-x^2-20x+50),

\end{align*}

for some $f_0\in\mathbb{R}$. We must now apply the condition of the $y$-intercept being $-50$, i.e. $f(0) = -50$. We can set $f_0$ to be a certain value for this to be true:

\begin{align*}

f_0 &= \frac{f(x)}{x^3-x^2-20x+50}, \\

&= \frac{f(0)}{(0)^3-(0)^2-20(0)+50}, \\

&= \frac{-50}{50}, \\

&= -1.

\end{align*}

Thus, $\boxed{f(x) = -1(x^3-x^2-20x+50) = -x^3+x^2+20x-50}$.

[Source: "Abstract Algebra: An Introduction" by Thomas W. Hungerford]

Answer 2

f(x) = a(x - (-5))[x - (3+i)][x - (3-i)] = a(x+5)[(x-3) - i][(x-3) + i] = a(x+5)(x² - 6x + 10)

Since f(0) = -50, a = -1.

f(x) = (x+5)(x² - 6x + 10) Multiply and simplify.