Question

Find an n-degree polynomial function with real coefficients satisfying the given condition.

1. n=3; 4 and 2i are zeros; f(-1)=50

2. n=3; 4 and -5+2i are zeros; f(2)= -636

3. n=4; -2, -1/2, and i are zeros; f(1)=18

4. n=4; -4, 1/3, and 2+3i are zeros; f(1)=100

5. n=4; 1+i and i are zeros; f(1)=2

Answer 1

In all cases, if $f$ has real coefficients, then any complex roots occur in conjugate pairs, so if $a+bi$ is a root, then so is $a-bi$. Also, by the fundamental theorem of algebra, if $r_1,\ldots,r_n$ are roots to $f$, then for some constant $a\in\mathbb R$,

$f(x)=a(x-r_1)\cdots(x-r_n)$

1. If $n=3$ and $f(3)=f(2i)=0$, then

$f(x)=a(x-3)(x-2i)(x+2i)=ax^3-3ax^2+4ax-12a$

Given that $f(-1)=50$, we have

$f(-1)=a(-1-3)(-1-2i)(-1+2i)=-20a=50\implies a=-\dfrac52$

$\implies\boxed{f(x)=-\dfrac52x^3+\dfrac{15}2x^2-10x+30}$

2.

$f(x)=a(x-4)(x-(-5+2i))(x-(-5-2i))=a x^3 + 6 a x^2 - 11 a x - 116 a$

With $f(2)=-636$, we have

$f(2)=a(2-4)(2+5-2i)(2+5+2i)=-106a=-636\implies a=6$

$\implies\boxed{f(x)=6x^3+36x^2-66x-696}$

The rest are done in the same exact way.