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=-\frac52`

`\implies\boxed{f(x)=-\frac52x^3+\frac{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.