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Write a polynomial from real and complex zeros

Each zero c supplies a factor x minus c

A polynomial with prescribed zeros $r_{1}, \dots, r_{n}$ can be written as $a \prod_{k = 1}^{n} (x - r_{k})$ , with $a \neq = 0$ . If no leading coefficient is specified, choosing $a = 1$ gives a simple monic example. Pairing conjugate roots makes the expansion easier.

Problem

Write one polynomial satisfying each condition:

1. Degree 3 with zeros $3$ , $5$ , and $- 2$ . 2. Degree 2 with zeros $211$ and $- 211$ . 3. Degree 3 with zeros $- 3$ , $4 i$ , and $- 4 i$ .

Answer

f (x) = x^{3} - 6 x^{2} - x + 30

. 2.

f (x) = x^{2} - 44

. 3.

f (x) = x^{3} + 3 x^{2} + 16 x + 48

These are monic examples. Multiplying any entire polynomial by a nonzero constant preserves its zeros and degree.

Step-by-step solution

1. Use the three integer zeros

The corresponding factors are $x - 3$ , $x - 5$ , and $x + 2$ . Choose leading coefficient 1:

f (x) = (x - 3) (x - 5) (x + 2) .

First multiply $(x - 3) (x - 5) = x^{2} - 8 x + 15$ . Then

(x^{2} - 8 x + 15) (x + 2) = x^{3} - 6 x^{2} - x + 30.

The three factors ensure that substituting 3, 5, or -2 gives zero.

2. Pair the radical zeros

Use a difference of squares:

f (x) = (x - 211) (x + 211) = x^{2} - (211)^{2} = x^{2} - 44.

The middle terms cancel. Its degree is 2, and both prescribed roots have square 44.

3. Pair the complex conjugates

The factors are $x + 3$ , $x - 4 i$ , and $x + 4 i$ . Since $i^{2} = - 1$ ,

(x - 4 i) (x + 4 i) = x^{2} - (4 i)^{2} = x^{2} + 16.

Now multiply the remaining factor:

f (x) = (x + 3) (x^{2} + 16) = x^{3} + 3 x^{2} + 16 x + 48.

The complex factors cancel the imaginary coefficients, producing a polynomial with real coefficients.

4. Check degree and leading coefficient

There are three, two, and three linear factors, respectively, so the degrees match the requested values. Because each example uses leading coefficient 1, none of the highest-degree terms vanishes.

5. Understand why answers may vary

If only zeros and degree are specified, every nonzero constant multiple of each displayed polynomial is another valid answer. Zero is not an allowed multiplier because it would remove the specified degree.

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