Polynomial with zeros 3 and 1 + i

Step by step

The question

Question: write a polynomial with zeros 3 and 1 + i.

A zero of f is an x that makes f(x) equal 0.

1 Each zero gives a bracket

If c is a zero, then x minus c is a factor.

Put c in, and that bracket is 0, so all of f(x) is 0.

The zero 3 gives x − 3.

A complex zero always comes with its partner: the same number with the sign of the i part turned.

So 1 − i is a zero too.

Careful!

Don't forget the partner 1 − i.

Without it, the numbers in f(x) would not be real.

The pair 1 + i and 1 − i gives two brackets.

Group x − 1 together in each bracket.

Then they look like A minus B, times A plus B.

A minus B, times A plus B, is A squared minus B squared.

And i squared is −1.

So we get (x − 1) squared, minus i squared.

i squared is −1.

Minus a minus is plus: x² − 2x + 2.

No i is left!

2 Write f(x)

So f(x) = (x − 3)(x² − 2x + 2).

3 Multiply out

Each part of the first bracket times each part of the second.

(x − 3) times (x² − 2x + 2).

Each part times each part.

Like terms together: x³ − 5x² + 8x − 6.

So f(x) = x³ − 5x² + 8x − 6.

4 Check

Let's check: put each zero in.

First x = 3.

We get 0.

The pair: put x = 1 + i into x² − 2x + 2.

Remember i² = −1.

The i parts cancel, and the numbers make 0.

It works!

Let's look back.

First: each zero gives a bracket.

Then: write f(x).

Then: multiply out.

Last: check.