Steply

Polynomial with zeros 2, i and -i

Step by step, with the reason for every step · Writing polynomials from zeros · Algebra 2

Step by step

The question

Question: write a polynomial with zeros 2, i and −i.

zeros : 2, i, −i

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

zero: an x where f(x) = 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.

zero c → bracket (x − c)

The zero 2 gives x − 2.

x = 2 → x − 2

Careful!

The sign flips.

The zero 2 gives x − 2, because that is 0 when x is 2.

Careful: not zero 2 → (x + 2) ✗

The pair i and −i gives two brackets.

x = i, −i → (x − i)(x − (−i))

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

And i squared is −1.

(A − B)(A + B) = A² − B²
i² = −1

So we get x squared, minus i squared.

= x2 − i2

i squared is −1.

= x2 − (−1)

Minus a minus is plus: x² + 1.

No i is left!

= x2 + 1

2 Write f(x)

So f(x) = (x − 2)(x² + 1).

f(x) = (x − 2)(x2 + 1)

3 Multiply out

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

each part × each part

(x − 2) times (x² + 1).

Each part times each part.

= x3 + x − 2x2 − 2

Like terms together: x³ − 2x² + x − 2.

= x3 − 2x2 + x − 2

So f(x) = x³ − 2x² + x − 2.

f(x) = x3 − 2x2 + x − 2

4 Check

Let's check: put each zero in.

First x = 2.

We get 0.

f(2) = 23 − 2(2)2 + 2 − 2 = 8 − 8 + 2 − 2 = 0

The pair: x² + 1 = 0 gives x squared is −1.

So x is plus or minus i.

It works!

x² + 1 = 0 → x2 = −1 → x = ±√−1 = ±i

Let's look back.

First: each zero gives a bracket.

Then: write f(x).

Then: multiply out.

Last: check.

Answer
f(x) = x3 − 2x2 + x − 2

Your turn

Try this one yourself, then check:

polynomial with zeros 1, 2i, −2i
Show the answer
f(x) = x3 − x2 + 4x − 4

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