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Quadratic with zeros 2 and 5 through (0, 20)

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

Step by step

The question

Question: write a quadratic with zeros 2 and 5 through (0, 20).

zeros : 2, 5
through : (0, 20)

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

Next, 5 gives x − 5.

x = 5 → x − 5

Careful!

The sign flips.

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

Careful: not zero 2 → (x + 2) ✗

2 Write f(x)

The zeros fix the brackets, but f can be stretched: a times the brackets.

The point tells us a.

f(x) = a(x − 2)(x − 5)

3 Find a

Put x = 0 and f(x) = 20 in.

20 = a(0 − 2)(0 − 5)

Work out each bracket: −2 and −5.

20 = a(−2)(−5)

−2 times −5 is 10.

So 20 = 10a.

20 = 10a

To remove times 10, we divide both sides by 10.

2010 = 10a10

10 divided by 10 is 1.

So just a is left.

So a = 2.

a = 2

We put a in front of the brackets.

f(x) = 2(x − 2)(x − 5)

4 Multiply out

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

each part × each part

(x − 2) times (x − 5).

Each part times each part.

= 2[x2 − 5x − 2x + 10]

Like terms together: x² − 7x + 10.

= 2(x2 − 7x + 10)

2 times each part.

= 2x2 − 14x + 20

So f(x) = 2x² − 14x + 20.

f(x) = 2x2 − 14x + 20

5 Check

Let's check: put each zero in.

First x = 2.

We get 0.

f(2) = 2(2)2 − 14(2) + 20 = 8 − 28 + 20 = 0

And x = 5.

We get 0.

f(5) = 2(5)2 − 14(5) + 20 = 50 − 70 + 20
= 0

And the point: put x = 0 in.

We get 20, as it should be.

f(0) = 2(0)2 − 14(0) + 20 = 0 + 0 + 20 = 20

Let's look back.

First: each zero gives a bracket.

Then: write f(x).

Then: find a.

Then: multiply out.

Last: check.

Answer
f(x) = 2x2 − 14x + 20

Your turn

Try this one yourself, then check:

quadratic with zeros 3, 4 through (2, 6)
Show the answer
f(x) = 3x2 − 21x + 36

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