Worked examples
The question
Question: write a quadratic with zeros −2 and 5.
zeros : −2, 5
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).
Minus a minus is plus: x + 2.
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)
So f(x) = (x + 2)(x − 5).
f(x) = (x + 2)(x − 5)
3 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.
= x2 − 5x + 2x − 10
Like terms together: x² − 3x − 10.
= x2 − 3x − 10
So f(x) = x² − 3x − 10.
f(x) = x2 − 3x − 10
4 Check
Let's check: put each zero in.
First x = −2.
We get 0.
f(−2) = (−2)2 − 3(−2) − 10 = 4 + 6 − 10 = 0
And x = 5.
We get 0.
f(5) = 52 − 3(5) − 10 = 25 − 15 − 10 = 0
Let's look back.
First: each zero gives a bracket.
Then: write f(x).
Then: multiply out.
Last: check.
Answer f(x) = x2 − 3x − 10
▶ Watch it solved
The question
Question: write a quadratic with zeros 3 and −1 through (1, −8).
zeros : 3, −1
through : (1, −8)
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 3 gives x − 3.
x = 3 → x − 3
Next, −1 gives x − (−1).
Minus a minus is plus: x + 1.
x = −1 → x − (−1) = x + 1
Careful!
The sign flips.
The zero 3 gives x − 3, because that is 0 when x is 3.
Careful: not zero 3 → (x + 3) ✗
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 − 3)(x + 1)
3 Find a
Put x = 1 and f(x) = −8 in.
−8 = a(1 − 3)(1 + 1)
Work out each bracket: −2 and 2.
−8 = a(−2)(2)
−2 times 2 is −4.
So −8 = −4a.
−8 = −4a
To remove times −4, we divide both sides by −4.
−8−4 = −4a−4
−4 divided by −4 is 1.
So just a is left.
So a = 2.
a = 2
We put a in front of the brackets.
f(x) = 2(x − 3)(x + 1)
4 Multiply out
Each part of the first bracket times each part of the second.
each part × each part
(x − 3) times (x + 1).
Each part times each part.
= 2[x2 + x − 3x − 3]
Like terms together: x² − 2x − 3.
= 2(x2 − 2x − 3)
2 times each part.
= 2x2 − 4x − 6
So f(x) = 2x² − 4x − 6.
f(x) = 2x2 − 4x − 6
5 Check
Let's check: put each zero in.
First x = 3.
We get 0.
f(3) = 2(3)2 − 4(3) − 6 = 18 − 12 − 6 = 0
And x = −1.
We get 0.
f(−1) = 2(−1)2 − 4(−1) − 6 = 2 + 4 − 6 = 0
And the point: put x = 1 in.
We get −8, as it should be.
f(1) = 2(1)2 − 4(1) − 6 = 2 − 4 − 6 = −8
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 − 4x − 6
▶ Watch it solved
The question
Question: write a polynomial with zeros 3 and 1 + i.
zeros : 3, 1 + 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 3 gives x − 3.
x = 3 → 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.
1 + i → its partner 1 − i
Careful!
Don't forget the partner 1 − i.
Without it, the numbers in f(x) would not be real.
Careful: complex zeros come in pairs
The pair 1 + i and 1 − i gives two brackets.
x = 1 + i, 1 − i → (x − (1 + i))(x − (1 − i))
Group x − 1 together in each bracket.
Then they look like A minus B, times A plus B.
= [(x − 1) − i][(x − 1) + 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 − 1) squared, minus i squared.
= (x − 1)2 − i2
i squared is −1.
= x2 − 2x + 1 − (−1)
Minus a minus is plus: x² − 2x + 2.
No i is left!
= x2 − 2x + 2
2 Write f(x)
So f(x) = (x − 3)(x² − 2x + 2).
f(x) = (x − 3)(x2 − 2x + 2)
3 Multiply out
Each part of the first bracket times each part of the second.
each part × each part
(x − 3) times (x² − 2x + 2).
Each part times each part.
= x3 − 2x2 + 2x − 3x2 + 6x − 6
Like terms together: x³ − 5x² + 8x − 6.
= x3 − 5x2 + 8x − 6
So f(x) = x³ − 5x² + 8x − 6.
f(x) = x3 − 5x2 + 8x − 6
4 Check
Let's check: put each zero in.
First x = 3.
We get 0.
f(3) = 33 − 5(3)2 + 8(3) − 6
= 27 − 45 + 24 − 6 = 0
The pair: put x = 1 + i into x² − 2x + 2.
Remember i² = −1.
x = 1 + i: (1 + i)² − 2(1 + i) + 2
The i parts cancel, and the numbers make 0.
It works!
= 1 + 2i − 1 − 2 − 2i + 2 = 0
Let's look back.
First: each zero gives a bracket.
Then: write f(x).
Then: multiply out.
Last: check.
Answer f(x) = x3 − 5x2 + 8x − 6
▶ Watch it solved
