Worked examples
The question
Question: find all the zeros of f(x) = x³ − 2x² − 5x + 6.
f(x) = x3 − 2x2 − 5x + 6
A zero of f is an x that makes f(x) equal 0.
A power 3 has up to three zeros.
a zero: an x where f(x) = 0
1 List the candidates
The rational root theorem: a whole or fraction zero is a factor of the last number, over a factor of the first number.
whole or fraction zero = factor of the last numberfactor of the first number
The last number is 6.
Its factors are 1, 2, 3, 6, each with plus and minus.
factors of 6 : ±1, ±2, ±3, ±6
Careful!
Don't forget the negative ones.
−1, −2 and the others can be zeros too.
Careful: every candidate has + and −
The first number is 1.
So the candidates are just these.
candidates : ±1, ±2, ±3, ±6
2 Test the candidates
We put each candidate into f.
When we get 0, we found a zero.
Then x minus that number is a factor, like in the factor theorem lesson.
f(c) = 0 → (x − c) is a factor
Start with 1: put it in place of every x.
f(1) = 13 − 2(1)2 − 5(1) + 6
We get 0.
So x = 1 is a zero!
= 1 − 2 − 5 + 6 = 0 zero!
3 Divide by x − 1
x = 1 is a zero, so (x − 1) is a factor.
We divide it out, with synthetic division.
x − 1 is 0 when x is 1.
So we use 1.
Then we write just the numbers in front: 1, −2, −5, 6.
The first number comes straight down: 1.
1 times 1 is 1.
Write it under the next number.
Add: −2 + 1 is −1.
−1 times 1 is −1.
Write it under the next number.
Add: −5 + (−1) is −6.
−6 times 1 is −6.
Write it under the next number.
Add: 6 + (−6) is 0.
The last number is the remainder: 0.
The other numbers are the answer's numbers: 1, −1, −6.
The answer starts one power lower than x³.
So f(x) = (x − 1)(x² − x − 6).
The other zeros come from x² − x − 6 = 0.
f(x) = (x − 1)(x² − x − 6)
4 Solve the quadratic
Now x² − x − 6 = 0, with the quadratic lesson.
x2 − x − 6 = 0
5 Factor
We need two numbers.
They multiply to −6, and they add to −1.
? × ? = −6
? + ? = −1
They multiply to a minus number, so one number is plus and one is minus.
They add to a minus number, so the bigger one is minus.
× is − → one +, one −; bigger one −
We list the pairs that make 6, smallest first.
We take the smaller from the bigger.
We need 1.
2 and 3 give 1.
Yes!
With the signs: 2 and −3.
1 × 6: 6 − 1 = 5
2 × 3: 3 − 2 = 1 ✓
So it is (x + 2)(x − 3) = 0.
(x + 2)(x − 3) = 0
6 One bracket is 0
Two numbers multiply to 0.
That only happens if one of them is 0.
So one bracket must be 0.
A × B = 0 → A = 0 or B = 0
We make each bracket 0.
Two small equations.
x + 2 = 0 or x − 3 = 0
In x + 2 = 0, we take away 2 on both sides.
In x − 3 = 0, we add 3 on both sides.
x + 2 − 2 = 0 − 2 or x − 3 + 3 = 0 + 3
So x = −2 or x = 3.
x = −2 or x = 3
7 Write all the zeros
So the zeros are 1, −2 and 3.
1 from the test, the others from the quadratic.
zeros : x = 1 or x = −2 or x = 3
8 Check
Let's check: put −2 into f.
We get 0.
x = −2: f = (−2)3 − 2(−2)2 − 5(−2) + 6
= −8 − 8 + 10 + 6 = 0
And put 3 into f.
We get 0 too.
x = 3: f = 33 − 2(3)2 − 5(3) + 6
= 27 − 18 − 15 + 6 = 0
Let's look back.
First: list the candidates.
Then: test the candidates.
Then: divide by x − 1.
Then: solve the quadratic.
Then: factor.
Then: one bracket is 0.
Then: write all the zeros.
Last: check.
Answer zeros : x = 1 or x = −2 or x = 3
▶ Watch it solved
The question
Question: find all the zeros of f(x) = x³ + 2x² − 5x − 6.
f(x) = x3 + 2x2 − 5x − 6
A zero of f is an x that makes f(x) equal 0.
A power 3 has up to three zeros.
a zero: an x where f(x) = 0
1 List the candidates
The rational root theorem: a whole or fraction zero is a factor of the last number, over a factor of the first number.
whole or fraction zero = factor of the last numberfactor of the first number
The last number is −6.
Its factors are 1, 2, 3, 6, each with plus and minus.
factors of 6 : ±1, ±2, ±3, ±6
Careful!
Don't forget the negative ones.
−1, −2 and the others can be zeros too.
Careful: every candidate has + and −
The first number is 1.
So the candidates are just these.
candidates : ±1, ±2, ±3, ±6
2 Test the candidates
We put each candidate into f.
When we get 0, we found a zero.
Then x minus that number is a factor, like in the factor theorem lesson.
f(c) = 0 → (x − c) is a factor
Start with 1: put it in place of every x.
f(1) = 13 + 2(1)2 − 5(1) − 6
We get −8, not 0.
So 1 is not a zero.
= 1 + 2 − 5 − 6 = −8 ✗
Next −1: put it in place of every x.
f(−1) = (−1)3 + 2(−1)2 − 5(−1) − 6
We get 0.
So x = −1 is a zero!
= −1 + 2 + 5 − 6 = 0 zero!
3 Divide by x + 1
x = −1 is a zero, so (x + 1) is a factor.
We divide it out, with synthetic division.
x + 1 is 0 when x is −1.
So we use −1.
Then we write just the numbers in front: 1, 2, −5, −6.
The first number comes straight down: 1.
1 times −1 is −1.
Write it under the next number.
Add: 2 + (−1) is 1.
1 times −1 is −1.
Write it under the next number.
Add: −5 + (−1) is −6.
−6 times −1 is 6.
Write it under the next number.
Add: −6 + 6 is 0.
The last number is the remainder: 0.
The other numbers are the answer's numbers: 1, 1, −6.
The answer starts one power lower than x³.
So f(x) = (x + 1)(x² + x − 6).
The other zeros come from x² + x − 6 = 0.
f(x) = (x + 1)(x² + x − 6)
4 Solve the quadratic
Now x² + x − 6 = 0, with the quadratic lesson.
x2 + x − 6 = 0
5 Factor
We need two numbers.
They multiply to −6, and they add to 1.
? × ? = −6
? + ? = 1
They multiply to a minus number, so one number is plus and one is minus.
They add to a plus number, so the bigger one is plus.
× is − → one +, one −; bigger one +
We list the pairs that make 6, smallest first.
We take the smaller from the bigger.
We need 1.
2 and 3 give 1.
Yes!
With the signs: −2 and 3.
1 × 6: 6 − 1 = 5
2 × 3: 3 − 2 = 1 ✓
So it is (x − 2)(x + 3) = 0.
(x − 2)(x + 3) = 0
6 One bracket is 0
Two numbers multiply to 0.
That only happens if one of them is 0.
So one bracket must be 0.
A × B = 0 → A = 0 or B = 0
We make each bracket 0.
Two small equations.
x − 2 = 0 or x + 3 = 0
In x − 2 = 0, we add 2 on both sides.
In x + 3 = 0, we take away 3 on both sides.
x − 2 + 2 = 0 + 2 or x + 3 − 3 = 0 − 3
So x = 2 or x = −3.
x = 2 or x = −3
7 Write all the zeros
So the zeros are −1, 2 and −3.
−1 from the test, the others from the quadratic.
zeros : x = −1 or x = 2 or x = −3
8 Check
Let's check: put 2 into f.
We get 0.
x = 2: f = 23 + 2(2)2 − 5(2) − 6
= 8 + 8 − 10 − 6 = 0
And put −3 into f.
We get 0 too.
x = −3: f = (−3)3 + 2(−3)2 − 5(−3) − 6
= −27 + 18 + 15 − 6 = 0
Let's look back.
First: list the candidates.
Then: test the candidates.
Then: divide by x + 1.
Then: solve the quadratic.
Then: factor.
Then: one bracket is 0.
Then: write all the zeros.
Last: check.
Answer zeros : x = −1 or x = 2 or x = −3
▶ Watch it solved
The question
Question: find all the zeros of f(x) = 2x³ − 3x² − 3x + 2.
f(x) = 2x3 − 3x2 − 3x + 2
A zero of f is an x that makes f(x) equal 0.
A power 3 has up to three zeros.
a zero: an x where f(x) = 0
1 List the candidates
The rational root theorem: a whole or fraction zero is a factor of the last number, over a factor of the first number.
whole or fraction zero = factor of the last numberfactor of the first number
The last number is 2.
Its factors are 1, 2, each with plus and minus.
factors of 2 : ±1, ±2
The first number is 2.
Its factors are 1, 2.
factors of 2 : 1, 2
Careful!
Don't forget the negative ones.
−1, −2 and the others can be zeros too.
Careful: every candidate has + and −
Each top factor over each bottom factor.
These are all the candidates.
candidates : ±1, ±2, ±12
2 Test the candidates
We put each candidate into f.
When we get 0, we found a zero.
Then x minus that number is a factor, like in the factor theorem lesson.
f(c) = 0 → (x − c) is a factor
Start with 1: put it in place of every x.
f(1) = 2(1)3 − 3(1)2 − 3(1) + 2
We get −2, not 0.
So 1 is not a zero.
= 2 − 3 − 3 + 2 = −2 ✗
Next −1: put it in place of every x.
f(−1) = 2(−1)3 − 3(−1)2 − 3(−1) + 2
We get 0.
So x = −1 is a zero!
= −2 − 3 + 3 + 2 = 0 zero!
3 Divide by x + 1
x = −1 is a zero, so (x + 1) is a factor.
We divide it out, with synthetic division.
x + 1 is 0 when x is −1.
So we use −1.
Then we write just the numbers in front: 2, −3, −3, 2.
The first number comes straight down: 2.
2 times −1 is −2.
Write it under the next number.
Add: −3 + (−2) is −5.
−5 times −1 is 5.
Write it under the next number.
Add: −3 + 5 is 2.
2 times −1 is −2.
Write it under the next number.
Add: 2 + (−2) is 0.
The last number is the remainder: 0.
The other numbers are the answer's numbers: 2, −5, 2.
The answer starts one power lower than x³.
So f(x) = (x + 1)(2x² − 5x + 2).
The other zeros come from 2x² − 5x + 2 = 0.
f(x) = (x + 1)(2x² − 5x + 2)
4 Solve the quadratic
Now 2x² − 5x + 2 = 0, with the quadratic lesson.
2x2 − 5x + 2 = 0
5 Find two numbers
There is a number in front of x², but it can still factor.
The number in front of x² is 2.
So we multiply 2 by 2: that is 4.
We need two numbers that multiply to 4 and add to −5.
2 × 2 = 4
? × ? = 4, ? + ? = −5
They multiply to a plus number, so the two numbers have the same sign.
They add to a minus number, so both are minus.
× is + → same signs: both −
We list the pairs that make 4, smallest first.
We add each pair.
We need 5.
1 and 4 give 5.
Yes!
With the signs: −1 and −4.
1 × 4: 1 + 4 = 5 ✓
6 Split the middle
We write −5x as −x − 4x.
It is the same amount.
2x2 − x − 4x + 2 = 0
7 Group in pairs
We put the first two together, and the last two together.
(2x2 − x) + (−4x + 2) = 0
From the first pair we take out x.
From the second pair we take out −2.
Both leave the same bracket.
(2x2 − x) = x(2x − 1)
(−4x + 2) = −2(2x − 1)
Now both parts have 2x − 1 in them.
x(2x − 1) − 2(2x − 1) = 0
8 Same bracket out
2x − 1 is in both parts, so we take it out once.
What is left, x − 2, goes in the other bracket.
ab + cb = b(a + c)
So we get (2x − 1)(x − 2).
(2x − 1)(x − 2) = 0
9 One bracket is 0
Two numbers multiply to 0.
That only happens if one of them is 0.
So one bracket must be 0.
A × B = 0 → A = 0 or B = 0
We make each bracket 0.
Two small equations.
2x − 1 = 0 or x − 2 = 0
In 2x − 1 = 0, we add 1 on both sides.
In x − 2 = 0, we add 2 on both sides.
2x − 1 + 1 = 0 + 1 or x − 2 + 2 = 0 + 2
Now we have 2x = 1 or x = 2.
2x = 1 or x = 2
To remove times 2, we divide both sides by 2.
So x = 1/2 or x = 2.
x = 12 or x = 2
10 Write all the zeros
So the zeros are −1, 1/2 and 2.
−1 from the test, the others from the quadratic.
zeros : x = −1 or x = 12 or x = 2
11 Check
Let's check: put 1/2 into f.
We get 0.
x = 12: f = 2(12)3 − 3(12)2 − 3(12) + 2
= 14 − 34 − 32 + 2 = 0
And put 2 into f.
We get 0 too.
x = 2: f = 2(2)3 − 3(2)2 − 3(2) + 2
= 16 − 12 − 6 + 2 = 0
Let's look back.
First: list the candidates.
Then: test the candidates.
Then: divide by x + 1.
Then: solve the quadratic.
Then: find two numbers.
Then: split the middle.
Then: group in pairs.
Then: same bracket out.
Then: one bracket is 0.
Then: write all the zeros.
Last: check.
Answer zeros : x = −1 or x = 12 or x = 2
▶ Watch it solved
