Step by step
The question
Question: find all the zeros of f(x) = 2x³ − 5x² − x + 6.
A zero of f is an x that makes f(x) equal 0.
A power 3 has up to three zeros.
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.
The last number is 6.
Its factors are 1, 2, 3, 6, each with plus and minus.
The first number is 2.
Its factors are 1, 2.
Careful!
Don't forget the negative ones.
−1, −2 and the others can be zeros too.
Each top factor over each bottom factor.
These are all the candidates.
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.
Start with 1: put it in place of every x.
We get 2, not 0.
So 1 is not a zero.
Next −1: put it in place of every x.
We get 0.
So x = −1 is a 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, −5, −1, 6.
The first number comes straight down: 2.
2 times −1 is −2.
Write it under the next number.
Add: −5 + (−2) is −7.
−7 times −1 is 7.
Write it under the next number.
Add: −1 + 7 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: 2, −7, 6.
The answer starts one power lower than x³.
So f(x) = (x + 1)(2x² − 7x + 6).
The other zeros come from 2x² − 7x + 6 = 0.
4 Solve the quadratic
Now 2x² − 7x + 6 = 0, with the quadratic lesson.
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 6: that is 12.
We need two numbers that multiply to 12 and add to −7.
They multiply to a plus number, so the two numbers have the same sign.
They add to a minus number, so both are minus.
We list the pairs that make 12, smallest first.
We add each pair.
We need 7.
3 and 4 give 7.
Yes!
With the signs: −3 and −4.
6 Split the middle
We write −7x as −3x − 4x.
It is the same amount.
7 Group in pairs
We put the first two together, and the last two together.
From the first pair we take out x.
From the second pair we take out −2.
Both leave the same bracket.
Now both parts have 2x − 3 in them.
8 Same bracket out
2x − 3 is in both parts, so we take it out once.
What is left, x − 2, goes in the other bracket.
So we get (2x − 3)(x − 2).
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.
We make each bracket 0.
Two small equations.
In 2x − 3 = 0, we add 3 on both sides.
In x − 2 = 0, we add 2 on both sides.
Now we have 2x = 3 or x = 2.
To remove times 2, we divide both sides by 2.
So x = 3/2 or x = 2.
10 Write all the zeros
So the zeros are −1, 3/2 and 2.
−1 from the test, the others from the quadratic.
11 Check
Let's check: put 3/2 into f.
We get 0.
And put 2 into f.
We get 0 too.
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.