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Find all zeros of f(x) = x^3 + 3x^2 - 4x - 12

Step by step, with the reason for every step · Finding all zeros · Algebra 2

Step by step

The question

Question: find all the zeros of f(x) = x³ + 3x² − 4x − 12.

f(x) = x3 + 3x2 − 4x − 12

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 −12.

Its factors are 1, 2, 3, 4, 6, 12, each with plus and minus.

factors of 12 : ±1, ±2, ±3, ±4, ±6, ±12

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, ±4, ±6, ±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) = 13 + 3(1)2 − 4(1) − 12

We get −12, not 0.

So 1 is not a zero.

= 1 + 3 − 4 − 12 = −12 ✗

Next −1: put it in place of every x.

f(−1) = (−1)3 + 3(−1)2 − 4(−1) − 12

We get −6, not 0.

So −1 is not a zero.

= −1 + 3 + 4 − 12 = −6 ✗

Next 2: put it in place of every x.

f(2) = 23 + 3(2)2 − 4(2) − 12

We get 0.

So x = 2 is a zero!

= 8 + 12 − 8 − 12 = 0 zero!

3 Divide by x − 2

x = 2 is a zero, so (x − 2) is a factor.

We divide it out, with synthetic division.

x − 2 is 0 when x is 2.

So we use 2.

Then we write just the numbers in front: 1, 3, −4, −12.

The first number comes straight down: 1.

1 times 2 is 2.

Write it under the next number.

Add: 3 + 2 is 5.

5 times 2 is 10.

Write it under the next number.

Add: −4 + 10 is 6.

6 times 2 is 12.

Write it under the next number.

Add: −12 + 12 is 0.

The last number is the remainder: 0.

The other numbers are the answer's numbers: 1, 5, 6.

The answer starts one power lower than x³.

So f(x) = (x − 2)(x² + 5x + 6).

The other zeros come from x² + 5x + 6 = 0.

f(x) = (x − 2)(x² + 5x + 6)

4 Solve the quadratic

Now x² + 5x + 6 = 0, with the quadratic lesson.

x2 + 5x + 6 = 0

5 Factor

We need two numbers.

They multiply to 6, and they add to 5.

? × ? = 6
? + ? = 5

They multiply to a plus number, so the two numbers have the same sign.

They add to a plus number, so both are plus.

× is + → same signs: both +

We list the pairs that make 6, smallest first.

We add each pair.

We need 5.

2 and 3 give 5.

Yes!

1 × 6: 1 + 6 = 7
2 × 3: 2 + 3 = 5 ✓

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 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 2, −2 and −3.

2 from the test, the others from the quadratic.

zeros : x = 2 or x = −2 or x = −3

8 Check

Let's check: put −2 into f.

We get 0.

x = −2: f = (−2)3 + 3(−2)2 − 4(−2) − 12
= −8 + 12 + 8 − 12 = 0

And put −3 into f.

We get 0 too.

x = −3: f = (−3)3 + 3(−3)2 − 4(−3) − 12
= −27 + 27 + 12 − 12 = 0

Let's look back.

First: list the candidates.

Then: test the candidates.

Then: divide by x − 2.

Then: solve the quadratic.

Then: factor.

Then: one bracket is 0.

Then: write all the zeros.

Last: check.

Answer
zeros : x = 2 or x = −2 or x = −3

Your turn

Try this one yourself, then check:

find all zeros of f(x) = x3 + 3x2 − x − 3
Show the answer
x = 1 or x = −1 or x = −3

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