Math / Algebra

QuestionQuestion Factor $v^{3}-1$ completely.
Answer Attempt 1 out of 3 $\square$

Studdy Solution

STEP 1

What is this asking? We need to break down $v^3 - 1$ into smaller parts that are multiplied together, just like turning 12 into 2\2\3! Watch out! Remember the difference of cubes formula!

STEP 2

1. Recognize the difference of cubes
2. Apply the formula
3. Check if we can factor further

STEP 3

We're dealing with a difference of cubes here!
We have $v^3$ , which is obviously a cube, and 1, which is also a cube (since $1^3 = 1$ ).
Recognizing this pattern is super important because it unlocks a special factoring formula.

STEP 4

The difference of cubes formula says: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ .
In our case, $a = v$ and $b = 1$ .
Let's plug those values in!

STEP 5

Substituting $a = v$ and $b = 1$ into the formula gives us: $v^3 - 1^3 = (v - 1)(v^2 + v \cdot 1 + 1^2)$ .

STEP 6

Simplifying this gives us $(v - 1)(v^2 + v + 1)$ .
Look at that, nice and neat!

STEP 7

Now, we need to check if $v^2 + v + 1$ can be factored further.
Can we find two numbers that multiply to 1 and add up to 1?
Let's think...

STEP 8

Nope, we can't!
This quadratic is irreducible, meaning it can't be factored using real numbers.
So, we're done!

STEP 9

The completely factored form of $v^3 - 1$ is $(v - 1)(v^2 + v + 1)$ .

Was this helpful?

QuestionQuestion Factor $v^{3}-1$ completely.
Answer Attempt 1 out of 3 $\square$

Studdy Solution

STEP 1

What is this asking? We need to break down $v^3 - 1$ into smaller parts that are multiplied together, just like turning 12 into 2\2\3! Watch out! Remember the difference of cubes formula!

STEP 2

1. Recognize the difference of cubes
2. Apply the formula
3. Check if we can factor further

STEP 3

We're dealing with a difference of cubes here!
We have $v^3$ , which is obviously a cube, and 1, which is also a cube (since $1^3 = 1$ ).
Recognizing this pattern is super important because it unlocks a special factoring formula.

STEP 4

The difference of cubes formula says: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ .
In our case, $a = v$ and $b = 1$ .
Let's plug those values in!

STEP 5

Substituting $a = v$ and $b = 1$ into the formula gives us: $v^3 - 1^3 = (v - 1)(v^2 + v \cdot 1 + 1^2)$ .

STEP 6

Simplifying this gives us $(v - 1)(v^2 + v + 1)$ .
Look at that, nice and neat!

STEP 7

Now, we need to check if $v^2 + v + 1$ can be factored further.
Can we find two numbers that multiply to 1 and add up to 1?
Let's think...

STEP 8

Nope, we can't!
This quadratic is irreducible, meaning it can't be factored using real numbers.
So, we're done!

STEP 9

The completely factored form of $v^3 - 1$ is $(v - 1)(v^2 + v + 1)$ .

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionQuestion Factor v3−1v^{3}-1v3−1 completely.Answer Attempt 1 out of 3 □\square□

Studdy Solution

STEP 1

What is this asking? We need to break down v3−1v^3 - 1v3−1 into smaller parts that are multiplied together, just like turning 12 into 2\*2\*3! Watch out! Remember the difference of cubes formula!

STEP 2

1. Recognize the difference of cubes2. Apply the formula3. Check if we can factor further

STEP 3

We're dealing with a **difference of cubes** here!We have v3v^3v3, which is obviously a cube, and 1, which is also a cube (since 13=11^3 = 113=1).Recognizing this pattern is **super important** because it unlocks a special factoring formula.

STEP 4

The **difference of cubes formula** says: a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2)a3−b3=(a−b)(a2+ab+b2).In our case, a=va = va=v and b=1b = 1b=1.Let's **plug those values in**!

STEP 5

Substituting a=va = va=v and b=1b = 1b=1 into the formula gives us: v3−13=(v−1)(v2+v⋅1+12)v^3 - 1^3 = (v - 1)(v^2 + v \cdot 1 + 1^2)v3−13=(v−1)(v2+v⋅1+12).

STEP 6

**Simplifying** this gives us (v−1)(v2+v+1)(v - 1)(v^2 + v + 1)(v−1)(v2+v+1).Look at that, nice and neat!

STEP 7

Now, we need to check if v2+v+1v^2 + v + 1v2+v+1 can be factored further.Can we find two numbers that multiply to 1 and add up to 1?Let's think...

STEP 8

Nope, we can't!This **quadratic** is **irreducible**, meaning it can't be factored using real numbers.So, we're done!

STEP 9

The completely factored form of v3−1v^3 - 1v3−1 is (v−1)(v2+v+1)(v - 1)(v^2 + v + 1)(v−1)(v2+v+1).

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionQuestion Factor $v^{3}-1$ completely.
Answer Attempt 1 out of 3 $\square$

What is this asking? We need to break down $v^3 - 1$ into smaller parts that are multiplied together, just like turning 12 into 2\2\3! Watch out! Remember the difference of cubes formula!

1. Recognize the difference of cubes
2. Apply the formula
3. Check if we can factor further

We're dealing with a difference of cubes here!
We have $v^3$ , which is obviously a cube, and 1, which is also a cube (since $1^3 = 1$ ).
Recognizing this pattern is super important because it unlocks a special factoring formula.

The difference of cubes formula says: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ .
In our case, $a = v$ and $b = 1$ .
Let's plug those values in!

Substituting $a = v$ and $b = 1$ into the formula gives us: $v^3 - 1^3 = (v - 1)(v^2 + v \cdot 1 + 1^2)$ .

Simplifying this gives us $(v - 1)(v^2 + v + 1)$ .
Look at that, nice and neat!

Now, we need to check if $v^2 + v + 1$ can be factored further.
Can we find two numbers that multiply to 1 and add up to 1?
Let's think...

Nope, we can't!
This quadratic is irreducible, meaning it can't be factored using real numbers.
So, we're done!

The completely factored form of $v^3 - 1$ is $(v - 1)(v^2 + v + 1)$ .