https://file.aiquickdraw.com/mathbot/file/1729477929112iaz91svx.jpg

The given expression is:

\[
x^{\frac{2}{5}} - 5x^{\frac{1}{5}} - 14
\]

We are asked to factor this expression using **\( u \)-substitution**. Let's go through the process step by step.

### Step 1: \( u \)-Substitution
We can simplify the expression by setting:

\[
u = x^{\frac{1}{5}}
\]

Thus, \( u^2 = x^{\frac{2}{5}} \). Now substitute \( u \) into the expression:

\[
u^2 - 5u - 14
\]

### Step 2: Factor the quadratic
We now need to factor \( u^2 - 5u - 14 \). First, we check if it can be factored easily, but this quadratic does not factor nicely with integers. Therefore, we'll use the **quadratic formula**:

\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For \( u^2 - 5u - 14 \), \( a = 1 \), \( b = -5 \), and \( c = -14 \). Plugging these values into the quadratic formula:

\[
u = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-14)}}{2(1)}
\]
\[
u = \frac{5 \pm \sqrt{25 + 56}}{2}
\]
\[
u = \frac{5 \pm \sqrt{81}}{2}
\]
\[
u = \frac{5 \pm 9}{2}
\]

So, the two solutions for \( u \) are:

\[
u = \frac{5 + 9}{2} = 7 \quad \text{and} \quad u = \frac{5 - 9}{2} = -2
\]

### Step 3: Write the factorization
Thus, the quadratic expression factors as:

\[
(u - 7)(u + 2)
\]

### Step 4: Substitute back \( u = x^{\frac{1}{5}} \)
Now, replace \( u \) with \( x^{\frac{1}{5}} \):

\[
(x^{\frac{1}{5}} - 7)(x^{\frac{1}{5}} + 2)
\]

This is the factored form of the given expression.

\[
x^{\frac{2}{5}} - 5x^{\frac{1}{5}} - 14 = (x^{\frac{1}{5}} - 7)(x^{\frac{1}{5}} + 2)
\]

### Final Answer:
\[
(x^{\frac{1}{5}} - 7)(x^{\frac{1}{5}} + 2)
\]

Would you like to see further details on any step, or have more questions?

Here are 5 relative questions for practice:
1. Can you factor \( x^{\frac{2}{3}} - 4x^{\frac{1}{3}} - 12 \) using \( u \)-substitution?
2. How would you solve \( x^{\frac{4}{5}} + 6x^{\frac{2}{5}} - 7 = 0 \)?
3. Can every quadratic expression be factored without using the quadratic formula?
4. What is the relationship between exponents and roots in the context of solving these equations?
5. How would the factoring change if the middle term were positive, i.e., \( x^{\frac{2}{5}} + 5x^{\frac{1}{5}} - 14 \)?

**Tip**: When factoring complicated expressions, always look for substitution opportunities to simplify the problem first!

Factor x^(2/5) - 5x^(1/5) - 14 using u-substitution.

Learn how to factor the expression x^(2/5) - 5x^(1/5) - 14 using u-substitution. This problem involves key algebraic concepts such as substitution and the quadratic formula, and is suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation