We are given the following expression to factor completely:

$\frac{1}{2} x^{-1/2} (3x + 4)^{1/2} + \frac{3}{2} x^{1/2} (3x + 4)^{-1/2}$

Let's factor step by step.

Step 1: Factor out the common terms

Both terms contain a factor of $\frac{1}{2}$, so we can factor that out:

$\frac{1}{2} \left[ x^{-1/2} (3x + 4)^{1/2} + 3 x^{1/2} (3x + 4)^{-1/2} \right]$

Step 2: Simplify the powers of $x$ and $(3x + 4)$

Notice that $x^{-1/2}$ and $x^{1/2}$ have different powers of $x$, as do $(3x + 4)^{1/2}$ and $(3x + 4)^{-1/2}$. Let's look for common factors:

• For $x$, the lowest power is $x^{-1/2}$, so we can factor out $x^{-1/2}$.
• For $(3x + 4)$, the lowest power is $(3x + 4)^{-1/2}$, so we can factor out $(3x + 4)^{-1/2}$.

Factoring these out, we get:

$\frac{1}{2} x^{-1/2} (3x + 4)^{-1/2} \left[ (3x + 4) + 3x \right]$

Step 3: Simplify the remaining terms inside the brackets

Inside the brackets, simplify the expression:

$(3x + 4) + 3x = 6x + 4$

Thus, the expression becomes:

$\frac{1}{2} x^{-1/2} (3x + 4)^{-1/2} (6x + 4)$

Step 4: Factor the remaining terms

Notice that $6x + 4$ can be factored as $2(3x + 2)$:

$\frac{1}{2} x^{-1/2} (3x + 4)^{-1/2} \cdot 2(3x + 2)$

The factor of 2 cancels with the $\frac{1}{2}$, leaving:

$x^{-1/2} (3x + 4)^{-1/2} (3x + 2)$

Final Factored Form:

The fully factored form of the given expression is:

$x^{-1/2} (3x + 4)^{-1/2} (3x + 2)$

Would you like further details or have any questions?

5 Related Questions:

1. How would the factoring change if different powers of $x$ were involved?
2. Can this expression be simplified further if $x$ takes a specific value?
3. What happens to the expression when $x = 0$?
4. How would you factor a similar expression involving higher powers of $x$?
5. How does factoring help in solving equations with these terms?

Tip:

Always look for the lowest power of common terms to factor them out efficiently, which simplifies complex expressions quickly.