Simplify Radical Products: A Math Breakdown

Hey guys! Today we're diving into the sometimes tricky, but totally doable, world of simplifying radical expressions. Specifically, we're going to tackle this beast: $(\sqrt{10 x^4}-x \sqrt{5 x^2})(2 \sqrt{15 x^4}+\sqrt{3 x^3})$, assuming that $x \geq 0$. We'll break it down step-by-step, making sure we get to the bottom of what this simplified product really is. Ready to get your math on?

Understanding the Basics: Radicals and Simplification

Before we jump into the main problem, let's quickly recap what we're dealing with. A radical expression is basically any expression that contains a root, most commonly a square root (that little $\sqrt{}$ symbol). Simplifying these expressions often involves a few key principles. One of the most important is understanding how to simplify terms inside the radical. For example, $\sqrt{x^2}$ simplifies to $x$ (when $x \geq 0$), and $\sqrt{x^3}$ can be written as $x\sqrt{x}$. The goal is to pull out as much as possible from under the radical sign. We also need to remember how to multiply radical expressions. When multiplying, we multiply the numbers outside the radicals together and the numbers inside the radicals together. For example, $\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$.

Another crucial part of simplifying radicals is dealing with variables. Since we're given the condition that $x \geq 0$, our lives are a bit easier. This means we don't have to worry about absolute values when we take square roots of variables raised to an even power. For instance, $\sqrt{x^4}$ is simply $x^2$ because $(x^2)^2 = x^4$. Similarly, $\sqrt{x^2}$ is just $x$. These simplifications are going to be super handy as we work through the problem. When we have terms like $\sqrt{10x^4}$, we can break it down: $\sqrt{10} \times \sqrt{x^4} = \sqrt{10} \times x^2$. This ability to separate and simplify components of the radical expression is key to making the overall problem less intimidating.

We'll also be using the distributive property, often remembered by the acronym FOIL (First, Outer, Inner, Last) when multiplying two binomials. In our case, we have two binomials, each with two terms involving radicals. So, we'll multiply each term in the first set of parentheses by each term in the second set of parentheses. This means we'll have four multiplication steps to perform. Each of these steps will likely involve simplifying radicals and combining terms where possible. It's essential to keep track of each multiplication and simplification to avoid errors. The process might look daunting at first, but by taking it one step at a time and applying these fundamental rules, we can definitely conquer it. Let's get ready to apply these concepts to our specific expression.

Step 1: Simplifying Terms Within the Radicals

Alright, let's get down to business with our expression: $(\sqrt{10 x^4}-x \sqrt{5 x^2})(2 \sqrt{15 x^4}+\sqrt{3 x^3})$. The very first thing we should do is simplify any terms inside the radicals that we can. This makes the subsequent multiplication much smoother. Remember, $x \geq 0$.

Let's look at the first parenthesis: $(\sqrt{10 x^4}-x \sqrt{5 x^2})$.

• For the first term, $\sqrt{10 x^4}$: We can rewrite $x^4$ as $(x^2)^2$. So, $\sqrt{10 x^4} = \sqrt{10} \times \sqrt{x^4} = \sqrt{10} \times x^2$. Since $x^2$ is already simplified, we leave it as $x^2 \sqrt{10}$.
• For the second term, $x \sqrt{5 x^2}$: We know $\sqrt{x^2} = x$ (since $x \geq 0$). So, $x \sqrt{5 x^2} = x \times \sqrt{5} \times \sqrt{x^2} = x \times \sqrt{5} \times x = x^2 \sqrt{5}$.

So, the first parenthesis simplifies to: $(x^2 \sqrt{10} - x^2 \sqrt{5})$. We can even factor out $x^2$ here if we want: $x^2(\sqrt{10} - \sqrt{5})$. But for now, let's keep it as is to make the multiplication clearer.

Now, let's look at the second parenthesis: $(2 \sqrt{15 x^4}+\sqrt{3 x^3})$.

• For the first term, $2 \sqrt{15 x^4}$: Similar to before, $\sqrt{x^4} = x^2$. So, $2 \sqrt{15 x^4} = 2 \times \sqrt{15} \times \sqrt{x^4} = 2 \times \sqrt{15} \times x^2 = 2x^2 \sqrt{15}$.
• For the second term, $\sqrt{3 x^3}$: We can rewrite $x^3$ as $x^2 \times x$. So, $\sqrt{3 x^3} = \sqrt{3} \times \sqrt{x^3} = \sqrt{3} \times \sqrt{x^2 \times x} = \sqrt{3} \times \sqrt{x^2} \times \sqrt{x} = \sqrt{3} \times x \times \sqrt{x} = x \sqrt{3x}$.

So, the second parenthesis simplifies to: $(2x^2 \sqrt{15} + x \sqrt{3x})$.

Now our expression looks much cleaner: $(x^2 \sqrt{10} - x^2 \sqrt{5})(2x^2 \sqrt{15} + x \sqrt{3x})$

This initial simplification is super important, guys. It takes those messy exponents and variables under the radical and pulls them out, making the next steps much more manageable. Don't skip this part!

Step 2: Applying the Distributive Property (FOIL)

With our simplified terms, we're ready to multiply. We'll use the distributive property, essentially FOILing our two binomials: $(x^2 \sqrt{10} - x^2 \sqrt{5})(2x^2 \sqrt{15} + x \sqrt{3x})$.

Let's do it term by term:

1. First (F): Multiply the first term of each parenthesis. $(x^2 \sqrt{10}) \times (2x^2 \sqrt{15})$ $= (x^2 \times 2x^2) \times (\sqrt{10} \times \sqrt{15})$ $= 2x^4 \times \sqrt{10 \times 15}$ $= 2x^4 \sqrt{150}$ Now, we need to simplify $\sqrt{150}$. $150 = 25 \times 6$. So, $\sqrt{150} = \sqrt{25 \times 6} = \sqrt{25} \times \sqrt{6} = 5 \sqrt{6}$. Our first term becomes: $2x^4 \times (5 \sqrt{6}) = 10x^4 \sqrt{6}$.

2. Outer (O): Multiply the outer terms. $(x^2 \sqrt{10}) \times (x \sqrt{3x})$ $= (x^2 \times x) \times (\sqrt{10} \times \sqrt{3x})$ $= x^3 \times \sqrt{10 \times 3x}$ $= x^3 \sqrt{30x}$. This term is already simplified as much as possible.

3. Inner (I): Multiply the inner terms. $(-x^2 \sqrt{5}) \times (2x^2 \sqrt{15})$ $= (-x^2 \times 2x^2) \times (\sqrt{5} \times \sqrt{15})$ $= -2x^4 \times \sqrt{5 \times 15}$ $= -2x^4 \sqrt{75}$ Now, simplify $\sqrt{75}$. $75 = 25 \times 3$. So, $\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25} \times \sqrt{3} = 5 \sqrt{3}$. Our inner term becomes: $-2x^4 \times (5 \sqrt{3}) = -10x^4 \sqrt{3}$.

4. Last (L): Multiply the last terms. $(-x^2 \sqrt{5}) \times (x \sqrt{3x})$ $= (-x^2 \times x) \times (\sqrt{5} \times \sqrt{3x})$ $= -x^3 \times \sqrt{5 \times 3x}$ $= -x^3 \sqrt{15x}$. This term is also already simplified.

So far, we have: $10x^4 \sqrt{6} + x^3 \sqrt{30x} - 10x^4 \sqrt{3} - x^3 \sqrt{15x}$.

This FOIL process can feel a bit long, but each step is just applying the same multiplication rules. If you do it carefully, you'll get there!

Step 3: Combining Like Terms (If Possible)

Now that we've done all the multiplication, we need to see if we can combine any of the resulting terms. Like terms in this context would need to have the same variable part (with the same exponents) and the same radical part. Let's look at our result from Step 2:

$10x^4 \sqrt{6} + x^3 \sqrt{30x} - 10x^4 \sqrt{3} - x^3 \sqrt{15x}$

Let's examine each term:

• $10x^4 \sqrt{6}$: Variable part is $x^4$, radical part is $\sqrt{6}$.
• $x^3 \sqrt{30x}$: Variable part is $x^3$, radical part is $\sqrt{30x}$.
• $-10x^4 \sqrt{3}$: Variable part is $x^4$, radical part is $\sqrt{3}$.
• $-x^3 \sqrt{15x}$: Variable part is $x^3$, radical part is $\sqrt{15x}$.

Comparing these terms, we see:

• The terms with $x^4$ are $10x^4 \sqrt{6}$ and $-10x^4 \sqrt{3}$. The radical parts ($\sqrt{6}$ and $\sqrt{3}$) are different, so we cannot combine these.
• The terms with $x^3$ are $x^3 \sqrt{30x}$ and $-x^3 \sqrt{15x}$. The radical parts ($\sqrt{30x}$ and $\sqrt{15x}$) are also different, so we cannot combine these either.

Since there are no like terms to combine, the expression we arrived at after the multiplication is our final simplified form.

Final Answer and Verification

After going through all the steps – simplifying inside the radicals, applying the distributive property (FOIL), and checking for like terms – our fully simplified product is:

$10x^4 \sqrt{6} + x^3 \sqrt{30x} - 10x^4 \sqrt{3} - x^3 \sqrt{15x}$

Let's compare this to the options given. The options provided were:

A. $10 x^4 \sqrt{6}+x^3 \sqrt{30 x}-10 x^4 \sqrt{3}+x^2 \sqrt{15 x}$ B. $10 x^4 \sqrt{6}+x^3 \sqrt{30 x}-x^4 \sqrt{75}+x^2 \sqrt{15 x}$

Our result is $10x^4 \sqrt{6} + x^3 \sqrt{30x} - 10x^4 \sqrt{3} - x^3 \sqrt{15x}$.

Looking at option A, the last term is $x^2 \sqrt{15x}$, whereas ours is $-x^3 \sqrt{15x}$. So, option A is incorrect.

Looking at option B, the third term is $-x^4 \sqrt{75}$. We simplified $\sqrt{75}$ to $5\sqrt{3}$, so $-x^4 \sqrt{75}$ would be $-x^4 (5\sqrt{3}) = -5x^4\sqrt{3}$. Our third term is $-10x^4 \sqrt{3}$. Also, the fourth term in B is $x^2 \sqrt{15x}$, while ours is $-x^3 \sqrt{15x}$. So, option B is also incorrect.

Wait a minute! It seems there might be a discrepancy between our derived answer and the provided options. Let's re-check our work very carefully, especially the multiplication steps and simplifications.

Revisiting Step 1: First parenthesis: $(\sqrt{10 x^4}-x \sqrt{5 x^2}) = (x^2\sqrt{10} - x^2\sqrt{5})$. This is correct. Second parenthesis: $(2 \sqrt{15 x^4}+\sqrt{3 x^3}) = (2x^2\sqrt{15} + x\sqrt{3x})$. This is also correct.

Revisiting Step 2 (FOIL):

1. F: $(x^2 \sqrt{10}) \times (2x^2 \sqrt{15}) = 2x^4 \sqrt{150} = 2x^4 (5\sqrt{6}) = 10x^4 \sqrt{6}$. Correct.
2. O: $(x^2 \sqrt{10}) \times (x \sqrt{3x}) = x^3 \sqrt{30x}$. Correct.
3. I: $(-x^2 \sqrt{5}) \times (2x^2 \sqrt{15}) = -2x^4 \sqrt{75} = -2x^4 (5\sqrt{3}) = -10x^4 \sqrt{3}$. Correct.
4. L: $(-x^2 \sqrt{5}) \times (x \sqrt{3x}) = -x^3 \sqrt{15x}$. Correct.

Combining these gives: $10x^4 \sqrt{6} + x^3 \sqrt{30x} - 10x^4 \sqrt{3} - x^3 \sqrt{15x}$.

It appears that our calculation is consistent. Let's re-examine the options provided, considering potential typos or alternative ways of writing the terms.

Option A: $10 x^4 \sqrt{6}+x^3 \sqrt{30 x}-10 x^4 \sqrt{3}+x^2 \sqrt{15 x}$ The last term $x^2 \sqrt{15 x}$ seems incorrect based on our multiplication of $(-x^2\sqrt{5}) \times (x\sqrt{3x})$ which should result in a term with $x^3$ and $\sqrt{15x}$.

Option B: $10 x^4 \sqrt{6}+x^3 \sqrt{30 x}-x^4 \sqrt{75}+x^2 \sqrt{15 x}$ Here, the third term is $-x^4 \sqrt{75}$. If we don't simplify $\sqrt{75}$, this matches our intermediate calculation before simplifying $\sqrt{75}$. However, the standard practice is to fully simplify. If we don't simplify $\sqrt{75}$ in the calculation, the third term is $-2x^4\sqrt{75}$. Option B has $-x^4\sqrt{75}$, which is also not matching. It seems option B might have intended to have $-2x^4\sqrt{75}$ and perhaps a typo in the last term too. Or maybe the entire calculation was done slightly differently.

Let's consider if there's a way the problem could lead to option A. The first three terms of option A match our derived expression exactly: $10x^4 \sqrt{6} + x^3 \sqrt{30x} - 10x^4 \sqrt{3}$. The only difference is the last term.

Could the original expression have been different? Assuming our initial simplification and multiplication are correct, the only way to get $x^2 \sqrt{15x}$ as the last term instead of $-x^3 \sqrt{15x}$ would imply a mistake in the initial setup or the multiplication.

Let's assume, for a moment, that option A is correct and work backwards or see if there's a subtle mistake. The last term in option A is $+x^2 \sqrt{15x}$. Our calculation for the