Question

what is the following simplified product? assume x ≥ 0.
\\(\left(\sqrt{10x^4} - x\sqrt{5x^2}\
ight)\left(2\sqrt{15x^4} + \sqrt{3x^3}\
ight)\\)
\\(\circ\\) \\(10x^4\sqrt{6} + x^3\sqrt{30x} - 10x^4\sqrt{3} + x^2\sqrt{15x}\\)
\\(\circ\\) \\(10x^4\sqrt{6} + x^3\sqrt{30x} - x^4\sqrt{75} + x^2\sqrt{15}\\)
\\(\circ\\) \\(10x^4\sqrt{6} + x^3\sqrt{30x} - 10x^4\sqrt{3} - x^2\sqrt{15}\\)
\\(\circ\\) \\(10x^4\sqrt{6} + x^3\sqrt{30x} - 10x^4\sqrt{3} - x^3\sqrt{15x}\\)

Explanation:

Step1: Simplify each radical term

First, simplify \(\sqrt{10x^4}\). Since \(x\geq0\), \(\sqrt{10x^4}=x^2\sqrt{10}\)? Wait, no, wait: \(\sqrt{x^4}=x^2\), so \(\sqrt{10x^4}=x^2\sqrt{10}\)? Wait, no, the first term is \(\sqrt{10x^4}\), let's do that again. Wait, actually, let's simplify each part:

First factor: \(\sqrt{10x^4} - x\sqrt{5x^2}\)

Simplify \(\sqrt{10x^4}\): \(x^4 = (x^2)^2\), so \(\sqrt{10x^4}=x^2\sqrt{10}\)? Wait, no, \(x^4=(x^2)^2\), so \(\sqrt{x^4}=x^2\), so \(\sqrt{10x^4}=x^2\sqrt{10}\)? Wait, no, wait the problem is \(\sqrt{10x^4}\), so that's \(x^2\sqrt{10}\)? Wait, no, maybe I made a mistake. Wait, let's check the second term: \(x\sqrt{5x^2}\). Since \(x\geq0\), \(\sqrt{x^2}=x\), so \(x\sqrt{5x^2}=x \cdot x\sqrt{5}=x^2\sqrt{5}\).

Wait, wait, the first factor is \(\sqrt{10x^4} - x\sqrt{5x^2}\). Let's simplify each term:

\(\sqrt{10x^4} = \sqrt{10} \cdot \sqrt{x^4} = \sqrt{10} \cdot x^2 = x^2\sqrt{10}\)? Wait, no, wait the problem is \(\sqrt{10x^4}\), but maybe I misread. Wait, no, the first term is \(\sqrt{10x^4}\), let's compute that: \(x^4 = (x^2)^2\), so \(\sqrt{10x^4} = x^2\sqrt{10}\)? Wait, no, wait the second factor has \(\sqrt{15x^4}\), which is \(x^2\sqrt{15}\), and \(\sqrt{3x^3}\) is \(x\sqrt{3x}\) (since \(x^3 = x^2 \cdot x\), so \(\sqrt{x^3}=x\sqrt{x}\) for \(x\geq0\)).

Wait, maybe I should start by simplifying each radical in the two binomials.

First binomial: \(\sqrt{10x^4} - x\sqrt{5x^2}\)

Simplify \(\sqrt{10x^4}\): \(x^4 = (x^2)^2\), so \(\sqrt{10x^4} = x^2\sqrt{10}\)? Wait, no, wait 10x^4: x^4 is (x^2)^2, so sqrt(10x^4) = x^2 sqrt(10). Wait, but maybe the problem is \(\sqrt{10x^4}\) is actually \(\sqrt{10} \cdot x^2\), but let's check the second term: \(x\sqrt{5x^2}\). Since x >=0, sqrt(x^2)=x, so this becomes x * x * sqrt(5) = x^2 sqrt(5). So first binomial: \(x^2\sqrt{10} - x^2\sqrt{5}\)? Wait, no, that can't be, because the second factor has \(\sqrt{15x^4}\) which is \(x^2\sqrt{15}\) and \(\sqrt{3x^3}\) which is x sqrt(3x). Wait, maybe I misread the original problem. Let me check again:

Original problem: \((\sqrt{10x^4} - x\sqrt{5x^2})(2\sqrt{15x^4} + \sqrt{3x^3})\)

Wait, maybe the first term is \(\sqrt{10x^4}\) which is \(x^2\sqrt{10}\), but let's compute 10x^4: x^4 is (x^2)^2, so sqrt(10x^4) = x^2 sqrt(10). Then the second term in the first binomial: x sqrt(5x^2) = x * x sqrt(5) = x^2 sqrt(5) (since x >=0, sqrt(x^2)=x). So first binomial: \(x^2\sqrt{10} - x^2\sqrt{5}\)? Wait, but that seems off. Wait, maybe the first term is \(\sqrt{10x^4}\) is actually \(\sqrt{10} \cdot x^2\), but let's check the second factor: 2 sqrt(15x^4) = 2 x^2 sqrt(15), and sqrt(3x^3) = x sqrt(3x) (since x^3 = x^2 * x, so sqrt(x^3)=x sqrt(x) for x >=0).

Wait, maybe I made a mistake in the first term. Let's re-express:

Wait, 10x^4: x^4 is (x^2)^2, so sqrt(10x^4) = x^2 sqrt(10). Then 5x^2: sqrt(5x^2) = x sqrt(5) (since x >=0), so x sqrt(5x^2) = x * x sqrt(5) = x^2 sqrt(5). So first binomial: \(x^2\sqrt{10} - x^2\sqrt{5}\).

Second binomial: 2 sqrt(15x^4) + sqrt(3x^3) = 2 x^2 sqrt(15) + x sqrt(3x) (since x^3 = x^2 * x, so sqrt(x^3)=x sqrt(x)).

Now, we need to multiply these two binomials: (a - b)(c + d) = ac + ad - bc - bd, where a = x^2 sqrt(10), b = x^2 sqrt(5), c = 2 x^2 sqrt(15), d = x sqrt(3x).

Compute ac: (x^2 sqrt(10))(2 x^2 sqrt(15)) = 2 x^4 sqrt(10*15) = 2 x^4 sqrt(150) = 2 x^4 sqrt(25*6) = 2 x^4 * 5 sqrt(6) = 10 x^4 sqrt(6).

Compute ad: (x^2 sqrt(10))(x sqrt(3x)) = x^3 sqrt(10*3x) = x^3 sqrt(30x).

Compute bc: (x^2 sqrt(5))(2 x^2 sqrt(15)) = 2 x^4 sqrt(5*15) = 2 x^4 sqrt(75) = 2 x^4 sqrt(25*3) = 2 x^4 * 5 sqrt(3) = 10 x^4 sqrt(3).

Compute bd: (x^2 sqrt(5))(x sqrt(3x)) = x^3 sqrt(5*3x) = x^3 sqrt(15x).

So putting it all together: ac + ad - bc - bd = 10x^4 sqrt(6) + x^3 sqrt(30x) - 10x^4 sqrt(3) - x^3 sqrt(15x).

Wait, let's check the options. The fourth option is: 10x⁴√6 + x³√30x − 10x⁴√3 − x³√15x. Which matches our result.

Wait, let's verify each step again.

First, simplify the first binomial:

\(\sqrt{10x^4}\): since x ≥ 0, \(\sqrt{x^4} = x^2\), so \(\sqrt{10x^4} = x^2\sqrt{10}\). Wait, no, wait 10x^4: x^4 is (x^2)^2, so sqrt(10x^4) = x^2 sqrt(10). Correct.

Then, \(x\sqrt{5x^2}\): since x ≥ 0, \(\sqrt{x^2} = x\), so this becomes x * x * \(\sqrt{5}\) = x²\(\sqrt{5}\). So first binomial: \(x^2\sqrt{10} - x^2\sqrt{5}\). Wait, but that seems different from what I thought earlier. Wait, maybe I misread the first term. Wait, the original problem is \(\sqrt{10x^4}\) or is it \(\sqrt{10x^4}\) as in 10x^4 under the square root? Yes, that's what's written.

Second binomial: \(2\sqrt{15x^4} + \sqrt{3x^3}\)

Simplify \(2\sqrt{15x^4}\): \(\sqrt{15x^4} = x^2\sqrt{15}\) (since x ≥ 0, \(\sqrt{x^4} = x^2\)), so this becomes 2x²\(\sqrt{15}\).

Simplify \(\sqrt{3x^3}\): \(\sqrt{3x^3} = \sqrt{3x \cdot x^2} = x\sqrt{3x}\) (since x ≥ 0, \(\sqrt{x^2} = x\)). So second binomial: \(2x^2\sqrt{15} + x\sqrt{3x}\).

Now, multiply the two binomials: (A - B)(C + D) = AC + AD - BC - BD, where:

A = \(x^2\sqrt{10}\), B = \(x^2\sqrt{5}\), C = \(2x^2\sqrt{15}\), D = \(x\sqrt{3x}\)

Compute AC: \(x^2\sqrt{10} \cdot 2x^2\sqrt{15} = 2x^4\sqrt{10 \cdot 15} = 2x^4\sqrt{150}\)

Simplify \(\sqrt{150}\): 150 = 25 * 6, so \(\sqrt{150} = 5\sqrt{6}\). Thus, AC = 2x^4 * 5\sqrt{6} = 10x^4\sqrt{6}

Compute AD: \(x^2\sqrt{10} \cdot x\sqrt{3x} = x^3\sqrt{10 \cdot 3x} = x^3\sqrt{30x}\)

Compute BC: \(x^2\sqrt{5} \cdot 2x^2\sqrt{15} = 2x^4\sqrt{5 \cdot 15} = 2x^4\sqrt{75}\)

Simplify \(\sqrt{75}\): 75 = 25 * 3, so \(\sqrt{75} = 5\sqrt{3}\). Thus, BC = 2x^4 * 5\sqrt{3} = 10x^4\sqrt{3}

Compute BD: \(x^2\sqrt{5} \cdot x\sqrt{3x} = x^3\sqrt{5 \cdot 3x} = x^3\sqrt{15x}\)

Now, putting it all together:

AC + AD - BC - BD = 10x^4\sqrt{6} + x^3\sqrt{30x} - 10x^4\sqrt{3} - x^3\sqrt{15x}

Which matches the fourth option: 10x⁴√6 + x³√30x − 10x⁴√3 − x³√15x

Step2: Match with the options

Now, let's check the options:

Option 1: 10x⁴√6 + x³√30x − 10x⁴√3 + x²√15x → The last term is x²√15x, but we have -x³√15x, so no.

Option 2: 10x⁴√6 + x³√30x − x⁴√75 + x²√15 → √75 is 5√3, so -x⁴√75 is -5x⁴√3, but we have -10x⁴√3, so no. Also, last term is x²√15, not x³√15x.

Option 3: 10x⁴√6 + x³√30x − 10x⁴√3 − x²√15 → Last term is x²√15, but we have -x³√15x, so no.

Option 4: 10x⁴√6 + x³√30x − 10x⁴√3 − x³√15x → Matches our result.

Answer:

D. \(10x^4 \sqrt{6} + x^3 \sqrt{30x} - 10x^4 \sqrt{3} - x^3 \sqrt{15x}\) (assuming D is the fourth option, as per the order: first option, second, third, fourth. So the fourth option is the correct one.)