Question

assuming ( x ) and ( y ) are both positive, write the following expression in simplest radical form.

( 4x^3y^2sqrt{125x^3y^4} )

Explanation:

Step1: Simplify the radical inside

First, factor \(125x^{3}y^{4}\) into perfect squares and remaining factors. We know that \(125 = 5^{3}=5^{2}\times5\), \(x^{3}=x^{2}\times x\), and \(y^{4}=(y^{2})^{2}\). So, \(\sqrt{125x^{3}y^{4}}=\sqrt{5^{2}\times5\times x^{2}\times x\times(y^{2})^{2}}\). Using the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (for \(a,b\geq0\)) and \(\sqrt{a^{2}} = a\) (for \(a\geq0\)), we get:
\(\sqrt{5^{2}}\cdot\sqrt{x^{2}}\cdot\sqrt{(y^{2})^{2}}\cdot\sqrt{5x}=5\cdot x\cdot y^{2}\cdot\sqrt{5x}\)
So, \(\sqrt{125x^{3}y^{4}} = 5xy^{2}\sqrt{5x}\)

Step2: Multiply with the outside term

Now, multiply \(4x^{3}y^{2}\) with \(5xy^{2}\sqrt{5x}\). First, multiply the coefficients and the variables separately.
- Coefficients: \(4\times5 = 20\)
- For \(x\) terms: \(x^{3}\times x=x^{3 + 1}=x^{4}\) (using \(a^{m}\cdot a^{n}=a^{m + n}\))
- For \(y\) terms: \(y^{2}\times y^{2}=y^{2+2}=y^{4}\)
- The radical term remains \(\sqrt{5x}\)

So, multiplying them together: \(20x^{4}y^{4}\sqrt{5x}\)

Wait, but let's check the \(x\) term again. Wait, in the radical, we had \(x^{3}=x^{2}\times x\), so when we took the square root, we had \(x\sqrt{x}\)? Wait no, \(\sqrt{x^{2}\times x}=x\sqrt{x}\), correct. Then the outside term is \(x^{3}\), so \(x^{3}\times x=x^{4}\), correct. Wait, but let's re - express the original radical simplification:

\(\sqrt{125x^{3}y^{4}}=\sqrt{5^{3}\times x^{3}\times y^{4}}=\sqrt{5^{2}\times5\times x^{2}\times x\times y^{4}} = 5y^{2}x\sqrt{5x}\) (since \(\sqrt{5^{2}} = 5\), \(\sqrt{x^{2}}=x\), \(\sqrt{y^{4}}=y^{2}\))

Then, the outside term is \(4x^{3}y^{2}\). So, multiplying \(4x^{3}y^{2}\) and \(5x y^{2}\sqrt{5x}\):

- Coefficients: \(4\times5 = 20\)
- \(x\) terms: \(x^{3}\times x=x^{4}\)
- \(y\) terms: \(y^{2}\times y^{2}=y^{4}\)
- Radical: \(\sqrt{5x}\)

But wait, let's check the exponent of \(x\) in the radical. The original \(x\) in the radical is \(x^{3}\), so when we simplify \(\sqrt{x^{3}}=\sqrt{x^{2}\times x}=x\sqrt{x}\), so the \(x\) from the radical is \(x^{1}\), and the outside \(x\) is \(x^{3}\), so total \(x\) is \(x^{3 + 1}=x^{4}\), correct.

Wait, but let's do the problem again step by step:

Original expression: \(4x^{3}y^{2}\sqrt{125x^{3}y^{4}}\)

Factor \(125 = 5^{3}=5^{2}\times5\), \(x^{3}=x^{2}\times x\), \(y^{4}=(y^{2})^{2}\)

So, \(\sqrt{125x^{3}y^{4}}=\sqrt{5^{2}\times5\times x^{2}\times x\times(y^{2})^{2}}\)

Using \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\), we have:

\(\sqrt{5^{2}}\cdot\sqrt{x^{2}}\cdot\sqrt{(y^{2})^{2}}\cdot\sqrt{5x}=5\cdot x\cdot y^{2}\cdot\sqrt{5x}\)

Now, multiply by \(4x^{3}y^{2}\):

\(4x^{3}y^{2}\times5xy^{2}\sqrt{5x}\)

Multiply the coefficients: \(4\times5 = 20\)

Multiply the \(x\) terms: \(x^{3}\times x=x^{3 + 1}=x^{4}\) (using the rule of exponents \(a^{m}\times a^{n}=a^{m + n}\))

Multiply the \(y\) terms: \(y^{2}\times y^{2}=y^{2+2}=y^{4}\)

So, the expression becomes \(20x^{4}y^{4}\sqrt{5x}\)

But wait, is there a mistake here? Let's check the exponent of \(x\) in the radical. The radical has \(x^{3}\), so when we take the square root, we get \(x\sqrt{x}\), so the \(x\) from the radical is \(x^{1}\), and the outside \(x\) is \(x^{3}\), so \(x^{3}\times x^{1}=x^{4}\), correct. The \(y\) terms: radical has \(y^{4}\), square root is \(y^{2}\), outside has \(y^{2}\), so \(y^{2}\times y^{2}=y^{4}\), correct. Coefficient: \(4\times5 = 20\), correct. Radical term: \(\sqrt{5x}\), correct.

Wait, but let's re - express the original expression:

\(4x^{3}y^{2}\sqrt{125x^{3}y^{4}}=4x^{3}y^{2}\sqrt{5^{3}x^{3}y^{4}}\)

\(=4x^{3}y^{2}\times5^{\frac{3}{2}}x^{\frac{3}{2}}y^{2}\) (using \(\sqrt{a^{m}}=a^{\frac{m}{2}}\))

\(=4\times5^{\frac{3}{2}}x^{3+\frac{3}{2}}y^{2 + 2}\)

\(=4\times5\times5^{\frac{1}{2}}x^{\frac{6 + 3}{2}}y^{4}\)

\(=20\sqrt{5}x^{\frac{9}{2}}y^{4}\)

\(=20x^{4}y^{4}\sqrt{5x}\) (since \(x^{\frac{9}{2}}=x^{4+\frac{1}{2}}=x^{4}\sqrt{x}\))

Yes, that's consistent. So the simplified form is \(20x^{4}y^{4}\sqrt{5x}\)

Answer:

\(20x^{4}y^{4}\sqrt{5x}\)