To simplify the expression \(5\sqrt{14z^2}\cdot 4\sqrt{21z^3}\), we will first multiply the coefficients and then the square roots.
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Multiply the coefficients: \[ 5 \cdot 4 = 20 \]
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Multiply the square roots: \[ \sqrt{14z^2} \cdot \sqrt{21z^3} = \sqrt{14 \cdot 21 \cdot z^2 \cdot z^3} = \sqrt{14 \cdot 21 \cdot z^{2+3}} = \sqrt{14 \cdot 21 \cdot z^5} \]
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Calculate \(14 \cdot 21\): \[ 14 \cdot 21 = 294 \] So, we have: \[ \sqrt{14 \cdot 21 \cdot z^5} = \sqrt{294z^5} \]
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Now combine the results: \[ 20 \cdot \sqrt{294z^5} \]
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Simplify \(\sqrt{294z^5}\):
- Factor \(z^5\) as \(z^4 \cdot z\), giving: \[ \sqrt{294z^5} = \sqrt{294} \cdot \sqrt{z^4} \cdot \sqrt{z} = \sqrt{294} \cdot z^2 \cdot \sqrt{z} \]
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Combine everything together: \[ 20 \cdot \sqrt{294} \cdot z^2 \cdot \sqrt{z} = 20z^2\sqrt{294z} \]
Now, \(294\) can be further simplified to remove any perfect squares:
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The prime factorization of \(294\) is: \[ 294 = 2 \cdot 3 \cdot 7^2 \]
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Therefore, \[ \sqrt{294} = \sqrt{2 \cdot 3 \cdot 7^2} = 7 \sqrt{6} \]
- Final expression: Replacing \(\sqrt{294}\) with \(7\sqrt{6}\): \[ 20z^2\sqrt{294z} = 20z^2 \cdot 7\sqrt{6z} = 140z^2\sqrt{6z} \]
Thus, the result of the simplification is: \[ \boxed{140z^2\sqrt{6z}} \]
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