directions: find the simplest radical form of each expression below by circling the matching answer.
1. simplify the radical expression below: √108
a 3√2 b 2√27 c 6√3 d 9√12 e 3√6
2. simplify the radical expression below: √240
a 4√6 b 2√60 c 2√15 d 4√15 e 15√6
3. simplify the radical expression below: √320
a 64√5 b 8√5 c 32√10 d 5√8 e 16√20
4. simplify the radical expression below: √392
a 14√2 b 7√56 c 14√28 d 2√14 e 56√7
5. simplify the radical expression below: 3√200
a 6√50 b 15√8 c 24√5 d 6√10 e 30√2
6. simplify the radical expression below: 4√192
a 24√5 b 16√12 c 8√48 d 32√3 e 20√6
7. simplify the radical expression below: √72x³
a 6x²√2x b 6x√2x c 3x√8x d 3x²√8x e 4x√18x
8. simplify the radical expression below: √288x²y⁵
a 6xy²√8y b 6x²y⁴√8y c 12xy²√2y d 12x²y⁴√2y e 2xy²√72y
9. simplify the radical expression below: √243x³y¹⁶
a 9xy⁸√3x b 9xy⁴√3x c 3xy⁸√27x d 3xy⁴√27x e 9x²y⁴√27x
10. simplify the radical expression below: √175a⁶b⁸c⁴
a 7b⁴c²√5ab b 7b⁴c²√5ab c 5b³c²√7ab d 5b⁴c²√7ab e 5b⁴c²√7ab

Question

Accepted Answer

1. **Simplify $\sqrt{108}$**:
   - **Step 1: Prime - factorize 108**
     - $108 = 2	imes54=2	imes2	imes27 = 2^{2}	imes3^{3}$.
   - **Step 2: Apply the square - root property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ ($a = 2^{2}$, $b = 3^{3}$)**
     - $\sqrt{108}=\sqrt{2^{2}	imes3^{3}}=\sqrt{2^{2}}	imes\sqrt{3^{3}}$.
     - Since $\sqrt{2^{2}} = 2$ and $\sqrt{3^{3}}=\sqrt{3^{2}	imes3}=3\sqrt{3}$, then $\sqrt{108}=2	imes3\sqrt{3}=6\sqrt{3}$.
2. **Simplify $\sqrt{240}$**:
   - **Step 1: Prime - factorize 240**
     - $240=2	imes120 = 2	imes2	imes60=2	imes2	imes2	imes30=2^{4}	imes3	imes5$.
   - **Step 2: Apply the square - root property**
     - $\sqrt{240}=\sqrt{2^{4}	imes3	imes5}=\sqrt{2^{4}}	imes\sqrt{3	imes5}$.
     - Since $\sqrt{2^{4}} = 4$, then $\sqrt{240}=4\sqrt{15}$.
3. **Simplify $\sqrt{320}$**:
   - **Step 1: Prime - factorize 320**
     - $320 = 2	imes160=2	imes2	imes80=2	imes2	imes2	imes40=2	imes2	imes2	imes2	imes20=2^{6}	imes5$.
   - **Step 2: Apply the square - root property**
     - $\sqrt{320}=\sqrt{2^{6}	imes5}=\sqrt{2^{6}}	imes\sqrt{5}$.
     - Since $\sqrt{2^{6}} = 8$, then $\sqrt{320}=8\sqrt{5}$.
4. **Simplify $\sqrt{392}$**:
   - **Step 1: Prime - factorize 392**
     - $392=2	imes196=2	imes2	imes98=2	imes2	imes2	imes49=2^{3}	imes7^{2}$.
   - **Step 2: Apply the square - root property**
     - $\sqrt{392}=\sqrt{2^{3}	imes7^{2}}=\sqrt{7^{2}}	imes\sqrt{2^{3}}$.
     - Since $\sqrt{7^{2}} = 7$ and $\sqrt{2^{3}}=\sqrt{2^{2}	imes2}=2\sqrt{2}$, then $\sqrt{392}=7	imes2\sqrt{2}=14\sqrt{2}$.
5. **Simplify $3\sqrt{200}$**:
   - **Step 1: Prime - factorize 200**
     - $200 = 2	imes100=2	imes2	imes50=2	imes2	imes2	imes25=2^{3}	imes5^{2}$.
   - **Step 2: Apply the square - root property to $\sqrt{200}$**
     - $\sqrt{200}=\sqrt{2^{3}	imes5^{2}}=\sqrt{5^{2}}	imes\sqrt{2^{3}}$.
     - Since $\sqrt{5^{2}} = 5$ and $\sqrt{2^{3}}=\sqrt{2^{2}	imes2}=2\sqrt{2}$, then $\sqrt{200}=10\sqrt{2}$.
   - **Step 3: Multiply by 3**
     - $3\sqrt{200}=3	imes10\sqrt{2}=30\sqrt{2}$.
6. **Simplify $4\sqrt{192}$**:
   - **Step 1: Prime - factorize 192**
     - $192=2	imes96=2	imes2	imes48=2	imes2	imes2	imes24=2	imes2	imes2	imes2	imes12=2	imes2	imes2	imes2	imes2	imes6=2^{6}	imes3$.
   - **Step 2: Apply the square - root property to $\sqrt{192}$**
     - $\sqrt{192}=\sqrt{2^{6}	imes3}=\sqrt{2^{6}}	imes\sqrt{3}$.
     - Since $\sqrt{2^{6}} = 8$, then $\sqrt{192}=8\sqrt{3}$.
   - **Step 3: Multiply by 4**
     - $4\sqrt{192}=4	imes8\sqrt{3}=32\sqrt{3}$.
7. **Simplify $\sqrt{72x^{3}}$**:
   - **Step 1: Prime - factorize 72 and consider the variable part**
     - $72=2	imes36=2	imes2	imes18=2	imes2	imes2	imes9=2^{3}	imes3^{2}$, and we have $x^{3}=x^{2}	imes x$.
   - **Step 2: Apply the square - root property**
     - $\sqrt{72x^{3}}=\sqrt{2^{3}	imes3^{2}	imes x^{2}	imes x}=\sqrt{3^{2}}	imes\sqrt{2^{3}}	imes\sqrt{x^{2}}	imes\sqrt{x}$.
     - Since $\sqrt{3^{2}} = 3$, $\sqrt{2^{3}}=\sqrt{2^{2}	imes2}=2\sqrt{2}$, $\sqrt{x^{2}} = x$, then $\sqrt{72x^{3}}=6x\sqrt{2x}$.
8. **Simplify $\sqrt{288x^{2}y^{5}}$**:
   - **Step 1: Prime - factorize 288 and consider the variable parts**
     - $288=2	imes144=2	imes2	imes72=2	imes2	imes2	imes36=2	imes2	imes2	imes2	imes18=2	imes2	imes2	imes2	imes2	imes9=2^{5}	imes3^{2}$, $x^{2}$ and $y^{5}=y^{4}	imes y$.
   - **Step 2: Apply the square - root property**
     - $\sqrt{288x^{2}y^{5}}=\sqrt{2^{5}	imes3^{2}	imes x^{2}	imes y^{4}	imes y}=\sqrt{3^{2}}	imes\sqrt{2^{5}}	imes\sqrt{x^{2}}	imes\sqrt{y^{4}}	imes\sqrt{y}$.
     - Since $\sqrt{3^{2}} = 3$, $\sqrt{2^{5}}=\sqrt{2^{4}	imes2}=4\sqrt{2}$, $\sqrt{x^{2}} = x$, $\sqrt{y^{4}} = y^{2}$, then $\sqrt{288x^{2}y^{5}}=12xy^{2}\sqrt{2y}$.
9. **Simplify $\sqrt{243x^{3}y^{16}}$**:
   - **Step 1: Prime - factorize 243 and consider the variable parts**
     - $243=3	imes81=3	imes3	imes27=3	imes3	imes3	imes9=3^{5}$, $x^{3}=x^{2}	imes x$ and $y^{16}=(y^{8})^{2}$.
   - **Step 2: Apply the square - root property**
     - $\sqrt{243x^{3}y^{16}}=\sqrt{3^{5}	imes x^{2}	imes x	imes y^{16}}=\sqrt{3^{4}	imes3}	imes\sqrt{x^{2}}	imes\sqrt{x}	imes\sqrt{y^{16}}$.
     - Since $\sqrt{3^{4}} = 9$, $\sqrt{x^{2}} = x$, $\sqrt{y^{16}} = y^{8}$, then $\sqrt{243x^{3}y^{16}}=9xy^{8}\sqrt{3x}$.
10. **Simplify $\sqrt{175a^{6}b^{8}c^{4}}$**:
   - **Step 1: Prime - factorize 175 and consider the variable parts**
     - $175=5	imes35=5	imes5	imes7$, $a^{6}=(a^{3})^{2}$, $b^{8}=(b^{4})^{2}$, $c^{4}=(c^{2})^{2}$.
   - **Step 2: Apply the square - root property**
     - $\sqrt{175a^{6}b^{8}c^{4}}=\sqrt{5^{2}	imes7	imes a^{6}	imes b^{8}	imes c^{4}}=\sqrt{5^{2}}	imes\sqrt{7}	imes\sqrt{a^{6}}	imes\sqrt{b^{8}}	imes\sqrt{c^{4}}$.
     - Since $\sqrt{5^{2}} = 5$, $\sqrt{a^{6}} = a^{3}$, $\sqrt{b^{8}} = b^{4}$, $\sqrt{c^{4}} = c^{2}$, then $\sqrt{175a^{6}b^{8}c^{4}}=5a^{3}b^{4}c^{2}\sqrt{7}$.

# Answer:
1. $6\sqrt{3}$
2. $4\sqrt{15}$
3. $8\sqrt{5}$
4. $14\sqrt{2}$
5. $30\sqrt{2}$
6. $32\sqrt{3}$
7. $6x\sqrt{2x}$
8. $12xy^{2}\sqrt{2y}$
9. $9xy^{8}\sqrt{3x}$
10. $5a^{3}b^{4}c^{2}\sqrt{7}$

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