Algebra 1 Worksheets | Radical Expressions Worksheets - Free Printable

Let's solve each of the radical expressions step by step. We'll simplify each one using properties of radicals, factoring, and rules for exponents.

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1) $\sqrt[3]{448b^7h^2}$

We want to simplify the cube root.

Break down $448$:
- $448 = 64 \times 7 = 4^3 \times 7$
- So: $448 = 8 \times 56 = 8 \times 8 \times 7 = 64 \times 7$, but better: $64 = 4^3$, so $448 = 64 \times 7 = 4^3 \times 7$

Wait: $64 = 4^3$, but $4^3 = 64$, yes. But $448 ÷ 64 = 7$, so:
$$
\sqrt[3]{448b^7h^2} = \sqrt[3]{64 \cdot 7 \cdot b^7 \cdot h^2}
= \sqrt[3]{64} \cdot \sqrt[3]{7} \cdot \sqrt[3]{b^7} \cdot \sqrt[3]{h^2}
$$

Now:
- $\sqrt[3]{64} = \sqrt[3]{4^3} = 4$
- $\sqrt[3]{b^7} = \sqrt[3]{b^6 \cdot b} = \sqrt[3]{(b^2)^3 \cdot b} = b^2 \sqrt[3]{b}$
- $\sqrt[3]{h^2}$ stays as is
- $\sqrt[3]{7}$ stays

So:
$$
= 4 \cdot b^2 \cdot \sqrt[3]{7bh^2}
$$

✔ Answer: $4b^2 \sqrt[3]{7bh^2}$

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2) $\sqrt{12p}$

Factor $12 = 4 \cdot 3 = 2^2 \cdot 3$

$$
\sqrt{12p} = \sqrt{4 \cdot 3p} = \sqrt{4} \cdot \sqrt{3p} = 2\sqrt{3p}
$$

✔ Answer: $2\sqrt{3p}$

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3) $\sqrt{36z^2}$

$$
\sqrt{36z^2} = \sqrt{36} \cdot \sqrt{z^2} = 6 \cdot |z| = 6|z|
$$

(Since square root gives non-negative result, we use absolute value for $z^2$)

✔ Answer: $6|z|$

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4) $\sqrt{8}$

$$
\sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2\sqrt{2}
$$

✔ Answer: $2\sqrt{2}$

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5) $\sqrt[3]{8}$

$$
\sqrt[3]{8} = \sqrt[3]{2^3} = 2
$$

✔ Answer: $2$

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6) $\sqrt[3]{64}$

$$
\sqrt[3]{64} = \sqrt[3]{4^3} = 4
$$

✔ Answer: $4$

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7) $\sqrt[3]{64z^3}$

$$
\sqrt[3]{64z^3} = \sqrt[3]{64} \cdot \sqrt[3]{z^3} = 4 \cdot z = 4z
$$

(Cube roots don’t require absolute values since odd roots can handle negative numbers)

✔ Answer: $4z$

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8) $\sqrt{112}$

Factor $112 = 16 \cdot 7 = 4^2 \cdot 7$

$$
\sqrt{112} = \sqrt{16 \cdot 7} = \sqrt{16} \cdot \sqrt{7} = 4\sqrt{7}
$$

✔ Answer: $4\sqrt{7}$

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9) $\sqrt{275b^4h^6}$

Break down $275 = 25 \cdot 11 = 5^2 \cdot 11$

And:
- $b^4 = (b^2)^2$
- $h^6 = (h^3)^2$

So:
$$
\sqrt{275b^4h^6} = \sqrt{25 \cdot 11 \cdot b^4 \cdot h^6}
= \sqrt{25} \cdot \sqrt{11} \cdot \sqrt{b^4} \cdot \sqrt{h^6}
= 5 \cdot \sqrt{11} \cdot b^2 \cdot h^3
$$

Note: Since even powers, no need for absolute value here because $b^2$ and $h^3$ are non-negative under square root.

✔ Answer: $5b^2h^3\sqrt{11}$

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10) $\sqrt[3]{135y^7d^2}$

Break down $135 = 27 \cdot 5 = 3^3 \cdot 5$

Now:
- $y^7 = y^6 \cdot y = (y^2)^3 \cdot y$
- $d^2$ stays

So:
$$
\sqrt[3]{135y^7d^2} = \sqrt[3]{27 \cdot 5 \cdot y^6 \cdot y \cdot d^2}
= \sqrt[3]{27} \cdot \sqrt[3]{5} \cdot \sqrt[3]{y^6} \cdot \sqrt[3]{y} \cdot \sqrt[3]{d^2}
= 3 \cdot \sqrt[3]{5} \cdot y^2 \cdot \sqrt[3]{y} \cdot \sqrt[3]{d^2}
$$

Combine:
$$
= 3y^2 \sqrt[3]{5yd^2}
$$

✔ Answer: $3y^2 \sqrt[3]{5yd^2}$

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✔ Final Answers:

1) $4b^2 \sqrt[3]{7bh^2}$
2) $2\sqrt{3p}$
3) $6|z|$
4) $2\sqrt{2}$
5) $2$
6) $4$
7) $4z$
8) $4\sqrt{7}$
9) $5b^2h^3\sqrt{11}$
10) $3y^2 \sqrt[3]{5yd^2}$

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