1. Express each radical as a mixed radical in simplest form: 1.a. Simplify $\sqrt{56}$: - Factor 56 into $7 \times 8$. - Since $8 = 4 \times 2$ and $\sqrt{4} = 2$, we can write: $$\sqrt{56} = \sqrt{7 \times 4 \times 2} = \sqrt{7} \times \sqrt{4} \times \sqrt{2} = 2\sqrt{14}$$ 1.b. Simplify $3\sqrt{75}$: - Factor 75 into $25 \times 3$. - Since $\sqrt{25} = 5$, we have: $$3\sqrt{75} = 3 \times \sqrt{25 \times 3} = 3 \times 5 \sqrt{3} = 15\sqrt{3}$$ 1.c. Simplify $3\sqrt{8m^4}$: - Factor 8 as $4 \times 2$. - $m^4$ is a perfect square since $\sqrt{m^4} = m^2$. - So: $$3\sqrt{8m^4} = 3 \times \sqrt{4 \times 2 \times m^4} = 3 \times \sqrt{4} \times \sqrt{2} \times \sqrt{m^4} = 3 \times 2 \times m^2 \sqrt{2} = 6m^2\sqrt{2}$$ 1.d. Simplify $\sqrt[3]{24q^5}$: - Factor 24 as $8 \times 3$. - $q^5 = q^3 \times q^2$. - Since $\sqrt[3]{8} = 2$ and $\sqrt[3]{q^3} = q$, we get: $$\sqrt[3]{24q^5} = \sqrt[3]{8 \times 3 \times q^3 \times q^2} = \sqrt[3]{8} \times \sqrt[3]{q^3} \times \sqrt[3]{3q^2} = 2q \sqrt[3]{3q^2}$$ 2. Simplify each expression: 2.a. Simplify $3\sqrt{75} - \sqrt{27}$: - From 1.b, $3\sqrt{75} = 15\sqrt{3}$. - Simplify $\sqrt{27}$: $27 = 9 \times 3$, $\sqrt{9} = 3$, so $\sqrt{27} = 3\sqrt{3}$. - Combine: $$15\sqrt{3} - 3\sqrt{3} = (15 - 3)\sqrt{3} = 12\sqrt{3}$$ 2.b. Simplify $2\sqrt{18} + 9\sqrt{7} - \sqrt{63}$: - Simplify $\sqrt{18}$: $18 = 9 \times 2$, $\sqrt{9} = 3$, so $\sqrt{18} = 3\sqrt{2}$. - Simplify $\sqrt{63}$: $63 = 9 \times 7$, $\sqrt{9} = 3$, so $\sqrt{63} = 3\sqrt{7}$. - Substitute: $$2 \times 3\sqrt{2} + 9\sqrt{7} - 3\sqrt{7} = 6\sqrt{2} + (9 - 3)\sqrt{7} = 6\sqrt{2} + 6\sqrt{7}$$ 2.c. Simplify $3\sqrt{2x} + 3\sqrt{8x} - \sqrt{x}$: - Simplify $\sqrt{8x}$: $8 = 4 \times 2$, $\sqrt{4} = 2$, so $\sqrt{8x} = 2\sqrt{2x}$. - Substitute: $$3\sqrt{2x} + 3 \times 2\sqrt{2x} - \sqrt{x} = 3\sqrt{2x} + 6\sqrt{2x} - \sqrt{x} = 9\sqrt{2x} - \sqrt{x}$$ - Cannot combine further because $\sqrt{2x}$ and $\sqrt{x}$ are unlike terms. 2.d. Simplify $5\sqrt{3}\sqrt{6}$: - Multiply under one radical: $$\sqrt{3} \times \sqrt{6} = \sqrt{3 \times 6} = \sqrt{18}$$ - Simplify $\sqrt{18} = 3\sqrt{2}$. - So: $$5 \times 3\sqrt{2} = 15\sqrt{2}$$ 2.e. Simplify $-2 \sqrt[3]{11} (4 \sqrt[3]{2} - 3\sqrt{3})$: - Distribute: $$-2 \sqrt[3]{11} \times 4 \sqrt[3]{2} + (-2 \sqrt[3]{11}) \times (-3\sqrt{3})$$ - First term: $$-8 \sqrt[3]{11} \sqrt[3]{2} = -8 \sqrt[3]{11 \times 2} = -8 \sqrt[3]{22}$$ - Second term: $$6 \sqrt[3]{11} \sqrt{3}$$ - Final expression: $$-8 \sqrt[3]{22} + 6 \sqrt[3]{11} \sqrt{3}$$ 2.f. Simplify $(4\sqrt{2} + 3)(\sqrt{7} - 5\sqrt{14})$: - Use distributive property: $$4\sqrt{2} \times \sqrt{7} - 4\sqrt{2} \times 5\sqrt{14} + 3 \times \sqrt{7} - 3 \times 5\sqrt{14}$$ - Calculate each term: 1) $4\sqrt{2} \times \sqrt{7} = 4\sqrt{14}$ 2) $4\sqrt{2} \times 5\sqrt{14} = 20 \sqrt{2 \times 14} = 20 \sqrt{28}$ 3) $3 \sqrt{7}$ 4) $15 \sqrt{14}$ - Simplify $\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}$, so: $$20 \sqrt{28} = 20 \times 2 \sqrt{7} = 40 \sqrt{7}$$ - Substitute back: $$4\sqrt{14} - 40\sqrt{7} + 3\sqrt{7} - 15\sqrt{14}$$ - Combine like terms: $$ (4\sqrt{14} - 15\sqrt{14}) + (-40\sqrt{7} + 3\sqrt{7}) = -11\sqrt{14} - 37\sqrt{7}$$ Final answers: 1.a. $2\sqrt{14}$ 1.b. $15\sqrt{3}$ 1.c. $6m^2\sqrt{2}$ 1.d. $2q \sqrt[3]{3q^2}$ 2.a. $12\sqrt{3}$ 2.b. $6\sqrt{2} + 6\sqrt{7}$ 2.c. $9\sqrt{2x} - \sqrt{x}$ 2.d. $15\sqrt{2}$ 2.e. $-8 \sqrt[3]{22} + 6 \sqrt[3]{11} \sqrt{3}$ 2.f. $-11\sqrt{14} - 37\sqrt{7}$