1. **Problem a:** Write $\sqrt{3} \times \sqrt{6}$ in the form $b\sqrt{2}$ where $b$ is an integer. 2. Use the property of square roots: $\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$. 3. Calculate: $$\sqrt{3} \times \sqrt{6} = \sqrt{3 \times 6} = \sqrt{18}$$ 4. Simplify $\sqrt{18}$ by factoring 18 into $9 \times 2$: $$\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2}$$ 5. So, $b = 3$ and the expression in the form $b\sqrt{2}$ is: $$3\sqrt{2}$$ 6. **Problem b:** Simplify fully $\sqrt{8} \times \sqrt{18}$. 7. Use the property of square roots again: $$\sqrt{8} \times \sqrt{18} = \sqrt{8 \times 18} = \sqrt{144}$$ 8. Simplify $\sqrt{144}$: $$\sqrt{144} = 12$$ 9. So, the fully simplified form is: $$12$$