1. **Problem statement:** We have two quadratic functions: $$A(x) = 9x^2 - 5$$ $$B(x) = x^2 + 2\sqrt{2} x + 2$$ We need to: - Factorize $A(x)$ and $B(x)$ into products of first-degree factors. - Solve the equations $A(x) = 0$ and $B(x) = 0$ in $\mathbb{R}$. - Solve the equation $6 \cdot 3x^2 - 7x = 0$ in $\mathbb{R}$. 2. **Factorize $A(x)$:** Given: $$A(x) = 9x^2 - 5$$ This is a difference of squares form: $$9x^2 - 5 = (3x)^2 - (\sqrt{5})^2$$ Using the identity $a^2 - b^2 = (a - b)(a + b)$: $$A(x) = (3x - \sqrt{5})(3x + \sqrt{5})$$ 3. **Factorize $B(x)$:** Given: $$B(x) = x^2 + 2\sqrt{2} x + 2$$ Check if it is a perfect square trinomial: $$\left(x + \sqrt{2}\right)^2 = x^2 + 2\sqrt{2} x + 2$$ So, $$B(x) = \left(x + \sqrt{2}\right)^2$$ 4. **Solve $A(x) = 0$:** Set each factor to zero: $$3x - \sqrt{5} = 0 \Rightarrow x = \frac{\sqrt{5}}{3}$$ $$3x + \sqrt{5} = 0 \Rightarrow x = -\frac{\sqrt{5}}{3}$$ 5. **Solve $B(x) = 0$:** Since $B(x) = \left(x + \sqrt{2}\right)^2$, set: $$x + \sqrt{2} = 0 \Rightarrow x = -\sqrt{2}$$ This root has multiplicity 2. 6. **Solve $6 \cdot 3x^2 - 7x = 0$:** Rewrite the equation: $$18x^2 - 7x = 0$$ Factor out $x$: $$x(18x - 7) = 0$$ Set each factor to zero: $$x = 0$$ $$18x - 7 = 0 \Rightarrow x = \frac{7}{18}$$ **Final answers:** - $A(x) = (3x - \sqrt{5})(3x + \sqrt{5})$ - $B(x) = \left(x + \sqrt{2}\right)^2$ - Solutions of $A(x) = 0$ are $x = \pm \frac{\sqrt{5}}{3}$ - Solution of $B(x) = 0$ is $x = -\sqrt{2}$ (double root) - Solutions of $18x^2 - 7x = 0$ are $x = 0$ and $x = \frac{7}{18}$