1. The problem gives four quadratic equations and partly factored forms with one missing factor in each. We will complete the factoring and then find the two values of $x$ for each. 2. For $x^2 + 6x + 8 = 0$, the given factor is $(x + 2)$. To find the missing factor, divide the quadratic by $(x + 2)$ or find the other root: The roots multiply to $8$ and add to $6$. Since $2$ is a root (from $x+2=0$), the other root is $4$ because $2 \times 4 = 8$ and $2 + 4 = 6$. So, missing factor is $(x + 4)$. Solutions: $$x = -2 \text{ or } x = -4$$ 3. For $x^2 - 4x + 3 = 0$ with factor $(x - 3)$ given, the roots must multiply to $3$ and sum to $4$ (since coefficient of $x$ is $-4$ and factoring signs show roots are positive). One root is $3$, so the other root is $1$. Missing factor: $$(x - 1)$$ Solutions: $$x = 3 \text{ or } x = 1$$ 4. For $x^2 + 4x - 21 = 0$, given factor is $(x - 3)$. Roots multiply to $-21$ and add to $4$. Since $3$ is a root, the other root $r$ satisfies $3 \times r = -21$, so $r = -7$. Missing factor: $$(x + 7)$$ Solutions: $$x = 3 \text{ or } x = -7$$ 5. For $x^2 - 8x - 9 = 0$, given factor is $(x + 1)$. Roots multiply to $-9$ and add to $8$ (opposite sign from middle term). Since root $-1$ from $(x + 1)$ is known, the other root $r$ satisfies $(-1) \times r = -9$, so $r = 9$. Missing factor: $$(x - 9)$$ Solutions: $$x = -1 \text{ or } x = 9$$