1. **Find the two prime factors of 323.** To factor 323, we test divisibility by prime numbers. Check divisibility by 17: $$323 \div 17 = 19$$ which is an integer. Both 17 and 19 are prime numbers. So, the two prime factors of 323 are **17** and **19**. 2. **Complete the statement:** $$18x^2 - 287x - 323 = (18x \quad)(x \quad)$$ We want to factor the quadratic into the form: $$(ax + b)(cx + d)$$ where $a \times c = 18$ and $b \times d = -323$. Since the factors of 323 are 17 and 19, and 18 factors as $18 = 18 \times 1 = 9 \times 2 = 6 \times 3$, we try to find integers $b$ and $d$ such that: $$ad + bc = -287$$ Try $a=18$, $c=1$: $$(18x + b)(x + d) = 18x^2 + (18d + b)x + bd$$ We want $bd = -323$ and $18d + b = -287$. Try $b = 17$, $d = -19$: $bd = 17 \times (-19) = -323$ $18d + b = 18 \times (-19) + 17 = -342 + 17 = -325$ (not -287) Try $b = -17$, $d = 19$: $bd = -17 \times 19 = -323$ $18d + b = 18 \times 19 - 17 = 342 - 17 = 325$ (not -287) Try $a=9$, $c=2$: $$(9x + b)(2x + d) = 18x^2 + (9d + 2b)x + bd$$ We want $bd = -323$ and $9d + 2b = -287$. Try $b = 17$, $d = -19$: $bd = 17 \times (-19) = -323$ $9d + 2b = 9 \times (-19) + 2 \times 17 = -171 + 34 = -137$ (not -287) Try $b = -17$, $d = 19$: $bd = -17 \times 19 = -323$ $9d + 2b = 9 \times 19 + 2 \times (-17) = 171 - 34 = 137$ (not -287) Try $a=6$, $c=3$: $$(6x + b)(3x + d) = 18x^2 + (6d + 3b)x + bd$$ We want $bd = -323$ and $6d + 3b = -287$. Try $b = 17$, $d = -19$: $bd = 17 \times (-19) = -323$ $6d + 3b = 6 \times (-19) + 3 \times 17 = -114 + 51 = -63$ (not -287) Try $b = -17$, $d = 19$: $bd = -17 \times 19 = -323$ $6d + 3b = 6 \times 19 + 3 \times (-17) = 114 - 51 = 63$ (not -287) Try $a=3$, $c=6$: $$(3x + b)(6x + d) = 18x^2 + (3d + 6b)x + bd$$ We want $bd = -323$ and $3d + 6b = -287$. Try $b = 17$, $d = -19$: $bd = 17 \times (-19) = -323$ $3d + 6b = 3 \times (-19) + 6 \times 17 = -57 + 102 = 45$ (not -287) Try $b = -17$, $d = 19$: $bd = -17 \times 19 = -323$ $3d + 6b = 3 \times 19 + 6 \times (-17) = 57 - 102 = -45$ (not -287) Try $a=2$, $c=9$: $$(2x + b)(9x + d) = 18x^2 + (2d + 9b)x + bd$$ We want $bd = -323$ and $2d + 9b = -287$. Try $b = 17$, $d = -19$: $bd = 17 \times (-19) = -323$ $2d + 9b = 2 \times (-19) + 9 \times 17 = -38 + 153 = 115$ (not -287) Try $b = -17$, $d = 19$: $bd = -17 \times 19 = -323$ $2d + 9b = 2 \times 19 + 9 \times (-17) = 38 - 153 = -115$ (not -287) Try $a=1$, $c=18$: $$(x + b)(18x + d) = 18x^2 + (d + 18b)x + bd$$ We want $bd = -323$ and $d + 18b = -287$. Try $b = 17$, $d = -19$: $bd = 17 \times (-19) = -323$ $d + 18b = -19 + 18 \times 17 = -19 + 306 = 287$ (not -287, but positive 287) Try $b = -17$, $d = 19$: $bd = -17 \times 19 = -323$ $d + 18b = 19 + 18 \times (-17) = 19 - 306 = -287$ (this matches!) So the factorization is: $$18x^2 - 287x - 323 = (x - 17)(18x + 19)$$ 3. **Solve the equation:** $$18x^2 - 287x - 323 = 0$$ Using the factorization: $$(x - 17)(18x + 19) = 0$$ Set each factor equal to zero: $$x - 17 = 0 \implies x = 17$$ $$18x + 19 = 0 \implies 18x = -19 \implies x = -\frac{19}{18}$$ **Final answers:** (i) Prime factors of 323 are **17** and **19**. (ii) Factorization: $$18x^2 - 287x - 323 = (x - 17)(18x + 19)$$ (iii) Solutions: $$x = 17$$ and $$x = -\frac{19}{18}$$