*This is one of the problems submitted. Reference to which one is deliberately
removed to prevent easy copying.*

*Factor the polynomial (f(x). Then solve the equation f(x) = 0.*

f(x) = x^{3} + 5x^{2} - 2x - 24

f(x) = x^{3}+ 5x^{2}- 2x - 24 Possible factors include ±1, ±2, ±3, ±4, ±6, ±8, ±12, and ±24. This is from taking the positive and negative factors of the constant (24) divided by the positive and negative factors of the lead coefficient of highest degree term (1). Try 3.3| 1 5 -2 -24 3 24 | 66 ----------------------- 1 8 22 | 42 The remainder is the answer to life, the universe, and everything, but does not indicate a factor. Try 2.2| 1 5 -2 -24 2 14 | 24 ----------------------- 1 7 12 | 0 Zero remainder means 2 is a factor, so factor the original equation. f(x) = x^{3}+ 5x^{2}- 2x - 24 f(x) = (x-2)(x^{2}+ 7x + 12) Factor the trinomial. f(x) = (x-2)(x^{2}+ 7x + 12) f(x) = (x-2)(x+3)(x+4) Solve f(x)=0. x-2=0 x+3=0 x+4=0 x=2 x=-3 x=-4 The solutions are -4, -3, and 2.Check.f(x) = x^{3}+ 5x^{2}- 2x - 24 f(-4) = (-4)^{3}+ 5((-4)^{2}) - 2(-4) - 24 f(-4) = -64 + 5(16) + 8 - 24 f(-4) = -64 + 80 + 8 - 24 f(-4) = 16 - 16 f(-4) = 0 f(x) = x^{3}+ 5x^{2}- 2x - 24 f(-3) = (-3)^{3}+ 5((-3)^{2}) - 2(-3) - 24 f(-3) = -27 + 5(9) + 6 - 24 f(-3) = -27 + 45 - 18 f(-3) = 18 - 18 f(-3) = 0 f(x) = x^{3}+ 5x^{2}- 2x - 24 f(2) = 2^{3}+ 5*(2^{2}) - 2*2 - 24 f(2) = 8 + 5*(4) - 4 - 24 f(2) = 8 + 20 - 28 f(2) = 28 - 28 f(2) = 0