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) = x3 + 5x2 - 2x - 24
f(x) = x3 + 5x2 - 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) = x3 + 5x2 - 2x - 24
f(x) = (x-2)(x2 + 7x + 12)
Factor the trinomial.
f(x) = (x-2)(x2 + 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) = x3 + 5x2 - 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) = x3 + 5x2 - 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) = x3 + 5x2 - 2x - 24
f(2) = 23 + 5*(22) - 2*2 - 24
f(2) = 8 + 5*(4) - 4 - 24
f(2) = 8 + 20 - 28
f(2) = 28 - 28
f(2) = 0