Write f(x) = x^3-3x^2-10x+24
The rational roots theorem tells us that any rational root p/q of f(x)=0, expressed in simplest terms will have the property that p is a divisor of the constant term (24) and q is a divisor of the coefficient (1) of the highest power of x.
So the only possible rational roots of f(x) = 0 are the integer factors of 24:
+-1, +-2, +-3, +-4, +-6, +-8, +-12, +-24
Aside: If all the roots are rational then they are 3 integers whose product is -24, so at least one of them must have absolute value less than or equal to root(3)(24) ~= 2.88, that is at least one of +-1 or +-2 must be a root.
Try the first few:
f(1) = 1-3-10+24 = 12
f(-1) = -1-3+10+24 = 30
f(2) = 8-12-20+24 = 0
So x=2 is a root of f(x) = 0 and (x-2) is a factor of f(x)
Next use, synthetic division to divide f(x) by (x-2)...
Choose the first multiplier of (x-2) to match the leading term x^3. The multiplier we need is color(red)(x^2)
x^2*(x-2) = x^3 - 2x^2
Subtract this from x^3-3x^2-10x+24 to give a remainder:
(x^3-3x^2-10x+24) - (x^3-2x^2)
=-x^2-10x+24
Choose the next multiplier of (x-2) to match the leading term -x^2 of this remainder. The multiplier we need is color(red)(-x)
-x*(x-2) = -x^2+2x
Subtract this from the previous remainder to get a new remainder:
(-x^2-10x+24) - (-x^2+2x)
=-12x+24
Choose the next multiplier of (x-2) to match the leading term -12x of this remainder. The multiplier we need is color(red)(-12)
-12(x-2) = -12x+24.
Since this matches the previous remainder, we are done. All that remains to complete the division is to add the multipliers we found together to get:
x^3-3x^2-10x+24 = (x-2)(x^2-x-12)
To factor (x^2-x-12) we need a pair of factors of 12 whose difference is 1. The pair 4, 3 works.
Hence (x-4)(x+3) = x^2-x-12
So, putting it all together:
x^3-3x^2-10x+24 = (x-2)(x-4)(x+3)
