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)#