How do you factor #y= x^3-5x^2-2x+24# ?

1 Answer
Jan 2, 2016

Answer:

Use the rational root theorem to get started, then factor the remaining quadratic to find:

#x^3-5x^2-2x+24 = (x+2)(x-4)(x-3)#

Explanation:

Let #f(x) = x^3-5x^2-2x+24#

By the rational root theorem, any rational zeros of #f(x)# must be expressible in the for #p/q# for integers #p#, #q# with #p# a divisor of the constant term #24# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are the factors of #24#, namely:

#+-1, +-2, +-3, +-4, +-6, +-12, +-24#

Try each in turn:

#f(1) = 1-5-2+24 = 18#

#f(-1) = -1-5+2+24 = 20#

#f(2) = 8-20-4+24 = 8#

#f(-2) = -8-20+4+24 = 0#

So #x=-2# is a zero and #(x+2)# is a factor.

#x^3-5x^2-2x+24 = (x+2)(x^2-7x+12)#

We can factor #x^2-7x+12# by noting that #4xx3 = 12# and #4+3=7#, so:

#x^2-7x+12 = (x-4)(x-3)#

Putting it all together:

#x^3-5x^2-2x+24 = (x+2)(x-4)(x-3)#