Is it possible to factor #y=x^4 - 2x^3 - 13x^2 + 14x + 24 #? If so, what are the factors?

1 Answer
George C.
Jan 19, 2016

Use the rational root theorem to help find the first two factors, then divide and factor the remaining quadratic to find:

#y = x^4-2x^3-13x^2+14x+24#

#=(x+1)(x-2)(x-4)(x+3)#

Explanation:

Let #f(x) = x^4-2x^3-13x^2+14x+24#

By the rational root theorem, any rational roots of #f(x) = 0# must be of the form #p/q# for some integers #p# and #q# with #p# as factor of the constant term #24# and #q# a factor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1#, #+-2#, #+-3#, #+-4#, #+-6#, #+-8#, #+-12#, #+-24#

Try the first few in turn:

#f(1) = 1-2-13+14+24 = 24#

#f(-1) = 1+2-13-14+24 = 0#

#f(2) = 16-16-52+28+24 = 0#

So #x=-1# and #x=2# are zeros and #(x+1)# and #(x-2)# are factors:

#x^4-2x^3-13x^2+14x+24#

#=(x+1)(x^3-3x^2-10x+24)#

#=(x+1)(x-2)(x^2-x-12)#

To factor the remaining quadratic, find a pair of factors of #12# that differ by #1#. The pair #4, 3# works, hence we find:

#=(x+1)(x-2)(x-4)(x+3)#

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