How do you factor #z^3-z^2-13z+4#?

1 Answer
George C.
Apr 14, 2016

#z^3-z^2-13z+4 = (z-4)(z + 3/2 - sqrt(5)/2)(z + 3/2 + sqrt(5)/2)#

Explanation:

By the rational root theorem, any rational zeros must be expressible in the form #p/q# for some integers #p# and #q# with #p# a divisor of the constant term #4# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1#, #+-2#, #+-4#

Let #f(z) = z^3-z^2-13z+4#

We find #f(4) = 64-16-52+4 = 0#, so #z=4# is a zero and #(z-4)# is a factor:

#z^3-z^2-13z+4 = (z-4)(z^2+3z-1)#

We can factor the remaining quadratic factor by completing the square:

#z^2+3z-1#

#= (z+3/2)^2-9/4+1#

#= (z+3/2)^2-5/4#

#= (z+3/2)^2-(sqrt(5)/2)^2#

#= (z + 3/2 - sqrt(5)/2)(z + 3/2 + sqrt(5)/2)#

Putting it all together:

#z^3-z^2-13z+4 = (z-4)(z + 3/2 - sqrt(5)/2)(z + 3/2 + sqrt(5)/2)#

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