Can #y=x^3+5x^2-12x-36 # be factored? If so what are the factors ?

1 Answer
George C.
Sep 19, 2016

#x^3+5x^2-12x-36 = (x-3)(x+2)(x+6)#

Explanation:

#f(x) = x^3+5x^2-12x-36#

By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-36# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-3, +-4, +-6, +-9, +-12, +-18, +-36#

In addition note that the signs of the coefficients are in the pattern: #+ + - -#. By Descartes' rule of signs we can deduce that #f(x)# has exactly #1# positive Real zero. It has #0# or #2# negative Real zeros.

So let's try the positive possibilities first:

#f(1) = 1+5-12-36 = -42#

#f(2) = 8+20-24-36 = -32#

#f(3) = 27+45-36-36 = 0#

So #x=3# is a zero and #(x-3)# a factor:

#x^3+5x^2-12x-36 = (x-3)(x^2+8x+12)#

To factor the remaining quadratic note that #2+6=8# and #2*6=12#

Hence:

#x^2+8x+12 = (x+2)(x+6)#

So our complete factorisation is:

#x^3+5x^2-12x-36 = (x-3)(x+2)(x+6)#

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