How do you find all the zeros of # f(x)=x^3+7x^2-6x-72 #?

1 Answer
May 21, 2016

Answer:

Use the rational root theorem to help find the first zero, then complete the square to find the other two, giving:

#3#, #-4#, #-6#

Explanation:

#f(x) = x^3+7x^2-6x-72#

By the rational root theorem, any rational zeros of #f(x)# must be expressible in the form #p/q# where #p# is a divisor of the constant term #-72# 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#, #+-12#, #+-18#, #+-24#, #+-36#, #+-72#

Trying each in turn, the first one that works is:

#f(3) = 3^3+(7*3^2)-(6*3)-72 = 27+63-18-72 = 0#

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

#x^3+7x^2-6x-72 = (x-3)(x^2+10x+24)#

One way of factoring the remaining quadratic factor #x^2+10x+24# is by completing the square:

#x^2+10x+24#

#=(x+5)^2-25+24#

#=(x+5)^2-1^2#

#=((x+5)-1)((x+5)+1)#

#=(x+4)(x+6)#

Hence zeros:

#x = -4# and #x=-6#