How do you find all the zeros of #f(x)=x^4+7x^3-x^2-67x-60#?

1 Answer
Jul 16, 2016

Answer:

Zeros: #-1, 3, -4, -5#

Explanation:

#f(x)=x^4+7x^3-x^2-67x-60#

By the rational root 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 #-60# and #q# a divisor of the coefficient #1# of the leading term.

So the only possible rational zeros are:

#+-1, +-2, +-3, +-4, +-5, +-6, +-10, +-12, +-15, +-20, +-30, +-60#

We find:

#f(-1) = 1-7-1+67-60 = 0#

So #-1# is a zero and #(x+1)# is a factor:

#x^4+7x^3-x^2-67x-60 = (x+1)(x^3+6x^2-7x-60)#

Let #g(x) = x^3+6x^2-7x-60#

We find:

#g(3) = 27+54-21-60 = 0#

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

#x^3+6x^2-7x-60 = (x-3)(x^2+9x+20)#

To factor the remaining quadratic, note that #4+5 = 9# and #4 xx 5 = 20#, so we find:

#x^2+9x+20 = (x+4)(x+5)#

Hence remaining two zeros: #-4# and #-5#.