How do you find all the zeros of #x^4 - 4x^3 - 20x^2 + 48x#?

1 Answer
Apr 16, 2016

Answer:

#x^4-4x^3-20x^2+48x = x(x-2)(x-6)(x+4)#

Explanation:

Firstly, since all of the factors are divisible by #x#, there is a zero #x=0#.

Dividing through by #x# we get a cubic:

#f(x) = x^3-4x^2-20x+48#

By the rational root theorem, any rational zeros of #f(x)# must be expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #48# 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, +-8, +-12, +-16, +-24, +-48#

We find #f(2) = 8-16-40+48 = 0#

So #x=2# is a zero and #(x-2)# a factor:

#x^3-4x^2-20x+48 = (x-2)(x^2-2x-24)#

To factor the remaining quadratic, note that #6-4 = 2# and #6 xx 4 = 24#, hence:

#x^2-2x-24 = (x-6)(x+4)#

Putting it all together:

#x^4-4x^3-20x^2+48x = x(x-2)(x-6)(x+4)#