How do you find the zeroes of #p(x)= x^4-4x^3-x^2+16x-12#?

Question

9536 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-23T11:54:06

By observation we can note that
if #p(x)=x^4-4x^3-x^2+16x-12#
then
#p(1) = 0#

Therefore #(x-1)# is a factor of #p(x)#

By synthetic division we obtain
#p(x) = (x-1)(x^3-3x^2-4x+12)#

Focusing on #(x^3-3x^2-4x+12)#
we notice the grouping
#x^2(x-3) -4(x-3)#

#=(x^2-4)(x-3)#

and using the difference of squares
#=(x+2)(x-2)(x-3)#

Therefore a complete factoring of #p(x)#
#=(x-1)(x+2)(x-2)(x-3)#

and the zeroes of #p(x)# are
#{1,-2, 2, 3}#