How do you find the real solutions of the polynomial #x^4+2x^3=12x^2+40x+32#?

1 Answer
Mar 26, 2017

Answer:

Solutions to #x^4+2x^3=12x^2+40x+32# are #x=-2# and #x=4#

Explanation:

#x^4+2x^3=12x^2+40x+32#

or #f(x)=x^4+2x^3-12x^2-40x-32=0#

This will have roots as #+-1#, #+-2#, #+-4#, #+-8#, #+-16# or #+-32#, although some of them may repeat.

Note that #x=-2# is a solution as #f(-2)=0# and #(x+2)# is a factor of #f(x)#

Similarly #x=4# is a solution as #f(4)=0# and #(x-4)# is a factor of #f(x)#

As #(x+2)(x-4)=x^2-2x-8#, dividing #x^4+2x^3-12x^2-40x-32# by #x^2-2x-8# we get

#x^2(x^2-2x-8)+4x(x^2-2x-8)+4(x^2-2x-8)# i.e.

our equation is #(x^2+4x+4)(x^2+4x+4)=0#

or #(x+2)(x-4)(x+2)^2=(x+2)^3(x-4)=0#

Hence solutions to #x^4+2x^3=12x^2+40x+32# are #x=-2# and #x=4#