What are the roots of the equation #x^3 +4x^2-4x- 16=0#?

2 Answers
Dec 18, 2016

Answer:

The roots are:

#x = 2#, #x = -2# and #x=-4#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We use this with #a=x# and #b=2# later.

Given:

#x^3+4x^2-4x-16 = 0#

Note that the ratio between the first and second terms is the same as the ratio between the third and fourth terms, so this cubic factors by grouping:

#0 = x^3+4x^2-4x-16#

#color(white)(0) = (x^3+4x^2)-(4x+16)#

#color(white)(0) = x^2(x+4)-4(x+4)#

#color(white)(0) = (x^2-4)(x+4)#

#color(white)(0) = (x^2-2^2)(x+4)#

#color(white)(0) = (x-2)(x+2)(x+4)#

Hence the roots are:

#x = 2#, #x = -2# and #x=-4#

Dec 18, 2016

Answer:

#x in {2,-4,-2}#

Explanation:

#x^3+4x^2-4x-16=0#

#rArr x^3+4x^2=4x+16#

#rarr x * x^2 +4 * x^2 = x * 2^2 + 4 * 2^2#

#rarr #(at least one possible solution is) # x=2#
(i.e. #(x-2)# is a factor)

Dividing #x^3+4x^2-4x-16# by #(x-2)# (using either synthetic or long division) gives:
#color(white)("XXX")x^2+6x+8#
which factors using standard operations as:
#color(white)("XXX")(x+4)(x+2)#

Therefore
#x^3+4x^2-4x-16=0#
#rArr (x-2)(x+4)(x+2)=0#

#rArr x in {2,-4,-2}#