How do you find the roots of #x^3-12x+16=0#?

2 Answers
May 29, 2017

Answer:

#x=2" with multiplicity 2"#
#x=-4" with multiplicity 1"#

Explanation:

#"note that " 2^3-12(2)+16=0#

#rArrx=2" is a root and " (x-2)" is a factor"#

#rArrx^2(x-2)+2x(x-2)-8(x-2)#

#=(x-2)(x^2+2x-8)#

#=(x-2)(x+4)(x-2)#

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

#rArrx=2" multiplicity of 2"#

#"and " x=-4" multiplicity of 1"#

May 29, 2017

Answer:

The roots are #4# and #2#(double root)

Explanation:

Let #f(x)=x^3-12x+16#

Then,

We see that

#f(2)=2^3-12*2+16=8-24+16=0#

Therefore,

#(x-2)# is a factor of #f(x)#

To find the other factors, we perform a long division

#color(white)(aaaa)##x-2##color(white)(aaaa)##|##x^3+0x^2-12x+16##color(white)(aaaa)##|##x^2+2x-8#

#color(white)(aaaaaaaaaaaaaa)##x^3-2x^2#

#color(white)(aaaaaaaaaaaaaaa)##0+2x^2-12x#

#color(white)(aaaaaaaaaaaaaaaaa)##+2x^2-4x#

#color(white)(aaaaaaaaaaaaaaaaaaa)##+0-8x+16#

#color(white)(aaaaaaaaaaaaaaaaaaaaaaa)##-8x+16#

#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaa)##-0+0#

The quotient is

#x^2+2x-8=(x-2)(x+4)#

Therefore,

#x^3-12x+16=(x-2)^2(x+4)# graph{x^3-12x+16 [-16.15, 15.88, -3.97, 12.05]}