#x^3-6x^2+16=0# How about #x#?.

2 Answers
Cem Sentin
Feb 12, 2018

#x_1=2#, #x_2=2+2sqrt3# and #x_3=2-2sqrt3#

Explanation:

#x^3-6x^2+16=0#

#(x^3-8)-(6x^2-24)=0#

#(x^3-8)-6*(x^2-4)=0#

#(x-2)(x^2+2x+4)-6*(x-2)(x+2)=0#

#(x-2)*[(x^2+2x+4)-6(x+2)]=0#

#(x-2)*(x^2-4x-8)=0#

From first multiplier, #x_1=2#. From second one #x_2=2+2sqrt3# and #x_3=2-2sqrt3#

Jim G.
Feb 12, 2018

#x=2,x=2+-2sqrt3#

Explanation:

#"note that for x = 2"#

#2^3-6(2)^2+16=0#

#rArr(x-2)" is a factor"#

#"dividing "x^3-6x^2+16" by "(x-2)#

#color(red)(x^2)(x-2)color(magenta)(+2x^2)-6x^2+16#

#=color(red)(x^2)(x-2)color(red)(-4x)(x-2)color(magenta)(-8x)+16#

#=color(red)(x^2)(x-2)color(red)(-4x)(x-2)color(red)(-8)(x-2)cancel(color(magenta)(-16))cancel(+16)#

#=color(red)(x^2)(x-2)color(red)(-4x)(x-2)color(red)(-8)(x-2)+0#

#rArrx^3-6x^2+16=0#

#rArr(x-2)(x^2-4x-8)=0#

#"solve "x^2-4x-8" using the "color(blue)"quadratic formula"#

#"with "a=1,b=-4" and "c=-8#

#x=(4+-sqrt(16+32))/2#

#color(white)(x)=(4+-sqrt48)/2=(4+-4sqrt3)/2=2+-2sqrt3#

#rArr(x-2)(x^2-4x-8)=0#

#"has solutions "x=2,x=2+-2sqrt3#

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