How do you factor completely # 2x^3 + 4x^2 + 6x + 12#?

2 Answers
kubik98
Dec 31, 2017

#(x^2+3)(2x+4)# or #(x+isqrt3)(x-isqrt3)(2x+4)#

Explanation:

#y=2x^3+4x^2+6x+12#

Moving #6x# closer to #2x^3#

#y=2x^3+6x+4x^2+12#

#y=2x(x^2+3)+4(x^2+3)#

Take out #x^2+3#

#y=(x^2+3)(2x+4)#

we can continue factoring #x^2+3# using complex numbers: #i^2=-1#

#y=(x^2-(3i^2))(2x+4)#

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

#y=(x+isqrt3)(x-isqrt3)(2x+4)#

Jim G.
Dec 31, 2017

#2(x+2)(x+sqrt3i)(x-sqrt3i)#

Explanation:

#"take out "color(blue)"common factor of 2"#

#rArr2(x^3+2x^2+3x+6)#

#"when "x=-2tox^3+2x^2+3x+6=0#

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

#color(red)(x^2)(x+2)cancel(color(magenta)(-2x^2))cancel(+2x^2)+3x+6#

#=color(red)(x^2)(x+2)color(red)(+3)(x+2)cancel(color(magenta)(-6))cancel(+6)#

#rArrx^3+2x^2+3x+6=(x+2)(x^2+3)#

#"factor "x^2+3#

#x^2+3=0rArrx^2=-3rArrx=+-sqrt3i#

#rArr2x^3+4x^2+6x+12#

#=2(x+2)(x+sqrt3i)(x-sqrt3i)#

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