How do you factor #4y=2x^3 + 4x^2 + 8x #?

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Answer 1 · 2018-05-13T10:05:53

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

See explanation. Taken it to complex numbers as well.

Given: #4y=2x^3+4x^2+8x#

Divide both sides by 2

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

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

#y=x/2(x^2+2x+4)# We could stop hear
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Set #y=0 => x=0 and x=(-2+-sqrt(2^2-4(1)(4)))/(2(1))#

#x=-1+-sqrt(-12)/2#

#x=-1+-sqrt(3) color(white)(.)i#

Giving: #y=x(x+1+sqrt3 i)( x+1-sqrt3 i)#

Tony B