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

2 Answers
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)

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)