oo=infinite limx->(+/-)oo(2x2+3)=oo
y=2x2+3
y'=4x=0
x=0 <-Location of absolute minimum
y(0)-2(0)2+3
y(0)=3
[3,oo)
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The answer is (2x^2+3)(4x+1)
(2x-3)(x-1)
2x2+5x+3 = (2x+3)(x+1) when factored
(2x - 1)(x + 3)
8x3 + 4x2 + 6x + 3 = 4x2(2x + 1) + 3(2x + 1) = (2x + 1)(4x2 + 3)