Factor completely e^-x - xe^-x + 2x^2e^-x

someone please help me!!

factor out the common e^-x, get
e^-x ( 1 – x + 2x²) =
e^-x ( 2x² – x + 1) =
e^-x (2x + 1)(x – 1)

Tom V changed signs on his 2x² term.

(2x + 1)(x - 1) and 2x^2 - x + 1 are not equal.

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Notice that e^(-x) is a factor of all three terms (e^(-x), x*e^(-x), and 2x^2*e^(-x)), so factoring it out gives:
e^(-x) - x*e^(-x) + 2x^2*e^(-x) = e^(-x)(1 - x + 2x^2) = e^(-x)(2x^2 - x + 1).

At this point, the only way this could factor further is if 2x^2 - x + 1 factored, but it doesn't since its discriminant:
b^2 - 4ac = (-1)^2 - 4(2)(1) = 1 - 8 = -7,

is not a perfect square.

Therefore, the final factorization is e^(-x)(2x^2 - x + 1).

I hope this helps!

Take e^(-x) out...
If y=e^(-x) - x*e^(-x)+2*x^(2)*e^(-x)
y= e^(-x){1-x+2*x^(2)}
=e^(-x){2*x^(2)-x+1}
=e^(-x)(2x+1)(x-1)

Philo has this right...Tom had his signs changed for the 2x part

e^(-x) (2x^2-x+1)

e^-x - xe^-x + 2x^2e^-x = (1-x-2x²)e^(-x)
= (1-2x)(1+x)e^(-x)