What is expand (-e^xx-e^x)(x+e^x) ?

The solution to expand (-e^xx-e^x)(x+e^x) is -e^xx^2-e^{2x}x-e^xx-e^{2x}

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expand (-e^xx-e^x)(x+e^x)

expand

(- e^{x} x - e^{x}) (x + e^{x})

- e^{x} x^{2} - e^{2 x} x - e^{x} x - e^{2 x}

Solution steps

(- e^{x} x - e^{x}) (x + e^{x})

Apply FOIL method:

(a + b) (c + d) = a c + a d + b c + b d

a = - e^{x} x, b = - e^{x}, c = x, d = e^{x}

= (- e^{x} x) x + (- e^{x} x) e^{x} + (- e^{x}) x + (- e^{x}) e^{x}

Apply minus-plus rules

+ (- a) = - a

= - e^{x} xx - e^{x} e^{x} x - e^{x} x - e^{x} e^{x}

Simplify

- e^{x} xx - e^{x} e^{x} x - e^{x} x - e^{x} e^{x} : - e^{x} x^{2} - e^{2 x} x - e^{x} x - e^{2 x}

= - e^{x} x^{2} - e^{2 x} x - e^{x} x - e^{2 x}

(3 x e^{y} + 2 y) d x + (x^{2} e^{y} + x) d y

\frac{( x - 2 ) ( 3 x + 1 )}{3}

\frac{d ^{2}}{d t ^{2}} ((v) (a v (t) - b v (t) ln (v)))

\frac{( 2 a ^{2} - 1 ) ( 4 a ^{4} - 2 a ^{2} + 1 )}{8}

- \frac{- l ( - 2 l ( 3 a - l ^{2} ) - a ) + 3 a}{2}

What is expand (-e^xx-e^x)(x+e^x) ?
The solution to expand (-e^xx-e^x)(x+e^x) is -e^xx^2-e^{2x}x-e^xx-e^{2x}