What is (1+x+(x^2)/2)(-x-(x^2)/2-(x^3)/3) ?

The solution to (1+x+(x^2)/2)(-x-(x^2)/2-(x^3)/3) is ((2+2x+x^2)(-6x-3x^2-2x^3))/(12)

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simplify (1+x+(x^2)/2)(-x-(x^2)/2-(x^3)/3)

simplify

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

\frac{( 2 + 2 x + x ^{2} ) ( - 6 x - 3 x ^{2} - 2 x ^{3} )}{12}

Solution steps

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

Simplify

1 + x + \frac{x ^{2}}{2} : \frac{2 + 2 x + x ^{2}}{2}

= \frac{2 + 2 x + x ^{2}}{2} (- x - \frac{x ^{2}}{2} - \frac{x ^{3}}{3})

Apply the fraction rule:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 2 + 2 x + x ^{2} ) ( - x - \frac{x ^{2}}{2} - \frac{x ^{3}}{3} )}{2}

Simplify

(2 + 2 x + x^{2}) (- x - \frac{x ^{2}}{2} - \frac{x ^{3}}{3}) : \frac{( 2 + 2 x + x ^{2} ) ( - x \cdot 6 - x ^{2} \cdot 3 - x ^{3} \cdot 2 )}{6}

= \frac{\frac{( 2 + 2 x + x ^{2} ) ( - x \cdot 6 - x ^{2} \cdot 3 - x ^{3} \cdot 2 )}{6}}{2}

Apply the fraction rule:

\frac{\frac{a}{b}}{c} = \frac{a}{b \cdot c}

= \frac{( 2 + 2 x + x ^{2} ) ( - x \cdot 6 - x ^{2} \cdot 3 - x ^{3} \cdot 2 )}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{( 2 + 2 x + x ^{2} ) ( - x \cdot 6 - x ^{2} \cdot 3 - x ^{3} \cdot 2 )}{12}

= \frac{( 2 + 2 x + x ^{2} ) ( - 6 x - 3 x ^{2} - 2 x ^{3} )}{12}

\frac{1}{2} (1 - cos (θ))

125 e^{(l n (\frac{37}{125}) - 4) (\frac{1}{3})}

- e^{x} (x + 1)

\frac{4}{( 4 x - 3 ) ( 4 n + 1 )}

\frac{z ( z - 3 )}{z ^{2} - 6 z + 8}

What is (1+x+(x^2)/2)(-x-(x^2)/2-(x^3)/3) ?
The solution to (1+x+(x^2)/2)(-x-(x^2)/2-(x^3)/3) is ((2+2x+x^2)(-6x-3x^2-2x^3))/(12)