What is expand 3(3Ae^xx^2+6Ae^xx+4Ae^{x^4}x^3) ?

The solution to expand 3(3Ae^xx^2+6Ae^xx+4Ae^{x^4}x^3) is 9Ae^xx^2+18Ae^xx+12Ae^{x^4}x^3

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expand 3(3Ae^xx^2+6Ae^xx+4Ae^{x^4}x^3)

expand

3 (3 A e^{x} x^{2} + 6 A e^{x} x + 4 A e^{x^{4}} x^{3})

9 A e^{x} x^{2} + 18 A e^{x} x + 12 A e^{x^{4}} x^{3}

Solution steps

3 (3 A e^{x} x^{2} + 6 A e^{x} x + 4 A e^{x^{4}} x^{3})

Distribute parentheses

= 3 \cdot 3 A e^{x} x^{2} + 3 \cdot 6 A e^{x} x + 3 \cdot 4 A e^{x^{4}} x^{3}

Simplify

3 \cdot 3 A e^{x} x^{2} + 3 \cdot 6 A e^{x} x + 3 \cdot 4 A e^{x^{4}} x^{3} : 9 A e^{x} x^{2} + 18 A e^{x} x + 12 A e^{x^{4}} x^{3}

= 9 A e^{x} x^{2} + 18 A e^{x} x + 12 A e^{x^{4}} x^{3}

\frac{x}{( 0.04 - x ) ( 0.12 - x )}

(x - 1 - 22) (x - 1 + 22)

lo g_{3} (9 x^{2} (y + 3) - (x^{2} + 2)^{\frac{3}{2}})

\frac{1}{( 1 - x ) ( 1 - x ^{3} ) ( 1 - x ^{5} )}

\frac{1}{8} (1 - cos (8 x))

What is expand 3(3Ae^xx^2+6Ae^xx+4Ae^{x^4}x^3) ?
The solution to expand 3(3Ae^xx^2+6Ae^xx+4Ae^{x^4}x^3) is 9Ae^xx^2+18Ae^xx+12Ae^{x^4}x^3