What is expand (2e^{3x}-e^{2x})(6e^{3x}-2e^{2x}) ?

The solution to expand (2e^{3x}-e^{2x})(6e^{3x}-2e^{2x}) is 12e^{6x}-10e^{5x}+2e^{4x}

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expand (2e^{3x}-e^{2x})(6e^{3x}-2e^{2x})

expand

(2 e^{3 x} - e^{2 x}) (6 e^{3 x} - 2 e^{2 x})

12 e^{6 x} - 10 e^{5 x} + 2 e^{4 x}

Solution steps

(2 e^{3 x} - e^{2 x}) (6 e^{3 x} - 2 e^{2 x})

Apply FOIL method:

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

a = 2 e^{3 x}, b = - e^{2 x}, c = 6 e^{3 x}, d = - 2 e^{2 x}

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

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

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

Simplify

2 \cdot 6 e^{3 x} e^{3 x} - 2 \cdot 2 e^{3 x} e^{2 x} - 6 e^{2 x} e^{3 x} + 2 e^{2 x} e^{2 x} : 12 e^{6 x} - 10 e^{5 x} + 2 e^{4 x}

= 12 e^{6 x} - 10 e^{5 x} + 2 e^{4 x}

\frac{( 2 + 3 s + s ^{2} )}{( s ^{2} ( s + 3 ) )}

cos (x^{2} + 2) \cdot 2 x (x^{2} + 2)

(5 cos (θ) - 3 sin (x))^{2}

6 (\frac{x ^{3}}{3} + 2 x^{2} + 4 x)

(2 e^{4 t} - 5)^{2}

What is expand (2e^{3x}-e^{2x})(6e^{3x}-2e^{2x}) ?
The solution to expand (2e^{3x}-e^{2x})(6e^{3x}-2e^{2x}) is 12e^{6x}-10e^{5x}+2e^{4x}