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

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

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

expand

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

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

Solution steps

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

Apply FOIL method:

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

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

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

Apply minus-plus rules

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

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

Simplify

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

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

A (1, - 4) B (4, 4)

e^{2 (l n (c o s (c_{1} + x)) + c_{2})}

e^{x^{2} - y} (5 - 2 x + y)

π (\frac{35}{4} + 2)

\frac{20 ( 10 + 3 22 )}{49}

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