What is expand f(x)(3x-1)^2e^{2x+1} ?

The solution to expand f(x)(3x-1)^2e^{2x+1} is 9e^{2x+1}fx^3-6e^{2x+1}fx^2+e^{2x+1}fx

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expand f(x)(3x-1)^2e^{2x+1}

expand

f (x) (3 x - 1)^{2} e^{2 x + 1}

9 e^{2 x + 1} f x^{3} - 6 e^{2 x + 1} f x^{2} + e^{2 x + 1} f x

Solution steps

f (x) (3 x - 1)^{2} e^{2 x + 1}

Remove parentheses:

(a) = a

= f x (3 x - 1)^{2} e^{2 x + 1}

(3 x - 1)^{2} = 9 x^{2} - 6 x + 1

= e^{2 x + 1} f x (9 x^{2} - 6 x + 1)

= e^{2 x + 1} f x (9 x^{2} - 6 x + 1)

Distribute parentheses

= f x e^{2 x + 1} \cdot 9 x^{2} + f x e^{2 x + 1} (- 6 x) + f x e^{2 x + 1} \cdot 1

Apply minus-plus rules

+ (- a) = - a

= 9 e^{2 x + 1} f x^{2} x - 6 e^{2 x + 1} f xx + 1 \cdot e^{2 x + 1} f x

Simplify

9 e^{2 x + 1} f x^{2} x - 6 e^{2 x + 1} f xx + 1 \cdot e^{2 x + 1} f x : 9 e^{2 x + 1} f x^{3} - 6 e^{2 x + 1} f x^{2} + e^{2 x + 1} f x

= 9 e^{2 x + 1} f x^{3} - 6 e^{2 x + 1} f x^{2} + e^{2 x + 1} f x

(4 x^{5} - 5) d x

sin (2 x - 9 0^{\circ})

e^{(x - 2)}

(3 i, - 6 j) - (4 - 6) + (- 3.3)

r (5)

What is expand f(x)(3x-1)^2e^{2x+1} ?
The solution to expand f(x)(3x-1)^2e^{2x+1} is 9e^{2x+1}fx^3-6e^{2x+1}fx^2+e^{2x+1}fx