What is expand 3(2x^3+1)(x^4+2x)^{1/2} ?

The solution to expand 3(2x^3+1)(x^4+2x)^{1/2} is 6x^3(x^4+2x)^{1/2}+3(x^4+2x)^{1/2}

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expand 3(2x^3+1)(x^4+2x)^{1/2}

expand

3 (2 x^{3} + 1) (x^{4} + 2 x)^{\frac{1}{2}}

6 x^{3} (x^{4} + 2 x)^{\frac{1}{2}} + 3 (x^{4} + 2 x)^{\frac{1}{2}}

Solution steps

3 (2 x^{3} + 1) (x^{4} + 2 x)^{\frac{1}{2}}

= 3 (x^{4} + 2 x)^{\frac{1}{2}} (2 x^{3} + 1)

Apply the distributive law:

a (b + c) = ab + a c

3 (x^{4} + 2 x)^{\frac{1}{2}} (2 x^{3} + 1) = 3 (x^{4} + 2 x)^{\frac{1}{2}} \cdot 2 x^{3} + 3 (x^{4} + 2 x)^{\frac{1}{2}} \cdot 1

= 3 (x^{4} + 2 x)^{\frac{1}{2}} \cdot 2 x^{3} + 3 (x^{4} + 2 x)^{\frac{1}{2}} \cdot 1

3 (x^{4} + 2 x)^{\frac{1}{2}} \cdot 2 x^{3} = 6 x^{3} (x^{4} + 2 x)^{\frac{1}{2}}

3 (x^{4} + 2 x)^{\frac{1}{2}} \cdot 1 = 3 (x^{4} + 2 x)^{\frac{1}{2}}

= 6 x^{3} (x^{4} + 2 x)^{\frac{1}{2}} + 3 (x^{4} + 2 x)^{\frac{1}{2}}

(X^{4} - 2 X^{2} + 2) l e f t (X^{2} - 2 X + 3)

- \frac{- sin ( θ ) + x cos ( θ )}{( cos ( θ ) + x sin ( θ ) ) ^{2}}

e^{(1 + 2 i) x}

(1 - t) (4, 6, 6) + t (6, 7, 2)

(3 x - 5 \circ - 4 x - 2) (- 5)

What is expand 3(2x^3+1)(x^4+2x)^{1/2} ?
The solution to expand 3(2x^3+1)(x^4+2x)^{1/2} is 6x^3(x^4+2x)^{1/2}+3(x^4+2x)^{1/2}