What is expand (2x^3-x^2+2x-8)(x+2) ?

The solution to expand (2x^3-x^2+2x-8)(x+2) is x(2x^3-x^2+2x-8)+2(2x^3-x^2+2x-8)

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expand (2x^3-x^2+2x-8)(x+2)

expand

(2 x^{3} - x^{2} + 2 x - 8)^{\circ} (x + 2)

x (2 x^{3} - x^{2} + 2 x - 8)^{\circ} + 2 (2 x^{3} - x^{2} + 2 x - 8)^{\circ}

Solution steps

(2 x^{3} - x^{2} + 2 x - 8)^{\circ} (x + 2)

= (x + 2) (2 x^{3} - x^{2} + 2 x - 8)^{\circ}

Apply the distributive law:

a (b + c) = ab + a c

a = (2 x^{3} - x^{2} + 2 x - 8)^{\circ}, b = x, c = 2

= (2 x^{3} - x^{2} + 2 x - 8)^{\circ} x + (2 x^{3} - x^{2} + 2 x - 8)^{\circ} \cdot 2

= x (2 x^{3} - x^{2} + 2 x - 8)^{\circ} + 2 (2 x^{3} - x^{2} + 2 x - 8)^{\circ}

(x + y^{2} - x y) \frac{d y}{d x} - y

x (u d x + x d u)

\frac{( - 9 + m ) ^{2}}{9}

- \frac{3 ( - 6 x + 5 )}{4 ( 14 + 5 x - 3 x ^{2} ) ^{\frac{7}{4}}}

I (R + (2 π F L - \frac{1}{2 π FC}))

What is expand (2x^3-x^2+2x-8)(x+2) ?
The solution to expand (2x^3-x^2+2x-8)(x+2) is x(2x^3-x^2+2x-8)+2(2x^3-x^2+2x-8)