What is 4n(3n^2+n-4)-n(3-n) ?

The solution to 4n(3n^2+n-4)-n(3-n) is 12n^3+5n^2-19n

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simplify 4n(3n^2+n-4)-n(3-n)

simplify

4 n (3 n^{2} + n - 4) - n (3 - n)

12 n^{3} + 5 n^{2} - 19 n

Solution steps

4 n (3 n^{2} + n - 4) - n (3 - n)

Expand

4 n (3 n^{2} + n - 4) : 12 n^{3} + 4 n^{2} - 16 n

= 12 n^{3} + 4 n^{2} - 16 n - n (3 - n)

Expand

- n (3 - n) : - 3 n + n^{2}

= 12 n^{3} + 4 n^{2} - 16 n - 3 n + n^{2}

Add similar elements:

- 16 n - 3 n = - 19 n

= 12 n^{3} + 4 n^{2} - 19 n + n^{2}

Group like terms

= 12 n^{3} + 4 n^{2} + n^{2} - 19 n

Add similar elements:

4 n^{2} + n^{2} = 5 n^{2}

= 12 n^{3} + 5 n^{2} - 19 n

\frac{c + 2}{c - 5} - \frac{c}{c + 1}

y^{2} - 5 y + 3 = 17

∣ 2 x + 3 ∣ + 1 < 11

∣ u ∣ + 5 < 12

∣ 7 - 2 x ∣ = 15

What is 4n(3n^2+n-4)-n(3-n) ?
The solution to 4n(3n^2+n-4)-n(3-n) is 12n^3+5n^2-19n