What is expand 16n(2k)^4-32(2k)^3+24(2k)^2-8(2k) ?

The solution to expand 16n(2k)^4-32(2k)^3+24(2k)^2-8(2k) is 256nk^4-256k^3+96k^2-16k

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expand 16n(2k)^4-32(2k)^3+24(2k)^2-8(2k)

expand

16 n (2 k)^{4} - 32 (2 k)^{3} + 24 (2 k)^{2} - 8 (2 k)

256 n k^{4} - 256 k^{3} + 96 k^{2} - 16 k

Solution steps

16 n (2 k)^{4} - 32 (2 k)^{3} + 24 (2 k)^{2} - 8 (2 k)

Remove parentheses:

(a) = a

= 16 n (2 k)^{4} - 32 (2 k)^{3} + 24 (2 k)^{2} - 8 \cdot 2 k

16 n (2 k)^{4} = 2^{4} \cdot 16 n k^{4}

32 (2 k)^{3} = 2^{3} \cdot 32 k^{3}

24 (2 k)^{2} = 2^{2} \cdot 24 k^{2}

8 \cdot 2 k = 16 k

= 2^{4} \cdot 16 n k^{4} - 2^{3} \cdot 32 k^{3} + 2^{2} \cdot 24 k^{2} - 16 k

16 n \cdot 2^{4} k^{4} = 256 n k^{4}

32 \cdot 2^{3} k^{3} = 256 k^{3}

24 \cdot 2^{2} k^{2} = 96 k^{2}

= 256 n k^{4} - 256 k^{3} + 96 k^{2} - 16 k

\frac{5 x + 10 - 4 x ^{2}}{x ^{2} ( x + 2 )}

(x + 2) e^{2}

(e^{s})^{2}

(arctan (x) + 5)^{6}

\frac{1}{( x - 1 ) ^{2} ( x + 1 )}

What is expand 16n(2k)^4-32(2k)^3+24(2k)^2-8(2k) ?
The solution to expand 16n(2k)^4-32(2k)^3+24(2k)^2-8(2k) is 256nk^4-256k^3+96k^2-16k