expandir (2z-3k)^4

Solução

expandir

(2 z - 3 k)^{4}

16 z^{4} - 96 z^{3} k + 216 z^{2} k^{2} - 216 z k^{3} + 81 k^{4}

Passos da solução

(2 z - 3 k)^{4}

Aplicar o teorema do binômio:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = 2 z, b = - 3 k

= i = 0 \sum 4 (i 4) (2 z)^{(4 - i)} (- 3 k)^{i}

Expandir a soma

= \frac{4 !}{0 ! ( 4 - 0 ) !} (2 z)^{4} (- 3 k)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (2 z)^{3} (- 3 k)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (2 z)^{2} (- 3 k)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (2 z)^{1} (- 3 k)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (2 z)^{0} (- 3 k)^{4}

Simplificar

\frac{4 !}{0 ! ( 4 - 0 ) !} (2 z)^{4} (- 3 k)^{0} : 16 z^{4}

Simplificar

\frac{4 !}{1 ! ( 4 - 1 ) !} (2 z)^{3} (- 3 k)^{1} : - 96 z^{3} k

Simplificar

\frac{4 !}{2 ! ( 4 - 2 ) !} (2 z)^{2} (- 3 k)^{2} : 216 z^{2} k^{2}

Simplificar

\frac{4 !}{3 ! ( 4 - 3 ) !} (2 z)^{1} (- 3 k)^{3} : - 216 z k^{3}

Simplificar

\frac{4 !}{4 ! ( 4 - 4 ) !} (2 z)^{0} (- 3 k)^{4} : 81 k^{4}

= 16 z^{4} - 96 z^{3} k + 216 z^{2} k^{2} - 216 z k^{3} + 81 k^{4}

s im pl i f y \frac{x ^{n}}{x ^{n - 1}}

s im pl i f y - 4 (- 3 t - 10 t - 7) - t

s im pl i f y \frac{32}{64}

f a c t or sin (x) - sin^{2} (x)

f a c t or - 2 x^{2} - 3 x + 5