expandir (x-4)^5

Solução

expandir

(x - 4)^{5}

x^{5} - 20 x^{4} + 160 x^{3} - 640 x^{2} + 1280 x - 1024

Passos da solução

(x - 4)^{5}

Aplicar o teorema do binômio:

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

a = x, b = - 4

= i = 0 \sum 5 (i 5) x^{(5 - i)} (- 4)^{i}

Expandir a soma

= \frac{5 !}{0 ! ( 5 - 0 ) !} x^{5} (- 4)^{0} + \frac{5 !}{1 ! ( 5 - 1 ) !} x^{4} (- 4)^{1} + \frac{5 !}{2 ! ( 5 - 2 ) !} x^{3} (- 4)^{2} + \frac{5 !}{3 ! ( 5 - 3 ) !} x^{2} (- 4)^{3} + \frac{5 !}{4 ! ( 5 - 4 ) !} x^{1} (- 4)^{4} + \frac{5 !}{5 ! ( 5 - 5 ) !} x^{0} (- 4)^{5}

Simplificar

\frac{5 !}{0 ! ( 5 - 0 ) !} x^{5} (- 4)^{0} : x^{5}

Simplificar

\frac{5 !}{1 ! ( 5 - 1 ) !} x^{4} (- 4)^{1} : - 20 x^{4}

Simplificar

\frac{5 !}{2 ! ( 5 - 2 ) !} x^{3} (- 4)^{2} : 160 x^{3}

Simplificar

\frac{5 !}{3 ! ( 5 - 3 ) !} x^{2} (- 4)^{3} : - 640 x^{2}

Simplificar

\frac{5 !}{4 ! ( 5 - 4 ) !} x^{1} (- 4)^{4} : 1280 x

Simplificar

\frac{5 !}{5 ! ( 5 - 5 ) !} x^{0} (- 4)^{5} : - 1024

= x^{5} - 20 x^{4} + 160 x^{3} - 640 x^{2} + 1280 x - 1024

s im pl i f y (4 + i) (2 - i)

f a c t or 3 x^{3} + 9 x^{2} - 6 x

s im pl i f y \frac{60}{6}

e x p an d a (b + c)

e x p an d (2 a + 1) (2 a - 1)