What is expand (2a-z)^4 ?

The solution to expand (2a-z)^4 is 16a^4-32a^3z+24a^2z^2-8az^3+z^4

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expand (2a-z)^4

expand

(2 a - z)^{4}

16 a^{4} - 32 a^{3} z + 24 a^{2} z^{2} - 8 a z^{3} + z^{4}

Solution steps

(2 a - z)^{4}

Apply binomial theorem:

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

a = 2 a, b = - z

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

Expand summation

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

Simplify

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

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (2 a)^{3} (- z)^{1} : - 32 a^{3} z

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (2 a)^{2} (- z)^{2} : 24 a^{2} z^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (2 a)^{1} (- z)^{3} : - 8 a z^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (2 a)^{0} (- z)^{4} : z^{4}

= 16 a^{4} - 32 a^{3} z + 24 a^{2} z^{2} - 8 a z^{3} + z^{4}

\frac{x}{x + 5}

(x^{2} - 4 x)^{2} - 3 (x^{2} - 4 x) - 10

(c^{2} - 0.7 c^{3})^{2}

j + j + 2 k

(x^{2} - 3)^{2} - (x - 9)^{2}

What is expand (2a-z)^4 ?
The solution to expand (2a-z)^4 is 16a^4-32a^3z+24a^2z^2-8az^3+z^4