What is expand (2x-7)^5 ?

The solution to expand (2x-7)^5 is 32x^5-560x^4+3920x^3-13720x^2+24010x-16807

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expand (2x-7)^5

expand

(2 x - 7)^{5}

32 x^{5} - 560 x^{4} + 3920 x^{3} - 13720 x^{2} + 24010 x - 16807

Solution steps

(2 x - 7)^{5}

Apply binomial theorem:

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

a = 2 x, b = - 7

= i = 0 \sum 5 (i 5) (2 x)^{(5 - i)} (- 7)^{i}

Expand summation

= \frac{5 !}{0 ! ( 5 - 0 ) !} (2 x)^{5} (- 7)^{0} + \frac{5 !}{1 ! ( 5 - 1 ) !} (2 x)^{4} (- 7)^{1} + \frac{5 !}{2 ! ( 5 - 2 ) !} (2 x)^{3} (- 7)^{2} + \frac{5 !}{3 ! ( 5 - 3 ) !} (2 x)^{2} (- 7)^{3} + \frac{5 !}{4 ! ( 5 - 4 ) !} (2 x)^{1} (- 7)^{4} + \frac{5 !}{5 ! ( 5 - 5 ) !} (2 x)^{0} (- 7)^{5}

Simplify

\frac{5 !}{0 ! ( 5 - 0 ) !} (2 x)^{5} (- 7)^{0} : 32 x^{5}

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} (2 x)^{4} (- 7)^{1} : - 560 x^{4}

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} (2 x)^{3} (- 7)^{2} : 3920 x^{3}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} (2 x)^{2} (- 7)^{3} : - 13720 x^{2}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} (2 x)^{1} (- 7)^{4} : 24010 x

Simplify

\frac{5 !}{5 ! ( 5 - 5 ) !} (2 x)^{0} (- 7)^{5} : - 16807

= 32 x^{5} - 560 x^{4} + 3920 x^{3} - 13720 x^{2} + 24010 x - 16807

What is expand (2x-7)^5 ?
The solution to expand (2x-7)^5 is 32x^5-560x^4+3920x^3-13720x^2+24010x-16807