What is expand (x(x-2)^5-128)/(x-4) ?

The solution to expand (x(x-2)^5-128)/(x-4) is x^5-6x^4+16x^3-16x^2+16x+32

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expand (x(x-2)^5-128)/(x-4)

expand

\frac{x ( x - 2 ) ^{5} - 128}{x - 4}

x^{5} - 6 x^{4} + 16 x^{3} - 16 x^{2} + 16 x + 32

Solution steps

\frac{x ( x - 2 ) ^{5} - 128}{x - 4}

(x - 2)^{5} = x^{5} - 10 x^{4} + 40 x^{3} - 80 x^{2} + 80 x - 32

= \frac{x ( x ^{5} - 10 x ^{4} + 40 x ^{3} - 80 x ^{2} + 80 x - 32 ) - 128}{x - 4}

Expand

x (x^{5} - 10 x^{4} + 40 x^{3} - 80 x^{2} + 80 x - 32) - 128 : x^{6} - 10 x^{5} + 40 x^{4} - 80 x^{3} + 80 x^{2} - 32 x - 128

= \frac{x ^{6} - 10 x ^{5} + 40 x ^{4} - 80 x ^{3} + 80 x ^{2} - 32 x - 128}{x - 4}

Factor

x^{6} - 10 x^{5} + 40 x^{4} - 80 x^{3} + 80 x^{2} - 32 x - 128 : (x - 4) (x^{5} - 6 x^{4} + 16 x^{3} - 16 x^{2} + 16 x + 32)

= \frac{( x - 4 ) ( x ^{5} - 6 x ^{4} + 16 x ^{3} - 16 x ^{2} + 16 x + 32 )}{x - 4}

Cancel the common factor:

x - 4

= x^{5} - 6 x^{4} + 16 x^{3} - 16 x^{2} + 16 x + 32

(2, - 5) y

\frac{3 ( a + b )}{c}

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

\frac{3 y ( - 3 y + 3 ) ^{2}}{2}

\frac{a ^{2} + b}{a ( b + c )}

What is expand (x(x-2)^5-128)/(x-4) ?
The solution to expand (x(x-2)^5-128)/(x-4) is x^5-6x^4+16x^3-16x^2+16x+32