What is expand ((2x+1)^4)/8 ?

The solution to expand ((2x+1)^4)/8 is 2x^4+4x^3+3x^2+x+1/8

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expand ((2x+1)^4)/8

expand

\frac{( 2 x + 1 ) ^{4}}{8}

2 x^{4} + 4 x^{3} + 3 x^{2} + x + \frac{1}{8}

Solution steps

\frac{( 2 x + 1 ) ^{4}}{8}

(2 x + 1)^{4} = 16 x^{4} + 32 x^{3} + 24 x^{2} + 8 x + 1

= \frac{16 x ^{4} + 32 x ^{3} + 24 x ^{2} + 8 x + 1}{8}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{16 x ^{4} + 32 x ^{3} + 24 x ^{2} + 8 x + 1}{8} = \frac{16 x ^{4}}{8} + \frac{32 x ^{3}}{8} + \frac{24 x ^{2}}{8} + \frac{8 x}{8} + \frac{1}{8}

= \frac{16 x ^{4}}{8} + \frac{32 x ^{3}}{8} + \frac{24 x ^{2}}{8} + \frac{8 x}{8} + \frac{1}{8}

\frac{16 x ^{4}}{8} = 2 x^{4}

\frac{32 x ^{3}}{8} = 4 x^{3}

\frac{24 x ^{2}}{8} = 3 x^{2}

\frac{8 x}{8} = x

= 2 x^{4} + 4 x^{3} + 3 x^{2} + x + \frac{1}{8}

\frac{10}{x ( x - 5 )}

\frac{( x ^{2} + 8 ) 2}{x ^{4}}

\frac{( x - 2 ) ( x + 8 )}{( x + 1 ) ( x - 1 )}

A (- 5, - 6) B (- 2, - 2)

(3 x + 7 y = - 12) (5)

What is expand ((2x+1)^4)/8 ?
The solution to expand ((2x+1)^4)/8 is 2x^4+4x^3+3x^2+x+1/8