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

The solution to (2x+1/2)(4x^2-x+1/4) is ((4x+1)(16x^2-4x+1))/8

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simplify (2x+1/2)(4x^2-x+1/4)

simplify

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

\frac{( 4 x + 1 ) ( 16 x ^{2} - 4 x + 1 )}{8}

Solution steps

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

Join

2 x + \frac{1}{2} : \frac{4 x + 1}{2}

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

Apply the fraction rule:

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

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

Simplify

(4 x + 1) (4 x^{2} - x + \frac{1}{4}) : \frac{( 4 x + 1 ) ( 16 x ^{2} - x \cdot 4 + 1 )}{4}

= \frac{\frac{( 4 x + 1 ) ( 16 x ^{2} - x \cdot 4 + 1 )}{4}}{2}

Apply the fraction rule:

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

= \frac{( 4 x + 1 ) ( 16 x ^{2} - x \cdot 4 + 1 )}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{( 4 x + 1 ) ( 16 x ^{2} - x \cdot 4 + 1 )}{8}

= \frac{( 4 x + 1 ) ( 16 x ^{2} - 4 x + 1 )}{8}

(sin (x) + cos (y))^{2} + (cos (x) + sin (y))^{2}

\frac{( x + h ) ^{\frac{3}{2}} - x ^{\frac{3}{2}}}{h}

\frac{e ^{- 2 s}}{( s ^{2} + 2 s + 2 ) s}

(x - 1) 2 x + 1

(1 - x)^{2}

What is (2x+1/2)(4x^2-x+1/4) ?
The solution to (2x+1/2)(4x^2-x+1/4) is ((4x+1)(16x^2-4x+1))/8