What is (4y^3-8y^2-3x+1)(y+1/2) ?

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

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simplify (4y^3-8y^2-3x+1)(y+1/2)

simplify

(4 y^{3} - 8 y^{2} - 3 x + 1) (y + \frac{1}{2})

\frac{( 4 y ^{3} - 8 y ^{2} - 3 x + 1 ) ( 2 y + 1 )}{2}

Solution steps

(4 y^{3} - 8 y^{2} - 3 x + 1) (y + \frac{1}{2})

Simplify

y + \frac{1}{2} : \frac{2 y + 1}{2}

= (4 y^{3} - 8 y^{2} - 3 x + 1) \frac{2 y + 1}{2}

Apply the fraction rule:

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

= \frac{( 4 y ^{3} - 8 y ^{2} - 3 x + 1 ) ( 2 y + 1 )}{2}

f (x) = \frac{3 ( x ^{2} + 1 ) ^{2} ( x ^{2} - 2 x - 1 )}{( x - 1 ) ^{4}}

(x + 3 x^{3} sin (y)) d x + x^{4} cos (y) d y

cos (x) (sec (x) + tan (x))

(2 - 3 x)^{2}

2 (ln (\frac{2}{3}) - 1)

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