What is (2x^3-3x^2+4x)(7x+2/9) ?

The solution to (2x^3-3x^2+4x)(7x+2/9) is ((2x^3-3x^2+4x)(63x+2))/9

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simplify (2x^3-3x^2+4x)(7x+2/9)

simplify

(2 x^{3} - 3 x^{2} + 4 x) (7 x + \frac{2}{9})

\frac{( 2 x ^{3} - 3 x ^{2} + 4 x ) ( 63 x + 2 )}{9}

Solution steps

(2 x^{3} - 3 x^{2} + 4 x) (7 x + \frac{2}{9})

Join

7 x + \frac{2}{9} : \frac{63 x + 2}{9}

= (2 x^{3} - 3 x^{2} + 4 x) \frac{63 x + 2}{9}

Apply the fraction rule:

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

= \frac{( 2 x ^{3} - 3 x ^{2} + 4 x ) ( 63 x + 2 )}{9}

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

3 \cdot (9 + 75)^{7}

\frac{( 24 x - 7 y ) ^{2}}{625}

\frac{3 s ^{3} + s ^{2} - 3 s + 6}{s ^{2} ( s ^{2} + 16 )}

\frac{1}{16} (- 4 e^{- 4 x} x - e^{- 4 x}) - x

What is (2x^3-3x^2+4x)(7x+2/9) ?
The solution to (2x^3-3x^2+4x)(7x+2/9) is ((2x^3-3x^2+4x)(63x+2))/9