What is 2(x+1/3)(x+1/5)(x+1/4)(x+1/2) ?

The solution to 2(x+1/3)(x+1/5)(x+1/4)(x+1/2) is ((3x+1)(5x+1)(4x+1)(2x+1))/(60)

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simplify 2(x+1/3)(x+1/5)(x+1/4)(x+1/2)

simplify

2 (x + \frac{1}{3}) (x + \frac{1}{5}) (x + \frac{1}{4}) (x + \frac{1}{2})

\frac{( 3 x + 1 ) ( 5 x + 1 ) ( 4 x + 1 ) ( 2 x + 1 )}{60}

Solution steps

2 (x + \frac{1}{3}) (x + \frac{1}{5}) (x + \frac{1}{4}) (x + \frac{1}{2})

Simplify

x + \frac{1}{3} : \frac{3 x + 1}{3}

= 2 \cdot \frac{3 x + 1}{3} (x + \frac{1}{5}) (x + \frac{1}{4}) (x + \frac{1}{2})

Apply the fraction rule:

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

= \frac{2 ( 3 x + 1 ) ( x + \frac{1}{5} ) ( x + \frac{1}{4} ) ( x + \frac{1}{2} )}{3}

Simplify

2 (3 x + 1) (x + \frac{1}{5}) (x + \frac{1}{4}) (x + \frac{1}{2}) : \frac{( 3 x + 1 ) ( x \cdot 5 + 1 ) ( x \cdot 4 + 1 ) ( x \cdot 2 + 1 )}{20}

= \frac{\frac{( 3 x + 1 ) ( x \cdot 5 + 1 ) ( x \cdot 4 + 1 ) ( x \cdot 2 + 1 )}{20}}{3}

Apply the fraction rule:

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

= \frac{( 3 x + 1 ) ( x \cdot 5 + 1 ) ( x \cdot 4 + 1 ) ( x \cdot 2 + 1 )}{20 \cdot 3}

Multiply the numbers:

20 \cdot 3 = 60

= \frac{( 3 x + 1 ) ( x \cdot 5 + 1 ) ( x \cdot 4 + 1 ) ( x \cdot 2 + 1 )}{60}

= \frac{( 3 x + 1 ) ( 5 x + 1 ) ( 4 x + 1 ) ( 2 x + 1 )}{60}

(2 x^{3} y^{4} + 5 x) d x + (2 x^{m + 1} y^{n} + y^{3}) d y

(2) (5 x + e^{x})

so l v e (x - 2) (x + 3) (x^{2} - 4 x - 1)

(t - 3) e (- s t)

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

What is 2(x+1/3)(x+1/5)(x+1/4)(x+1/2) ?
The solution to 2(x+1/3)(x+1/5)(x+1/4)(x+1/2) is ((3x+1)(5x+1)(4x+1)(2x+1))/(60)