What is (-1/2-x)(-1/8-x) ?

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

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simplify (-1/2-x)(-1/8-x)

simplify

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

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

Solution steps

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

Simplify

- \frac{1}{2} - x : \frac{- 2 x - 1}{2}

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

Apply the fraction rule:

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

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

Simplify

(- 2 x - 1) (- \frac{1}{8} - x) : \frac{( - 2 x - 1 ) ( - x \cdot 8 - 1 )}{8}

= \frac{\frac{( - 2 x - 1 ) ( - x \cdot 8 - 1 )}{8}}{2}

Apply the fraction rule:

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

= \frac{( - 2 x - 1 ) ( - x \cdot 8 - 1 )}{8 \cdot 2}

Multiply the numbers:

8 \cdot 2 = 16

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

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

7 (6 - 10)

(2 - 3 i) x^{3} + (\frac{3}{2} + 5 i) x^{2} - 6

(r cos^{2} (θ) + r sin^{2} (θ))^{2}

(1 + e^{- 3 t} + e^{2 t}) \cdot (e^{t} + 2 e^{2 t} - e^{- t})

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

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