Put each term in 1/(2 x^2 - 4 x) - 2/x over the common denominator x (2 x - 4): 1/(2 x^2 - 4 x) - 2/x = ((x (2 x - 4))/(2 x^2 - 4 x))/(x (2 x - 4)) - (2 (2 x - 4))/(x (2 x - 4)):

((x (2 x - 4))/(2 x^2 - 4 x))/(x (2 x - 4)) - (2 (2 x - 4))/(x (2 x - 4))

A common factor of 2 x - 4 and 2 x^2 - 4 x is 2 x - 4, so (x (2 x - 4))/(2 x^2 - 4 x) = (x (2 x - 4))/(x (2 x - 4)):

((x (2 x - 4))/(x (2 x - 4)))/(x (2 x - 4)) - (2 (2 x - 4))/(x (2 x - 4))

(x (2 x - 4))/(x (2 x - 4)) = 1:

1/(x (2 x - 4)) - (2 (2 x - 4))/(x (2 x - 4))

1/(x (2 x - 4)) - (2 (2 x - 4))/(x (2 x - 4)) = (1 - 2 (2 x - 4))/(x (2 x - 4)):

(1 - 2 (2 x - 4))/(x (2 x - 4))

-2 (2 x - 4) = 8 - 4 x:

(8 - 4 x + 1)/(x (2 x - 4))

Add like terms. 1 + 8 = 9:

Answer: (9 - 4 x)/(x (2 x - 4))

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4 years ago

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The polynomial p(x)=x^5-9x^4+13x^3+57x^2-86x-120 has how many zeros

natita [175]

It has a total of 5 zeros

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4 years ago

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