How do you factor completely $2x^5 - 7x^3 - 4x$ ?

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Answer 1 · 2016-07-17T08:39:04

$(2x^2 +1)(x+2)(x-2)$

Take out the common factor of $x$ first.

$x(2x^4 -7x^2 -4)$

We now have a disguised quadratic, but we treat it the same as a quadratic.

Find factors of 2 and 4 which subtract to make 7.

$"2 +1 "rArr " 1 x 1 = +1"$
$"1 -4 "rArr " 2 x 4 = -8 " rArr -8+1 =-7$

$(2x^2 +1)(x^2 -4)$
$(2x^2 +1)(x+2)(x-2)$

How do you factor completely 2x^5 - 7x^3 - 4x2x^5 - 7x^3 - 4x?