How do you factor $F(x)= -x^4 + 2x^3 + 4x^2 - 8x$ ?

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Answer 1 · 2015-08-03T22:09:33

$F(x) = -x * (x-2)^2 * (x+2)$

Notice that you can group the first two terms and the second two terms together and factor them to get

$F(x) = (-x^4 + 2x^3) + (4x^2 - 8x)$

$F(x) = -x^3(x-2) + 4x(x-2)$

You can factor these two terms by $(x-2)$ to get

$F(x) = (x-2) * (-x^3 + 4x)$

Notice that you can factor the second term of the expression by $x$

$F(x) = (x-2) * x * (4-x^2)$

Finally, you can factor $(4-x^2)$ as the difference of two squares

$color(blue)(a^2 - b^2 = (a-b)(a+b))$

In your case, you have

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

The expression will now become

$F(x) = (x-2) * x * (2-x) * (2+x)$

which is equivalent to

$F(x) = -x (x-2) * (x-2) * (2+x) = color(green)(-x(x-2)^2(x+2))$

How do you factor F(x)= -x^4 + 2x^3 + 4x^2 - 8xF(x)= -x^4 + 2x^3 + 4x^2 - 8x?