How do you factor $- x^{3} + 6 x^{2} - x = 5$ $-x^3 +6x^2 -x =5$ ?

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Answer 1 · 2015-10-06T20:16:49

$x (- x^{2} + 6 x - 1) = 5$ $x(-x^2+6x-1)=5$

Start by finding a variable/term that all of the numbers have in common.
Factor out that term and simplify.
$x$ $x$ is common in all of the terms, so we can factor this out of the equation.
Because all numbers have an exponent of one, factor out $x^{1}$ $x^1$ from all of the terms containing $x$ $x$ .
$- x^{3} - x 1 = - x^{2}$ $-x^3-x1=-x^2$
$6 x^{2} - x^{1} = 6 x$ $6x^2-x^1=6x$
$x$ $x$ can be taken out of itself once, and we use 1 as a place holder for any number that has been factored out of the equation.

How do you factor −x3+6x2−x=5-x^3 +6x^2 -x =5?