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How do you factor #y= x^3-2x^2+x-2#?

How do you factor #y= x^3-2x^2+x-2#?
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ReportAnswer #1

#(x-2)(x+i)(x-i)#

Explanation:

#"factor the terms "color(blue)"by grouping"#

#=color(red)(x^2)(x-2)color(red)(+1)(x-2)#

#"take out the "color(blue)"common factor "(x-2)#

#=(x-2)(color(red)(x^2+1))#

#x^2+1" can be factored using "color(blue)"difference of squares"#

#a^2-b^2=(a-b)(a+b)#

#"with "a=x" and "b=ito(i=sqrt-1)#

#=(x-2)(x+i)(x-i)larrcolor(red)"in factored form"#

answered by: Jim G.
ReportAnswer #2

#y=(x-2) xx (x^2+ 1)#

Explanation:

#x^3 -2x^2# can be written as
#x^2xx(x-2)#, so we can re-write the original equation as

#y=x^2xx(x-2) + (x-2)#
or
#y=x^2xx(x-2) + 1xx(x-2)#

and bringing the common factor #(x-2)# to the front, gives the factors:

#y=(x-2) xx (x^2+ 1)#

solving the equation for #y=0#, gives the solution #x=2#, meaning that the graph of the equation crosses the x-axis at the point (2,0)

solving the equation for #x=0# gives #y=-2#, meaning that the graph of the equation crosses the y-axis at (0,-2)

graph{x^3-2x^2+x-2 [-10, 10, -5, 5]}

answered by: anskat
ReportAnswer #3

#y=(x^2+1)(x-2)#

Explanation:

#y=x^3-2x^2+x-2#

We will use a factoring method called grouping. If we group the first and the third and the second and the fourth together, each grout has its own greatest common factor:

#y=x(x^2+1)-2(x^2+1)#

Now we see another CF between the two terms :#x^2+1#. We can factor it out.

#y=(x^2+1)(x-2)#

answered by: Boba
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