How do you simplify and divide #(b^4-2b^3+b^2-3b+2)(b-2)^-1#?

1 Answer
George C.
Dec 8, 2016

#(b^4-2b^3+b^2-3b+2)(b-2)^-1 = b^3+b-1#

Explanation:

#(b^4-2b^3+b^2-3b+2)(b-2)^-1 = (b^4-2b^3+b^2-3b+2)/(b-2)#

#color(white)((b^4-2b^3+b^2-3b+2)(b-2)^-1) = ((b^4-2b^3)+(b^2-2b)-(b-2))/(b-2)#

#color(white)((b^4-2b^3+b^2-3b+2)(b-2)^-1) = (b^3(b-2)+b(b-2)-1(b-2))/(b-2)#

#color(white)((b^4-2b^3+b^2-3b+2)(b-2)^-1) = ((b^3+b-1)color(red)(cancel(color(black)((b-2)))))/color(red)(cancel(color(black)((b-2))))#

#color(white)((b^4-2b^3+b^2-3b+2)(b-2)^-1) = b^3+b-1#

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