How do you factor #x^3 - x^2 - 2x + 2=0#?

2 Answers
Hoat V.
Mar 11, 2018

#x = sqrt(2), x = -sqrt(2), x = 1#

Explanation:

Rewrite the equation
#x^3 - x^2 - 2x + 2=0#
#x^2(x - 1) - 2(x - 1)=0#
#(x^2 - 2)(x - 1)=0#
#a^2 - b^2 = (a - b)(a+b)#
#x^2 - 2 = x^2 - (sqrt(2))^2 = (x - sqrt(2))(x + sqrt(2))#
#(x - sqrt(2))(x + sqrt(2))(x - 1) = 0#
#(x - sqrt(2)) = 0,(x + sqrt(2)) = 0, (x - 1) = 0 #
#x = sqrt(2), x = -sqrt(2), x = 1#

Jim G.
Mar 11, 2018

#(x-1)(x-sqrt2)(x+sqrt2)#

Explanation:

#"note that the coefficients "1-1-2+2=0#

#rArr(x-1)" is a factor"#

#"divide the polynomial by "(x-1)#

#color(red)(x^2)(x-1)cancel(color(magenta)(+x^2))cancel(-x^2)-2x+2#

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

#rArrx^3-x^2-2x+2=(x-1)(x^2-2)#

#x^2-2larrcolor(blue)"is a difference of squares"#

#•color(white)(x)a^2-b^2=(a-b)(a+b)#

#"with "a=x" and "b=sqrt2#

#rArrx^2-2=(x-sqrt2)(x+sqrt2)#

#rArrx^3-x^2-2x+2=0#

#rArr(x-1)(x-sqrt2)(x+sqrt2)=0#

#rArrx=1" or "x=+-sqrt2#

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