Question

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

Answer 1

`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`

Answer 2

`(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`