Question

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(x^3 - 3x +2):(x^2 -2x + 1)

Answer 1

\(\dfrac{x^3-3x+2}{x^2-2x+1}=\dfrac{x^3-2x^2+x+2x^2-4x+2}{x^2-2x+1}\)

\(=\dfrac{x\left(x^2-2x+1\right)+2\left(x^2-2x+1\right)}{x^2-2x+1}=\dfrac{\left(x+2\right)\left(x^2-2x+1\right)}{x^2-2x+1}\)

\(=x+2\)

Answer 2

Answer 3

(x³ - 3x + 2) : (x² - 2x + 1)

= \(\dfrac{x^3-3x+2}{x^2-2x+1}=\dfrac{x^3-x-2x+2}{x^2-x-x+1}=\dfrac{x\left(x^2-1\right)-2\left(x-1\right)}{x\left(x-1\right)-\left(x-1\right)}=\dfrac{x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)}{\left(x-1\right)\left(x-1\right)}=\dfrac{\left(x-1\right).\left[x\left(x+1\right)-2\right]}{\left(x-1\right)^2}=\dfrac{x^2+x-2}{x-1}=\dfrac{x^2-x+2x-2}{x-1}=\dfrac{x\left(x-1\right)+2\left(x-1\right)}{\left(x-1\right)}=\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-1\right)}=x+2\)