what is the quotient of $(x^{3}-3x^{2}+3x - 2)div(x^{2}-x + 1)$?\n$x - 2$\n$x + 2$\n$x - 4$\n$x + 1$
what is the quotient of $(x^{3}-3x^{2}+3x - 2)div(x^{2}-x + 1)$?\n$x - 2$\n$x + 2$\n$x - 4$\n$x + 1$
Answer
Explanation:
Step1: Use polynomial long - division
We perform long - division of the polynomial (x^{3}-3x^{2}+3x - 2) by (x^{2}-x + 1). First, divide the leading term of the dividend (x^{3}-3x^{2}+3x - 2) (which is (x^{3})) by the leading term of the divisor (x^{2}-x + 1) (which is (x^{2})). The result is (x). Multiply (x^{2}-x + 1) by (x): (x(x^{2}-x + 1)=x^{3}-x^{2}+x). Subtract this from the dividend: ((x^{3}-3x^{2}+3x - 2)-(x^{3}-x^{2}+x)=x^{3}-3x^{2}+3x - 2 - x^{3}+x^{2}-x=-2x^{2}+2x - 2).
Step2: Divide the new leading term
Divide the leading term of (-2x^{2}+2x - 2) (which is (-2x^{2})) by the leading term of (x^{2}-x + 1) (which is (x^{2})). The result is (- 2). Multiply (x^{2}-x + 1) by (-2): (-2(x^{2}-x + 1)=-2x^{2}+2x - 2). Subtract this from (-2x^{2}+2x - 2): ((-2x^{2}+2x - 2)-(-2x^{2}+2x - 2)=0).
Answer:
(x - 2)