If the polynomial 3x² − x³ − 3x + 5 is divided by another polynomial x − 1 − x², the remainder comes out to be 3, the quotient polynomial is:
x − 2
Since the dividend \(-x^3+3x^2-3x+5\) has degree 3 and the divisor \(-x^2+x-1\) has degree 2, the quotient must be linear: let it be \(ax+b\).
Using Dividend = Divisor × Quotient + Remainder: \(-x^3+3x^2-3x+5 = (-x^2+x-1)(ax+b)+3\).
Expanding, \((-x^2+x-1)(ax+b) = -ax^3+(a-b)x^2+(b-a)x-b\).
Comparing the coefficient of \(x^3\): \(-a=-1\), so \(a=1\).
Comparing the coefficient of \(x^2\): \(a-b=3\), so \(1-b=3\), giving \(b=-2\); this is confirmed by the constant term, since \(-b=5-3=2\) also gives \(b=-2\).
So the quotient is \(x-2\).
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