Question
By how much does $3 x^2-5 x+6$ exceed $x^3-x^2+4 x-1$ ?
✓
Answer
$=\left(3 x^2-5 x+6\right)-\left(x^3-x^2+4 x-1\right)$
$=3 x^2-5 x+6-x^3+x^2-4 x+1$
$=-x^3+4 x^2-9 x+7$
$=3 x^2-5 x+6-x^3+x^2-4 x+1$
$=-x^3+4 x^2-9 x+7$
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