How does polynomial long division work?

I get normal long division but this doesn't make sense. How can doing it by only dividing by the leading term work? The problem is $$\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3},$$ not $$\dfrac{3x^3 - 2x^2 + 4x - 3} {x^2}.$$

Solution 1:

As said, "it works the same way as euclidean division works for integers" (quoting Bernard). So, if you take $$A=\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3}$$ The division of $3x^3$ by $x^2$ (the highest terms) gives $3x$. So $$A=3x+\Big(\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3}-3x\Big)=3x -\frac{11 x^2+5x+3}{x^2 + 3x + 3}=3x-B$$ Now, consider $$B=\frac{11 x^2+5x+3}{x^2 + 3x + 3}$$ The division of $11x^2$ by $x^2$ (the highest terms) gives $11$. So $$B=\frac{11 x^2+5x+3}{x^2 + 3x + 3}=11+\Big(\frac{11 x^2+5x+3}{x^2 + 3x + 3}-11\Big)=11-\frac{28 x+30}{x^2 + 3x + 3}=11-C$$ Now, the highest degree in the numerator of $C$ is smaller that the highest degree in its denominator so you stop and have the remainder. So, by the end $$A=\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3}=3x-11-\frac{28 x+30}{x^2 + 3x + 3}$$

Solution 2:

Some trick on how to do this quickly: $3x^3 - 2x^2 + 4x - 3 = 3x(x^2+3x+3) - 11x^2 - 5x - 3 = 3x(x^2+3x+3) - 11(x^2+3x+3) + 28x + 30 = (3x-11)(x^2+3x+3) + 28x+30$.