# How do you divide (x^3-6x^2+9x-4) / (x-4)  using polynomial long division?

${\left(x - 1\right)}^{2}$
To get the initial term, namely ${x}^{3}$, one needs to multiply the divisor, namely $x - 4$ by ${x}^{2}$. Doing this multiplication, of $x - 4$ by ${x}^{2}$, one gets ${x}^{3} - 4 {x}^{2}$. Subtracting this from ${x}^{3} - 6 {x}^{2} + 9 x - 4$ yields the initial remainder $- 2 {x}^{2} + 9 x - 4$.
Once again one needs to divide this by $x - 4$. To get the initial term $- 2 {x}^{2}$ one needs to multiply $x - 4$ by $- 2 x$. Doing this multiplication one gets $- 2 {x}^{2} + 8 x$. Subtracting this from the initial remainder one gets $x - 4$.
One needs to divide this second remainder by $x - 4$. Clearly, $x - 4$ times 1 is $x - 4$ and there is no remainder. So, by adding up the multipliers, one gets ${x}^{2} - 2 x + 1$, which is equal to ${\left(x - 1\right)}^{2}$.