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

1 Answer
Feb 8, 2016

(x-1)^2

Explanation:

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^3-4x^2. Subtracting this from x^3-6x^2+9x-4 yields the initial remainder -2x^2+9x-4.

Once again one needs to divide this by x-4. To get the initial term -2x^2 one needs to multiply x-4 by -2x. Doing this multiplication one gets -2x^2+8x. 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^2-2x+1, which is equal to (x-1)^2.