How do you divide $(x^3+7x^2-4x-1)/(3x-1)$ ?

Question

1535 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-05T14:34:22

$y = 1/3 x^2 + 22/9 x - 14/27 -41/27 frac{1}{3x - 1}$

Divide $x^3$ by $3x$ , the quotient is $1/3 x^2$

$y = 1/3 x^2 + frac{1/3 x^2 + 7x^2 - 4x - 1}{3x - 1}$

$y = 1/3 x^2 + frac{22/3 x^2 - 4x - 1}{3x - 1}$

Divide $22/3 x^2$ by $3x$ , the quotient is $22/9 x$

$y = 1/3 x^2 + 22/9 x + frac{22/9 x - 4x - 1}{3x - 1}$

$y = 1/3 x^2 + 22/9 x + frac{-14/9 x - 1}{3x - 1}$

Divide $14/9 x$ by $3x$ , the quotient is $-14/27$

$y = 1/3 x^2 + 22/9 x - 14/27 + frac{-14/27 - 1}{3x - 1}$

Degree above < degree below.

How do you divide (x^3+7x^2-4x-1)/(3x-1) (x^3+7x^2-4x-1)/(3x-1) ?