Question

rewrite
$\frac{3x^4 - 6x^3 - 25x^2 + 46x - 12}{x^2 - 4x + 2}$
in the form
$q(x)+\frac{r(x)}{x^2 - 4x + 2}$.
$\frac{3x^4 - 6x^3 - 25x^2 + 46x - 12}{x^2 - 4x + 2}=\square+\frac{\square}{x^2 - 4x + 2}$
question help: video read written example

Explanation:

Step1: Divide leading terms

Divide $3x^4$ by $x^2$ to get $3x^2$. Multiply divisor by $3x^2$:
$3x^2(x^2-4x+2)=3x^4-12x^3+6x^2$
Subtract from dividend:
$(3x^4-6x^3-25x^2+46x-12)-(3x^4-12x^3+6x^2)=6x^3-31x^2+46x-12$

Step2: Divide new leading terms

Divide $6x^3$ by $x^2$ to get $6x$. Multiply divisor by $6x$:
$6x(x^2-4x+2)=6x^3-24x^2+12x$
Subtract:
$(6x^3-31x^2+46x-12)-(6x^3-24x^2+12x)=-7x^2+34x-12$

Step3: Divide next leading terms

Divide $-7x^2$ by $x^2$ to get $-7$. Multiply divisor by $-7$:
$-7(x^2-4x+2)=-7x^2+28x-14$
Subtract:
$(-7x^2+34x-12)-(-7x^2+28x-14)=6x+2$

Answer:

$\frac{3x^4 - 6x^3 - 25x^2 + 46x - 12}{x^2 - 4x + 2} = 3x^2+6x-7 + \frac{6x+2}{x^2 - 4x + 2}$