Question

a division problem is shown below:
$x^{2}+1\enclose{longdiv}{2x^{4}-3x^{3}+6x^{2}-3x+9}$
complete the quotient below, including the remainder, by filling in the boxes with the correct values.
$\square x^{2}+\square x+\square$ r. $\square$

Explanation:

Step1: Divide leading terms

$\frac{2x^4}{x^2}=2x^2$

Step2: Multiply divisor by $2x^2$

$2x^2(x^2+1)=2x^4+2x^2$

Step3: Subtract from dividend

$(2x^4-3x^3+6x^2-3x+9)-(2x^4+2x^2)=-3x^3+4x^2-3x+9$

Step4: Divide new leading terms

$\frac{-3x^3}{x^2}=-3x$

Step5: Multiply divisor by $-3x$

$-3x(x^2+1)=-3x^3-3x$

Step6: Subtract from current polynomial

$(-3x^3+4x^2-3x+9)-(-3x^3-3x)=4x^2+9$

Step7: Divide new leading terms

$\frac{4x^2}{x^2}=4$

Step8: Multiply divisor by 4

$4(x^2+1)=4x^2+4$

Step9: Subtract to find remainder

$(4x^2+9)-(4x^2+4)=5$

Answer:

$2x^2 + (-3)x + 4$ r. $5$