Divide $3x^{3}+x^{2}+2x+5$ by $1+2x+x^{2}$.

(N/A) We first arrange the terms of the dividend and the divisor in the decreasing order of their degrees. This is called writing the polynomials in standard form. The dividend is already in standard form,and the divisor,in standard form,is $x^{2}+2x+1$.
Step $1$: To obtain the first term of the quotient,divide the highest degree term of the dividend $(3x^{3})$ by the highest degree term of the divisor $(x^{2})$. This gives $3x$. Multiplying the divisor by $3x$,we get $3x(x^{2}+2x+1) = 3x^{3}+6x^{2}+3x$. Subtracting this from the dividend,we get $(3x^{3}+x^{2}+2x+5) - (3x^{3}+6x^{2}+3x) = -5x^{2}-x+5$.
Step $2$: To obtain the second term of the quotient,divide the highest degree term of the new dividend $(-5x^{2})$ by the highest degree term of the divisor $(x^{2})$. This gives $-5$. Multiplying the divisor by $-5$,we get $-5(x^{2}+2x+1) = -5x^{2}-10x-5$. Subtracting this from the current dividend,we get $(-5x^{2}-x+5) - (-5x^{2}-10x-5) = 9x+10$.
Step $3$: The degree of the remainder $9x+10$ is $1$,which is less than the degree of the divisor $(2)$. Thus,the division process stops.
Therefore,the quotient is $3x-5$ and the remainder is $9x+10$.