How do you find the quotient #(3q^2+20q+11)div(q+6)# using long division?

#" "3q^2+20q+11#
#color(magenta)(3q)(q+6)->ul(3q^2 +18q) larr" Subtract"#
#" "color(white)(.)0+color(white)(2)2q+11#
#color(magenta)(color(white)(2)2)(q+6)->" "ul(2q+12 ) larr" Subtract"#
#" "0color(magenta)(-1 larr" Remainder")#

#color(magenta)(3q+2-1/(q+6))#

Given:

#color(white)( (q + 6)/color(black)(q + 6))color(white)((d + e + f))/("|" color(white)(x)3q^2 + 20q + 11)#

Find the first term in the quotient by dividing first term in the dividend by the first term in the divisor:

#(3q^2)/q = 3q#

Write the first term in the quotient:

#color(white)( (q + 6)/color(black)(q + 6))(3qcolor(white)(d+ e + f))/("|" color(white)(x)3q^2 + 20q + 11)#

Multiply that term by the divisor #3q(q+6) = 3q^2+18q# subtract this from the dividend:

#color(white)( (q + 6)/color(black)(q + 6))(3qcolor(white)(d+ e + f))/("|" color(white)(x)3q^2 + 20q + 11)#
#color(white)("..........")ul(-3q^2-18q#
#color(white)(".........................")2q+ 11#

Find the next term in the quotient by dividing the first non-zero term in the results of the subtraction by the first term in the divisor:

#(2q)/q = 2#

Write this term into the quotient:

#color(white)( (q + 6)/color(black)(q + 6))(3q+2color(white)(e. + f))/("|" color(white)(x)3q^2 + 20q + 11)#
#color(white)("..........")ul(-3q^2-18q#
#color(white)(".........................")2q+ 11#

Multiply that term by the divisor #2(q+6) = 2q+12# subtract this from the dividend:

#color(white)( (q + 6)/color(black)(q + 6))(3q+2color(white)(e. + f))/("|" color(white)(x)3q^2 + 20q + 11)#
#color(white)("..........")ul(-3q^2-18q#
#color(white)(".........................")2q+ 11#
#color(white)("......................")ul(-2q-12#
#color(white)("................................")-1#

Because the power of the divisor is greater than the power of the results of the subtraction, we know that we are done and that #-1# is a remainder.

The quotient can be expressed as:

#3q+2-1/(q+6)#