How do you divide #(4n^2+7n-5)div(n+3)# and identify any restrictions on the variable?

1 Answer
Nov 1, 2017

Answer:

#4n-5+10/(n+3)#

Explanation:

#"one way is to use the divisor as a factor in the numerator"#

#"consider the numerator"#

#color(red)(4n)(n+3)color(magenta)(-12n)+7n-5#

#=color(red)(4n)(n+3)color(red)(-5)(n+3)color(magenta)(+15)-5#

#=color(red)(4n)(n+3)color(red)(-5)(n+3)+10#

#"quotient "=color(red)(4n-5)," remainder "=10#

#rArr(4n^2+7n-5)/(n+3)=4n-5+10/(n+3)#

#"with restriction "n!=-3#