How do you divide #(4n^2+7n-5)div(n+3)# and identify any restrictions on the variable?
1 Answer
Nov 1, 2017
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#
