How do you divide (4n^2+7n-5)div(n+3)(4n2+7n−5)÷(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"one way is to use the divisor as a factor in the numerator
"consider the numerator"consider the numerator
color(red)(4n)(n+3)color(magenta)(-12n)+7n-54n(n+3)−12n+7n−5
=color(red)(4n)(n+3)color(red)(-5)(n+3)color(magenta)(+15)-5=4n(n+3)−5(n+3)+15−5
=color(red)(4n)(n+3)color(red)(-5)(n+3)+10=4n(n+3)−5(n+3)+10
"quotient "=color(red)(4n-5)," remainder "=10quotient =4n−5, remainder =10
rArr(4n^2+7n-5)/(n+3)=4n-5+10/(n+3)⇒4n2+7n−5n+3=4n−5+10n+3
"with restriction "n!=-3with restriction n≠−3