Question 388968
{{{(x^(2)+7x+10)/(x+3)}}}  
     
     x +4
     ___________
x+3/x^(2)+7x+10
***-x^(2)-3x
****---------
***********4x+10
***********-4x-12
***********------
***************-2


 (The final answer is the quotient plus the remainder over the divisor)
{{{x+4-(2)/(x+3)}}}

doing more work to further the answer of x+4  :

x+4-(2)/(x+3)

Multiply each term by a factor of 1 that will equate all the denominators.  In this case, all terms need a denominator of (x+3).
x*(x+3)/(x+3)+4*(x+3)/(x+3)-(2)/(x+3)

Multiply the expression by a factor of 1 to create the least common denominator (LCD) of (x+3).
(x(x+3))/(x+3)+4*(x+3)/(x+3)-(2)/(x+3)

Multiply the expression by a factor of 1 to create the least common denominator (LCD) of (x+3).
(x(x+3))/(x+3)+(4(x+3))/(x+3)-(2)/(x+3)

The numerators of expressions that have equal denominators can be combined.  In this case, (x(x+3))/((x+3)) and ((4(x+3)))/((x+3)) have the same denominator of (x+3), so the numerators can be combined.
(x(x+3)+(4(x+3))-2)/(x+3)

Simplify the numerator of the expression.
(x^(2)+3x+4x+12-2)/(x+3)

Combine all similar terms in the polynomial x^(2)+3x+4x+12-2.
(x^(2)+7x+10)/(x+3)

In this problem 5*2=10 and 5+2=7, so insert 5 as the right hand term of one factor and 2 as the right-hand term of the other factor.
((x+5)(x+2))/(x+3)


Yet; your origional, long division answer= x+4 (ignore the ***'s)
I had to put them there to show you the work on an even scale.