How do you find the quotient of #(a^3+4a^2+7a+6) /( a+2)#?

Question

Eva Adamson · Accepted Answer

Start with dividing the term with the highest exponent in the numerator by the term with the highest exponent in the denominator, so here, divide #a^3# by #a#: 
#a^3 / a = a^2#. So, your first term needs to be #a^2(a+2)#. 
#=># Transform the numerator into #a^2(a+2) + 2a^2 + 7a + 6#.  Always pay attention that the value of the numerator doesn't change - here, you have "splitted" the term #4a^2# into the part in #a^2(a+2)# and the rest: #2a^2#. So far, you've got:
#(a^3 + 4a^2 + 7a + 6) / (a+2)#
# = (a^2(a+2) + 2a^2 + 7a + 6) / (a+2)#
# = (a^2(a+2)) / (a+2) + (2a^2 + 7a + 6) / (a+2)#
# = a^2 + (2a^2 + 7a + 6) / (a+2)# Your fraction is smaller now. Proceed in the same way:
- divide #2a^2# by #a#, result: #(2a^2) / a = 2a#
- create the expression #2a(a+2)# in the numerator. 
- take #4a# (part of your new expression) from #7a# Now you have:
# a^2 + (2a^2 + 7a + 6) / (a+2)#
#= a^2 + (2a(a+2) + 3a + 6) / (a+2)#
# = a^2 + (2a(a+2)) / (a+2) + (3a + 6) / (a+2)#
# = a^2 + 2a + (3a + 6) / (a+2)# The last part is easy: #3a# divided by #a# is #3#, and the last expression can be factorized cleanly in #3(a+2) = 3a + 6#. # a^2 + 2a + (3a + 6) / (a+2)#
# = a^2 + 2a + (3(a+2)) / (a+2)#
# = a^2 + 2a + 3# This is the outcome you ultimately achieved: #(a^3 + 4a^2 + 7a + 6) / (a+2) = a^2 + 2a + 3#

Isaiah Anthony · Answer

undefined To find the quotient of (a^3+4a^2+7a+6) /( a+2), you can use long division or synthetic division.

Using long division:
- Divide a^3 by a to get a^2.
- Multiply a^2 by (a+2) to get a^3+2a^2.
- Subtract a^3+2a^2 from a^3+4a^2 to get 2a^2.
- Bring down the 7a term.
- Divide 2a^2 by a to get 2a.
- Multiply 2a by (a+2) to get 2a^2+4a.
- Subtract 2a^2+4a from 2a^2+7a to get 3a.
- Bring down the 6 term.
- Divide 3a by a to get 3.
- Multiply 3 by (a+2) to get 3a+6.
- Subtract 3a+6 from 3a+6 to get 0.

Therefore, the quotient is a^2+2a+3.