Dividing Polynomials
Example 3 using a 0x as a place holder for a missing term
![]() |
divide the first term of the binomial outside
into the first term inside![]() |
multiply the x on top by the x + 4 outside and write the product of x( x + 4) under the trinominal
multiply EACH term in the new row by −1 to create a subtraction problem
combine the terms above
the underline and bring down the + 5
Start the process over again with the − 4 x + 5 term and the x + 4 outside
divide the first term
of the binomial outside into the first term of − 4 x +1
multiply the − 4 on top by the x + 4 outside − 4(x + 4) and write the product under the − 4 x + 5
multiply EACH term in the new row by −1 to create a subtraction problem
![]() |
|
|
the answer is
Example 4
![]() |
divide the first term of the binomial outside
into the first term inside![]() |
multiply the x on top by the 2x + 3 outside and write the product of 3x(2x + 3) under the trinominal
multiply EACH term in the new row by −1 to create a subtraction problem
combine the terms above
the underline and bring down the − 5
Start the process over again with the 6x − 5 term and the 2x + 3 outside
divide the first term of the binomial outside
into the first term of
multiply the 3 on top by the 2x + 3 outside 3(2x + 3) and write the product under the 6x −5
multiply EACH term in the new row by −1 to create a subtraction problem
![]() |
|
|
the answer is
divide the 2 outside
into the first number inside and put the results over the 9
multiply the 4 on top by the 2 outside and write the product 8 under the 9
multiply the 8 in the
new row by −1 to create a subtraction problem
combine the terms above
the underline and bring down the 7
Start the process over again
the answer is