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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 insideand put the results above the division bar

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

 

 
 

 


combine the terms above the underline

we are done and have a remainer of 21

the answer is

Example 4

divide the first term of the binomial outside into the first term insideand put the results above the division bar

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

 
 

 


combine the terms above the underline

we are done and have a remainer of −14

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