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### Grade 7 - Mathematics4.14 Division of Multinomial by Multinomial

Example: Divide (a4 - 12a2 + 5a + 17) by (a2 + 2a + 5).
Solution:  In the above problems we missed a3 term so we leave that place blank.
 a2-2a-13a2+5a+5 | a4 + ..... - 12a2 + 5a + 17                  a4 + 2a3 + 5a2                        -2a3 - 17a2 + 5a                        -2a3 - 4a2 - 10                                -13a2 + 15a + 17                                  -13a2 - 26a - 65                                               41a + 82  Therefore the degree of the remainder is less than the divisor. Quotient = a2 - 2a -13Remainder = 41a + 82

Directions: Solve the following division problems. Also write at least five examples of your own.

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### Grade 7 - Mathematics4.14 Division of Multinomial by Multinomial

 Q 1: Divide (a4+a3+6a2+9a+7) by (a2+a+2).Quotient=(a2+3) and Remainder=(5a-1)Quotient=(a2+4) and Remainder=(5a-1)Quotient=(a2+4) and Remainder=(5a+1) Q 2: Divide (4a4+2a3-12a2-9a+9) by (2a2-a+4).Quotient=(2a2+2a-9) and Remainder=(-26a-45)Quotient=(2a2+2a-9) and Remainder=(-26a+45)Quotient=(2a2+2a+9) and Remainder=(-26a+45) Q 3: Divide (6a4+4a3-12a2-9a+6) by (3a2+2a-2).Quotient=(2a2+2) and Remainder=(-2a2-5a+2)Quotient=(2a2-2) and Remainder=(-2a2-5a-2)Quotient=(2a2-2) and Remainder=(-2a2-5a+2) Q 4: Divide (4a4-8a2+6a-4) by (2a2+2a+1).Quotient=(2a2+2a+7) and Remainder=(18a+3)Quotient=(2a2-2a-3) and Remainder=(14a-1)Quotient=(2a2+2a-7) and Remainder=(18a+3)Quotient=(2a2+2a-7) and Remainder=(18a-3) Question 5: This question is available to subscribers only! Question 6: This question is available to subscribers only!