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Grade 7 - Mathematics
4.14 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+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 -13
Remainder = 41a + 82

Directions: Solve the following division problems. Also write at least five examples of your own.
Q 1: 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 2: 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)

Q 3: Divide (9a4-6a3+18a2-2a-5) by (3a2-2a+2).
Quotient=(3a2-4) and Remainder=(6a-13)
Quotient=(3a2+4) and Remainder=(6a+13)
Quotient=(3a2+4) and Remainder=(6a-13)

Q 4: Divide (a4-4a3-2a2-9a-4) by (a2+2a+1).
Quotient=(a2+6a+9) and Remainder=(-21a-13)
Quotient=(a2-6a+9) and Remainder=(-21a-13)
Quotient=(a2-6a+9) and Remainder=(-21a+13)

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