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Grade 8 - Mathematics
7.2 Division of Multinomial by Binomial

1. Arrange the terms of the dividend and divisor in decreasing order of powers leaving space for missing terms.
2. Divide the first term of the dividend by the first term of the divisor and write the result as the first term of the quotient.
3. Multiply the entire divisor by the first terms of the quotient and put the product under the dividend, keeping like terms under each other.
4. Subtract the product from the dividend and bring down the rest of the dividend.
5. Continue the division till the remainder becomes zero or the degree of the remainder becomes less than the divisor.

Example: Divide (8x2 - 14x + 3) by (2x - 3).
             4x - 1
2x - 3 |8x2 - 14x + 3
           8x2 - 12x
                   -2x + 3  
                   -2x + 3 

Therefore Quotient = 4x -1
Remainder = 0


Dividend = Quotient * Divisor + Remainder
In above problem quotient = 4x - 1, divisor = 2x - 3, Remainder = 0.
Therefore Dividend = (4x - 1)(2x - 3) + 0
= 8x2 - 12x - 2x + 3 + 0
= 8x2 - 14x + 3
= Dividend
Therefore the answer is correct.

Direction: Solve the following division problems. Also write at least ten examples of your own.

Name: ___________________


Grade 8 - Mathematics
7.2 Division of Multinomial by Binomial

Q 1: Divide (x3-9x2+27x-27) by (x-3).
Quotient=(x2-6x+9) and Remainder=0
Quotient=(x2-6x+9) and Remainder=6
Quotient=(x2-6x+9) and Remainder=2
None of these

Q 2: Divide (4x2-4x+1) by (2x-1).
None of these
Quotient=(2x-1) and Remainder=5
Quotient=(2x-1) and Remainder=1
Quotient=(2x-1) and Remainder=0

Q 3: Divide (x2+7x+2) by (x+3).
None of these
Quotient=(x+4) and Remainder=0
Quotient=(x+4) and Remainder=3
Quotient=(x+4) and Remainder=2

Q 4: Divide (16x3-46x2+39x-9) by (8x-3).
Quotient=(2x2-5x+3) and Remainder=9
Quotient=(2x2-5x+3) and Remainder=15
None of these
Quotient=(2x2-5x+3) and Remainder=0

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