Question

3a²b+6ab²-3ab+6a ÷ 3a By long division method with steps

Answer 1

you can also get the answer by removing common .

3a2b + 6ab2 - 3ab + 6a ÷ 3a
3a( ab + 2b2 - b + 2 ) ÷ 3a

so 3a from numerator and denominator get cancelled out.

Therefore the remainder is
$ab + 2 {b}^{2} - b + 2$

Answer 2

Given:

The given expression is 3a²b+6ab²-3ab+6a.

Find:

(3a²b+6ab²-3ab+6a)÷3a by long division method.

Answer:

(ab + 2b² - b + 2)

Solution:

Long Division Method:

Dividend = 3a²b+6ab²-3ab+6a

Divisor = 3a

3a $\overline{)3a^2b + 6ab^2 - 3ab + 6a(}$ab + 2b² - b + 2

    +3a²b            

           0 + 6ab²

              - 6ab²        

                     0   - 3ab

                          - 3ab

                          +                

                               0  +  6a

                                    -  6a

                                         0

Quotient = ab + 2b² - b + 2

Remainder = 0

Verification:

Dividend = Divisor × Quotient + Remainder

3a²b+6ab²-3ab+6a = 3a × (ab + 2b² - b + 2) + 0

3a²b+6ab²-3ab+6a = 3a²b+6ab²-3ab+6a + 0

3a²b+6ab²-3ab+6a = 3a²b+6ab²-3ab+6a

L.H.S. = R.H.S.

Hence verified.

Hence, (3a²b+6ab²-3ab+6a) ÷ 3a = (ab + 2b² - b + 2).

     

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