Grade 8 Division of Polynomials

Step 1: Write the division as a fraction.
$ \frac{6x^3 - 9x^2 + 12x}{3x} $

Step 2: Divide each term in the numerator (dividend) by the monomial in the denominator (divisor). Remember the rule for dividing powers with the same base: subtract the exponents ($ a^m \div a^n = a^{m-n} $).
$ \frac{6x^3}{3x} - \frac{9x^2}{3x} + \frac{12x}{3x} $

Step 3: Perform the division for each term.
$ \frac{6x^3}{3x} = \frac{6}{3} \times \frac{x^3}{x^1} = 2 \times x^{3-1} = 2x^2 $
$ \frac{9x^2}{3x} = \frac{9}{3} \times \frac{x^2}{x^1} = 3 \times x^{2-1} = 3x $
$ \frac{12x}{3x} = \frac{12}{3} \times \frac{x^1}{x^1} = 4 \times x^{1-1} = 4 \times x^0 = 4 \times 1 = 4 $

Step 4: Combine the results of each term's division.
$ 2x^2 - 3x + 4 $

Step 5: The quotient is $ 2x^2 - 3x + 4 $. The remainder is 0.

Step 1: Write the division as a fraction.
$ \frac{10y^4 + 15y^3 - 20y^2 + 7}{5y^2} $

Step 2: Divide each term in the numerator by the monomial in the denominator.
$ \frac{10y^4}{5y^2} + \frac{15y^3}{5y^2} - \frac{20y^2}{5y^2} + \frac{7}{5y^2} $

Step 3: Perform the division for each term.
$ \frac{10y^4}{5y^2} = \frac{10}{5} \times \frac{y^4}{y^2} = 2 \times y^{4-2} = 2y^2 $
$ \frac{15y^3}{5y^2} = \frac{15}{5} \times \frac{y^3}{y^2} = 3 \times y^{3-2} = 3y $
$ \frac{20y^2}{5y^2} = \frac{20}{5} \times \frac{y^2}{y^2} = 4 \times y^{2-2} = 4 \times y^0 = 4 \times 1 = 4 $
$ \frac{7}{5y^2} $ cannot be divided to get a whole number or a term with a non-negative integer exponent in the numerator. This term will contribute to the remainder.

Step 4: Combine the results. The terms that could be divided form the quotient. The term that could not be divided cleanly forms the remainder over the divisor.
Quotient: $ 2y^2 + 3y - 4 $
Remainder: $ 7 $
Divisor: $ 5y^2 $

Step 5: The result can be written as Quotient $ + \frac{\text{Remainder}}{\text{Divisor}} $.
$ 2y^2 + 3y - 4 + \frac{7}{5y^2} $
The quotient is $ 2y^2 + 3y - 4 $ and the remainder is $ 7 $.

Step 1: Set up the long division. Arrange both the dividend and the divisor in descending powers of the variable.
$ \qquad \qquad \quad x + 2 \overline{) x^2 + 5x + 6} $

Step 2: Divide the first term of the dividend ($ x^2 $) by the first term of the divisor ($ x $). This gives the first term of the quotient ($ \frac{x^2}{x} = x $). Write this term above the dividend.
$ \qquad \qquad \quad \quad x $
$ \qquad \qquad \quad x + 2 \overline{) x^2 + 5x + 6} $

Step 3: Multiply the first term of the quotient ($ x $) by the entire divisor ($ x + 2 $). Write the result below the dividend and subtract it.
$ \qquad \qquad \quad \quad x $
$ \qquad \qquad \quad x + 2 \overline{) x^2 + 5x + 6} $
$ \qquad \qquad \quad \quad -(x^2 + 2x) $
$ \qquad \qquad \quad \quad \overline{\qquad \quad 3x + 6} $ (Subtracting $ x^2 + 2x $ from $ x^2 + 5x $)

Step 4: Bring down the next term of the dividend ($ +6 $).
$ \qquad \qquad \quad \quad x $
$ \qquad \qquad \quad x + 2 \overline{) x^2 + 5x + 6} $
$ \qquad \qquad \quad \quad -(x^2 + 2x) $
$ \qquad \qquad \quad \quad \overline{\qquad \quad 3x + 6} $

Step 5: Repeat the process. Divide the first term of the new dividend ($ 3x $) by the first term of the divisor ($ x $). This gives the next term of the quotient ($ \frac{3x}{x} = 3 $). Write this term above the dividend.
$ \qquad \qquad \quad \quad x + 3 $
$ \qquad \qquad \quad x + 2 \overline{) x^2 + 5x + 6} $
$ \qquad \qquad \quad \quad -(x^2 + 2x) $
$ \qquad \qquad \quad \quad \overline{\qquad \quad 3x + 6} $

Step 6: Multiply the new term of the quotient ($ 3 $) by the entire divisor ($ x + 2 $). Write the result below the current dividend ($ 3x + 6 $) and subtract it.
$ \qquad \qquad \quad \quad x + 3 $
$ \qquad \qquad \quad x + 2 \overline{) x^2 + 5x + 6} $
$ \qquad \qquad \quad \quad -(x^2 + 2x) $
$ \qquad \qquad \quad \quad \overline{\qquad \quad 3x + 6} $
$ \qquad \qquad \quad \quad -(3x + 6) $
$ \qquad \qquad \quad \quad \overline{\qquad \quad \quad \quad 0} $ (Subtracting $ 3x + 6 $ from $ 3x + 6 $)

Step 7: The process stops when the remainder is 0 or its degree is less than the degree of the divisor.
The quotient is $ x + 3 $ and the remainder is $ 0 $.