we wish to divide

**Step – 1 **: Divide the first term of dividend i.e., x3 by the first term of divisor i.e., x to get \( x^3/x = x^2 \)

**Step – 2 **: Write the given polynomial on LIIS of the equality sign and on the right hand side write the product of x2 obtained in step I

and divisor (x —2) i.e., x2 (x —2)

**Step – 3** : Since \( x^2 (x —2) = x^3- 2x^3. \) So, we get the first term of the dividend and an additional term \( – 2x^2 .\) On the LHS of the equality, we need \( – 4x^2. \) To get \( – 4x^2, \) we require \( – 4×2 -(-2×2) = —2x^2 \) On RHS. So, we write — 2x (x — 2) as the second term on RHS.

**Step – 4 **: Since \( -2x (x -2) -2x^2 + 4x. \) This adjusts the second term on LHS and give an additional term 4x. We required 7x on LHS. To get 7x, on LHS, we require 7x — 4x = 3x on RHS. So, we write 3 (x —2) as the third term on RHS.

**Step – 5 **: Since \( 3 (x —2) = 3x —6. \) This adjusts the third term on LHS and gives an additional term – 6 on RHS. We require — 2 on LHS. So, we write -2-(-6)= 4 on RHS.

**Step – 6** : Take (x —2) common from first three terms to obtain x^2 — 2x ÷ 3 as the quotient and 4 as the remainder.

Hence, quotient \( x^2 — 2X + 3 \) and remainder = 4.

**Example**

**Divide** \( 12x^3 – 8x^2 – 6x + 10 by (3x – 2) . \) Also, write the quotient and the remainder.

**Solution** : \( 12x^3 – 8x^2 – 6x+l0

= 4x^2 (3x – 2) – 2(3x-2) + 6 \)

Hence, Quotient = \( 4x^2 – 2 \) and, Remainder = 6.