The division of polynomials p(x) and g(x) is expressed by the following “division algorithm” of algebra.

Let p(x) and g(x) be two polynomials such that degree of p(x) ≥ degree of g(x) and g(x) ≠ 0. Then there exists unique polynomials q(x) and r (x)

such that

p(x) = g(x) q(x) + r (x) ... (1)

where r (x) = 0 or degree of r (x) < degree of g(x) .

The polynomial p(x) is the dividend, g(x) is the divisor, q(x) is the quotient and r (x) is the remainder.

(1) ==> Dividend = (Divisor x Quotient) + Remainder

**Example 1 :**

**Divide the polynomial 2x ^{3}** - 6

**Solution : **

Let P(x) = 2**x ^{3}** - 6

To divide the given polynomial by x - 2, we have divide the first term of the polynomial P(x) by the first term of the polynomial g(x).

If we divide 2**x ^{3}** by x, we get 2

Now we have to subtract 2**x ^{3}** - 4

Now we have to subtract 2**x ^{3}** - 4

repeat this process until we get the degree of p(x) ≥ degree of g(x).

So,

Quotient = 2x^{2} - 2x + 1

Remainder = 6

**Example 2 :**

Find the quotient and remainder when 4x^{3} - 5x^{2} + 6x - 2 by x - 1.

**Solution :**

So,

Quotient = 4x^{2} - x + 5

Remainder = 3

**Example 3 :**

Find the quotient and remainder when x^{3} - 7x^{2} - x + 6 by x + 2.

**Solution :**

So,

Quotient = x^{2} - 9x + 17

Remainder = -28

**Example 4 :**

Find the quotient and the remainder when 10- 4x + 3x^{2} is divided by x - 2.

**Solution :**

Let us first write the terms of each polynomial in descending order ( or ascending order).

Thus, the given problem becomes (10- 4x + 3x^{2}) ÷ (x - 2)

f(x) = 10- 4x + 3x^{2}

= 3x^{2} - 4x + 10

g(x) = x - 2

**Step 1 :**

In the first step, we are going to divide the first term of the dividend by the first first term of the divisor.

After changing the signs, +3x^{2} and -3x^{2} will get canceled. By simplifying, we get 2x + 10.

**Step 2 :**

In the second step again we are going to divide the first term that is 2x by the first term of divisor that is x.

So,

Quotient = 3x + 2

Remainder = 14

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