**Division Algorithm For Polynomials**

If p(x) and g(x) are any two polynomials with

g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that

p(x) = q(x) × g(x) + r(x)

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

The result is called Division Algorithm for polynomials.

**Dividend = Quotient × Divisor + Remainder**

## Polynomials – Long Division

**Working rule to Divide a Polynomial ****by Another Polynomial:**

**Step 1:** First arrange the term of dividend and the divisor in the decreasing order of their degrees.

**Step 2:** To obtain the first term of quotient divide the highest degree term of the dividend by the highest degree term of the divisor.

**Step 3:** To obtain the second term of the quotient, divide the highest degree term of the new dividend obtained as remainder by the highest degree term of the divisor.

**Step 4:** Continue this process till the degree of remainder is less than the degree of divisor.

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**Division Algorithm For Polynomials With Examples**

**Example 1:** Divide 3x^{3} + 16x^{2} + 21x + 20 by x + 4.

**Sol. **

Quotient = 3x^{2} + 4x + 5

Remainder = 0

**Example 2:** Apply the division algorithm to find the quotient and remainder on dividing p(x) by g(x) as given below :

p(x) = x^{3} – 3x^{2} + 5x – 3 and g(x) = x^{2} – 2

**Sol. **We have,

p(x) = x^{3} – 3x^{2} + 5x – 3 and g(x) = x^{2} – 2

We stop here since

degree of (7x – 9) < degree of (x^{2} – 2)

So, quotient = x – 3, remainder = 7x – 9

Therefore,

Quotient × Divisor + Remainder

= (x – 3) (x^{2} – 2) + 7x – 9

= x3 – 2x – 3x^{2} + 6 + 7x – 9

= x^{3} – 3x^{2} + 5x – 3 = Dividend

Therefore, the division algorithm is verified.

**Example 3:** Apply the division algorithm to find the quotient and remainder on dividing p(x) by g(x) as given below

p(x) = x^{4} – 3x^{2} + 4x + 5, g (x) = x^{2} + 1 – x

**Sol. **We have,

p(x) = x^{4} – 3x^{2} + 4x + 5, g (x) = x^{2} + 1 – x

We stop here since

degree of (8) < degree of (x^{2} – x + 1).

So, quotient = x^{2} + x – 3, remainder = 8

Therefore,

Quotient × Divisor + Remainder

= (x^{2} + x – 3) (x^{2} – x + 1) + 8

= x^{4} – x^{3} + x^{2} + x^{3} – x^{2} + x – 3x^{2} + 3x – 3 + 8

= x^{4} – 3x^{2} + 4x + 5 = Dividend

Therefore the Division Algorithm is verified.

**Example 4:** Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm. t^{2} – 3; 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12.

**Sol. **We divide 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12 by t^{2} – 3

Here, remainder is 0, so t^{2} – 3 is a factor of 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12.

2t^{4} + 3t^{3} – 2t^{2} – 9t – 12 = (2t^{2} + 3t + 4) (t^{2} – 3)

**Example 5:** Obtain all the zeroes of 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5, if two of its zeroes are and .

**Sol. **Since two zeroes are and

x = , x =

Or 3x^{2} – 5 is a factor of the given polynomial.

Now, we apply the division algorithm to the given polynomial and 3x^{2} – 5.

So, 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5

= (3x^{2} – 5) (x^{2} + 2x + 1) + 0

Quotient = x^{2} + 2x + 1 = (x + 1)^{2}

Zeroes of (x + 1)^{2} are –1, –1.

Hence, all its zeroes are , , –1, –1.

**Example 6:** On dividing x^{3} – 3x^{2} + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x).

**Sol. **p(x) = x^{3} – 3x^{2} + x + 2 q(x) = x – 2 and r (x) = –2x + 4

By Division Algorithm, we know that

p(x) = q(x) × g(x) + r(x)

Therefore,

x^{3} – 3x^{2} + x + 2 = (x – 2) × g(x) + (–2x + 4)

⇒ x^{3} – 3x^{2} + x + 2 + 2x – 4 = (x – 2) × g(x)

On dividing x^{3} – 3x^{2} + x + 2 by x – 2,

we get g(x)

Hence, g(x) = x^{2} – x + 1.

**Example 7:** Give examples of polynomials p(x), q(x) and r(x), which satisfy the division algorithm and

(i) deg p(x) = deg q(x)

(ii) deg q(x) = deg r(x)

(iii) deg q(x) = 0

**Sol. **

**(i)** Let q(x) = 3x^{2} + 2x + 6, degree of q(x) = 2

p(x) = 12x^{2} + 8x + 24, degree of p(x) = 2

Here, deg p(x) = deg q(x)

**(ii)** p(x) = x^{5} + 2x^{4} + 3x^{3}+ 5x^{2} + 2

q(x) = x^{2} + x + 1, degree of q(x) = 2

g(x) = x^{3} + x^{2} + x + 1

r(x) = 2x^{2} – 2x + 1, degree of r(x) = 2

Here, deg q(x) = deg r(x)

**(iii) ** Let p(x) = 2x^{4} + x^{3} + 6x^{2} + 4x + 12

q(x) = 2, degree of q(x) = 0

g(x) = x^{4} + 4x^{3} + 3x^{2} + 2x + 6

r(x) = 0

Here, deg q(x) = 0

**Example 8:** If the zeroes of polynomial x^{3} – 3x^{2} + x + 1 are a – b, a , a + b. Find a and b.

**Sol. ∵ ** a – b, a, a + b are zeros

∴ product (a – b) a(a + b) = –1

⇒ (a^{2} – b^{2}) a = –1 …(1)

and sum of zeroes is (a – b) + a + (a + b) = 3

⇒ 3a = 3 ⇒ a = 1 …(2)

by (1) and (2)

(1 – b^{2})1 = –1

⇒ 2 = b^{2} ⇒ b = ± √2

∴ a = –1 & b = ± √2

**Example 9:** If two zeroes of the polynomial x^{4} – 6x^{3} –26x^{2} + 138x – 35 are 2 ± √3, find other zeroes.

**Sol. ****∵ **2 ± √3 are zeroes.

∴ x = 2 ± √3

⇒ x – 2 = ±(squaring both sides)

⇒ (x – 2)^{2} = 3 ⇒ x^{2} + 4 – 4x – 3 = 0

⇒ x^{2} – 4x + 1 = 0 , is a factor of given polynomial

∴ other factors

∴ other factors = x^{2} – 2x – 35

= x^{2} – 7x + 5x – 35 = x(x – 7) + 5(x – 7)

= (x – 7) (x + 5)

∴ other zeroes are (x – 7) = 0 ⇒ x = 7

x + 5 = 0 ⇒ x = – 5

**Example 10:** If the polynomial x^{4} – 6x^{3} + 16x^{2} –25x + 10 is divided by another polynomial x^{2} –2x + k, the remainder comes out to be x + a, find k & a.

**Sol. **

According to questions, remainder is x + a

∴ coefficient of x = 1

⇒ 2k – 9 = 1

⇒ k = (10/2) = 5

Also constant term = a

⇒ k^{2} – 8k + 10 = a ⇒ (5)^{2} – 8(5) + 10 = a

⇒ a = 25 – 40 + 10

⇒ a = – 5

∴ k = 5, a = –5

kipsang William says

imagine I have not understood how to calculate from example 2 ,3 &4 ? please is there any other easy method on how to calculate it ?

aakash says

Very easy questions