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# Value of Polynomial and Division Algorithm - 1 Mathematics Notes | EduRev

## Algebra for IIT JAM Mathematics

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## Mathematics : Value of Polynomial and Division Algorithm - 1 Mathematics Notes | EduRev

The document Value of Polynomial and Division Algorithm - 1 Mathematics Notes | EduRev is a part of the Mathematics Course Algebra for IIT JAM Mathematics.
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Value of Polynomial and Division Algorithm
Arithmetic operations like addition, subtraction, multiplication and division play a huge and most basic rule in Mathematics. Maths is made by these operations. All other operations go easy with the polynomials except the division operation, which gets complex when dealt with polynomials. But this section will explain to you the division of polynomials and the division algorithm related to it, from basics.
So, what’s the basic formula we are learning from the day we solved our first division problem? This is:

Dividend = Quotient × Divisor + Remainder
Example: Divide the polynomial 2x2+3x+1 by polynomial x+2.
Solution: Divisor= x+2
Dividend=2x2 + 3x + 1
Note: Put the dividend under the division sign and divisor outside the sign. Steps for Division of Polynomials

• Step 1: Firstly, Arrange the divisor as well as dividend individually in decreasing order of their degree of terms.
• Step 2: In case of division we seek to find the quotient. To find the very first term of the quotient, divide the first term of the dividend by the highest degree term in the divisor. In the current case,
2x2/x = 2x.
• Step 3: Write 2x in place of the quotient.
• Step 4: Multiply the divisor by the quotient obtained. Put the product underneath the dividend.
• Step 5: Subtract the product obtained as happens in case of a division operation.
• Step 6: Write the result obtained after drawing another bar to separate it from prior operations performed.
• Step 7: Bring down the remaining terms of the dividend.
• Step 8: Again divide the dividend by the highest degree term of the remaining divisor. Follow the same prior procedure until either the remainder becomes zero or its degree is less than the degree of the divisor.
• Step 9: At this stage, the quotient obtained is our answer.
Quotient Obtained = 2x + 1

Note: Division Algorithms for Polynomials is same as the Long Division Algorithm In Polynomials

Division Algorithm For Polynomials
Division algorithm for polynomials states that, suppose f(x) and g(x) are the two polynomials, where g(x)≠0, we can write:
f(x) = q(x) g(x) + r(x)
which is same as the Dividend = Divisor * Quotient + Remainder and where r(x) is the remainder polynomial and is equal to 0 and degree r(x) < degree g(x).
Verification of Division Algorithm

Take the above example and verify it.
Divisor = x+2
Dividend = 2x2 + 3x + 1
Quotient = 2x – 1
Remainder = 0

Applying the Algorithm:
2x2 + 3x + 1 = (x + 2) (2x + 1) + 0
2x2 + 3x + 1 = 2x2 + 3x + 1

Hence verified.

Finding Factors of Polynomials with Division Algorithm
Long division algorithm is used to find out factors of polynomials of degree greater than equal to two. We’ll be describing the steps to find out the factors along with an example.

Example: Find roots of cubic polynomial P(x)=3x3 – 5x2 – 11x – 3
Solution

• Step 1: Use the factor theorem to find a factor of the polynomial.
• Step 2: First divide the whole equation by the coefficient of the highest degree term of the dividend.
P(x)=3x3 – 5x2 – 11x – 3
On dividing the whole equation by 3,
P(x) =x3 – (5/3)x2 – (11/3)x – 1
• Step 3: Find out factors of the constant term so obtained. In the present case, factors of the constant term are 1 and -1
• Step 4: Put the value of x in P(x) = 3x3 – 5x2 – 11x – 3   equal to 1 and find the remainder. Again put the value of remainder equal to -1 in and find the remainder using remainder theorem. Find the value of x for which remainder is zero for the cubic polynomial.
P (1) = 3(1)3 – 5(1)2 – 11(1) – 3  = -16
P(-1) = 3( -1 )3 – 5( -1 )2 – 11( -1 ) – 3 =0
• Step 5: Remainder is zero for x = -1. So, (x + 1) is a root of the polynomial.
• Step 6: By Division Algorithm, find out the quotient. It comes out:  3x2 – 8x – 3
• Step 7: Now, Quotient = 3x2 – 8x – 3

Dividend = (Divisor) * (Quotient) + Remainder
In present case,
3x3 – 5x2 – 11x – 3 = (x + 1) (3x– 8x – 3) + 0

By factorizing the quadratic polynomial we shall be able to find out remaining factors of the cubic polynomial.

• Step 8: Break middle term in terms of a pair of numbers such that its product is equal to -9 and summation equal to -3.
• Step 9: On factorizing,  possible pair of number satisfying both conditions is (-9, 1). Breaking the middle term,
f(x) = 3x– 8x – 3
= 3x– 9x + x – 3
• Step 10: Form pairs of terms and factor out GCD of the two pairs separately. Then again factor out GCD of the remaining two products.
• Step 11:
f(x) = 3x– 8x – 3 = 3x– 9x + x – 3
= 3x(x – 3) + 1(x – 3) = (x – 3)(3x + 1)
Now,
3x3 – 5x2 – 11x – 3 = (x + 1) (3x– 8x – 3) + 0
= (x + 1) ( x – 3)(3x + 1)
Factors of cubic polynomial are -1, 3 and -1/3.

Factorisation of Polynomials
Do you know what factorization of polynomials means? Now that you have an idea of what polynomials are,  let’s learn how to factorize a polynomial. This is an important step towards solving an equation in mathematics. Let’s find out.
Before starting, let’s try to solve an example. If you are given 12 chocolates, choose the correct option in which you can divide them, such that you are left with no chocolates:

• Group of 3 people
• Group of 4 people
• Group of 6 people

Well, the answer is, any one of the options will do. Why so? Well, you know that 3, 4 and 6 are factors of 12 from your knowledge of number system, which you have learned long back. So, we can divide the 12 chocolates into these groups.

Factors
What are the factors? When it comes to integers, if a number has an integral value and that particular value get divided completely by another number(s) without leaving any remainder other than zero, then the dividend is said to be a factor(s) of that integer.

OR

When an integer could be written as a product of two or more integers each then such numbers will be its factor. Why not consider some examples to understand more. What can be the factors for 12?
12 = 6 × 2
12 = 4 × 3
12 = 12 × 1

So, you can see that a number can be factorized in more than one way. But it’s not applicable to all the numbers. For a number like 7, 3, 5, 11 (which are prime numbers), these numbers have only one factor other than itself.

7 = 7 × 1

So, it can be factorized in only one possible way.

Factorization of Polynomials
The same case of numbers also exists with Polynomials. A polynomial can be written as a product of two or more polynomials of degree less than or equal to that of it. Each polynomial involved in the product will be a factor of it. The process involved in breaking a polynomial into the product of its factors is known as the factorization of polynomials. As I’ve told in the previous sections that Monomial, Binomials are just the other names for Polynomials. Its always good to start with the smaller and easiest one. So, let’s hit the Monomials.

Monomials
Monomials can be factorized in the same way as integers, just by writing the monomial as the product of its constituent prime factors. In the case of monomials, these prime factors can be integers as well as other monomials which cannot be factorized further. Factorize:

• a3 = (a) × (a) × (a)
• 3abc = (3) × (a) × (b) × (c)

We will factorize monomials with binomials, trinomial and polynomial. But before doing that, we will brush up concepts on GCD or HCF.

GCD or HCF
For a given set of numbers, the Greatest Common Number that will divide each of the numbers will be the GCD of that particular set of numbers. It is also known as HCF of numbers .i.e. Highest Common Factor.

Steps To Calculate the GCD of Integers

• Step 1: Break the number into the product of its prime factors.
• Step 2: Identify the common factors for the given set of numbers.
• Step 3: The product of common factors will be gcd of the number set.
• Step 4: If no common factor is found choose 1 as a common factor.

Example: Find GCD of 15 and 24.

15 = 3 x 5
24 = 2 x 2 x 2 x 3

GCD of 15 and 24 is 3. This same procedure applies to polynomials.

Steps To Calculate the GCD of Polynomials

• Step 1: For a given set of polynomials, break the polynomial into its factors such that each factor polynomial cannot be factorized further.
• Step 2: Identify common terms or polynomials for a given set of polynomials.

Example: Find GCD of 15ab and 3bc.

15ab = 3 x (5) x (a) x (b)
23bc = 3 x (b) x (c)

Here, GCD of 15ab and 3bc is 3b.

Factoring Binomials
Follow the 4 easy steps to factorize Binomials:

• Step 1: Break each term into its prime factors.
• Step 2: Find common factors or GCD for all individual terms.
• Step 3: Factor out the common factor or GCD.
• Step 4: The remaining terms in individual terms will form another polynomial. Put them in separate bracket multiplied by the GCD factored out initially.
• Step 5: The resulting product of terms will be the factors of the initial polynomial (binomial).

Example: Factorize 15ab + 3bc
= 3 × (5) × (a) × (b) + 3 × (b) × (c) = 3b (5a + c)

Case 1: When the coefficient of x2 is unity.

The general form of an equation is x2+bx+c. Every quadratic equation can be expressed as: x2+bx+c= (x+d)(x+e). Here, b is the sum of d and e & c is the product of d and e. Example: (x+2)(x+3) = x2+ 2x + 3x + 6 = x2+ 5x + 6.

Here, 5 = 2 + 3 = d + e = b in general form and 6 = 2 × 3 = d × e = c in general form. To factorize quadratic polynomial, we shall be looking for numbers which on multiplication will get equal to c and on summation equal to b.

Example: Factorize x2+8x+12.
Solution: Below the steps are given for your understanding.

• Step 1: Factorize 12 as 12 = 2 × 6 or = 4 × 3. We have to find a pair, such that its product is equal to 12 and summation is equal to 8. Only one such pair is possible i.e. 2 and 6.
2 + 6 = 8
2 × 6 = 12
• Step 2: Break middle term in terms of the summation of a pair of numbers such that its product is equal to c i.e. 12 in above case. We will write 8=6+2
x2+ (6+2)x+ 12
= x2+ 6x +2x + 12
• Step 3: Form pairs of terms and factor out GCD of the two pairs separately.
= x2+ 6x +2x + 12 = (x2+ 6x) +(2x + 12) = x(x+6)+2(x+6)
• Step 4: Again factor out GCD of remaining sum of products. Follow factorization procedure of binomials as explained earlier. Factor out (x+6) from sum of product,
=x(x+6)+2(x+6) = (x+6)(x+2)

Example: Do the factorization of polynomials: x2-5x-6
Soluiton: Factors of -6 are 2 × -3; – 3 × 2; 1 × -6;  -1 × 6 and the pair summing up to -5 is (-6,1) as required. Hence,
x2-5x-6 = x2-6x+x-6 = x(x-6)+(x-6) = (x-6)(x+1)

Case 2:When the coefficient of x2 a is an integer other than 1 or -1

General form is given by ax2+bx+c. Any quadratic of form ax2+bx+c is expressible in the product of two linear polynomials:
ax2+bx+c= (a1x+b1)(a2x+b2)

where a is the product of a1 and a2, c is the product of b1 and b2 and b is the sum of the product of a1b2 and a2b1. Consider one more example,
(3x+2)(x+4)= 3x2+ 2x +12x +8 = 3x2+14x+ 8

Here, 3 = 3 × 1 = a1 × a2 =  a in general form. 14 = 2 × 1 + 4 × 3 = (a1b2) × (a2b1) = b in general form. 8 = 4 × 2 = b1 × b2= c in general form. Example: Factorize 3x2 – 5x – 2

• Step 1: find factors of 3 × – 2 = -6.
• Step 2: Product of factors of -6 are = -3 × 2; -2 × 3; 6 × -1; -1 × 6. Here, the possible pair of factor gives a summation of -1 is (-6,1).
•  Step 3: Break the middle term as the summation of two numbers such that its product is equal to -6. Calculated above such two numbers are -6 and 1.
• Step 4: Breaking the middle term: x2-6x+x-2
• Step 5: Making pairs of terms: (3x2-6x)+(x-2)
• Step 6: Factor out GCD for both the pairs: 3x(x-2)+1(x-2). Note: 1 is factored out from the second bracket as there was no other common factor.
• Step 7: Factor out GCD from the resulting summation of products: =3x(x-2)+1(x-2) =(x-2)(3x+1)

Factorization of Polynomials by Grouping
We can factorize polynomial containing even number of terms by forming pairs of terms.

• Step 1: Form pairs out of given even number of terms.
• Step 2: Try factoring out GCD from all the pairs separately.
• Step 3: Lastly, factor out the remaining common factor from the products formed.

Note: In case you do not get common factors for the pairs formed, try rearranging the terms and follow the same procedure again. Example: Factorize x2+ 4xy+4y+x
= (x2 + 4xy) + (4y + x)
=x(x + 4y) + 1( 4y + x)
=(x + 4y)(x + 1)

Factorization of Polynomials by Perfect Squares

A trinomial which can be factored, such that both the factors are same. Then, it will form a perfect square trinomial. For example, x2+2x+1 = (x + 1)(x + 1) = (x + 1)2. There are certain identities which are important for perfect square trinomials are as follows:
(a+b)2=(a2+2ab+b2)

Example:Do the factorization of polynomials, 4x2+12x+9
Solution: 4x2+12x+9 = (2x)2 + 2(2x)(3x) + 32 = (2x + 3)2

It can be seen from the identities and example above, that a trinomial with first and last terms as perfect squares and the middle term can be written as twice of the product of roots of first and last terms then the trinomial can be expressed as a perfect square.

Factorization of Polynomials by Difference of Squares

This applies mainly to the pair of two polynomial terms which are a perfect square and expressed as the difference between them.
a2-b2 = (a – b)(a + b)

Example: 9x2 – 4
Solution: 9x2 – 4 = (3x+2)(3x-2) { Using Identity }

Solved Example for You
Question: Factorise x2+11x+18
Solution: Consider the quadratic polynomial, x2+11x+18
=x2 + 9x + 2x + 18
=(x2 + 9x) + (2x + 18)
=x(x + 9) + 2(x + 9)
=(x + 9)(x + 2)

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