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**Definition of polynomials:**

Polynomials is the algebraic topic. An algebraic expression, which have only non negative powers of the variables is called polynomial.

In an expression, generally we have variables and constant terms.

**Example ** 3 x² + 2 x + 6

Here x is variable and 6 without any variant is called Constant.**Degree:**

The **degree of a algebraic equation** is the highest degree for a term with non-zero coefficient. The degree of a term is sum of the powers of each variable in the term.**Example**

4x⁷-5

In this expression we have only one term and the power of the term is 7, so the expression has deg 7

5 x³ y³ - 4 x² y² + 7 x y

Here, it has 3 terms and power of the first term is 3+3=6, the power of the second term is 2+2=4, and the power of the last term is 1+1=2, here the highest power is 6, so the expression has deg 3+3=6

We can do basic arithmetic operations with two algebraic equation. That is we can add, subtract, multiply or divide any two polynomial.**Addition:**

For adding any two polynomials we have to combine the like terms.

Add 4 x² + 7 x - 6, x² - 3 x - 2

For adding these two poly-nomials we have to combine the like terms. Here the like terms are 4x² and x², 7x and -3x , -6 and -2

If we combine 4x² and x² we will get 5x²

If we combine 7x and -3x we will get 4x

If we combine -6 and 2 we will get -4

So the final answer is 5x² + 4x - 4**Subtraction:**

(2 x³ - 2 x² + 4 x - 3)- (x³ + x² - 5 x + 4)

Step 1:In the first step we are going to multiply the negative with inner terms.= 2x³ -2x² + 4x -3 - x³-x²+5x-4

Step 2:In the second step we have to combine the like terms= 2x³ - x³ -2x²-x² + 4x + 5x - 3 - 4

Step 3:After combining the like terms we will get the answer= x³ + x² + 9x - 7

**Multiplication:**

there are two formats for this: horizontal and vertical, like in addition.

The simplest case of multiplication of polynomials is multiplication of monomials.**For instance:**

Simplify: ( 5 x² )(-2x ³)

For multiplying these two monomials we have to just multiply the numbers and add the powers using the exponent rule.

So (-6x²)(3x³) = -18 x² ⁺ ³= -18 x⁵**Division:**

Division of polynomials involves two cases, the first one is simplification,which is reducing the fraction and the second one is long division.**Polynomial in One Variable**

The degree of polynomials in one variable is the highest power of the variable in the algebraic expression. For example, in the following equation: x2+2x+4. The degree of the equation is 2 .i.e. the highest power of variable in the equation.**Multivariable polynomial**

For a multivariable polynomial, it the highest sum of powers of different variables in any of the terms in the expression. Take following example, x^{5}+3x^{4}y+2xy^{3}+4y^{2}-2y+1. It is a multivariable polynomial in x and y, and the degree of the polynomial is 5 – as you can see the degree in the terms x^{5} is 5, x^{4}y it is also 5 (4+1) and so the highest degree among these individual terms is 5.

A polynomial of two variable x and y, like ax^{r}y^{s} is the algebraic sum of several terms of the prior mentioned form, where r and s are possible integers. Here, the degree of the polynomial is r+s where r and s are whole numbers.**Note: **Exponents of variables of a polynomial .i.e. degree of polynomials should be whole numbers.**How to find the Degree of a Polynomial?**

There are 4 simple steps are present to find the degree of a polynomial:-

**Example:** *6x ^{5}+8x^{3}+3x^{5}+3x^{2}+4+2x+4*

Step 1:Combine all the like terms that are the terms of the variable terms.(6x

^{5}+3x^{5})+8x^{3}+3x^{2}+2x+(4+4)

Step 2:Ignore all the coefficientsx

^{5}+x^{3}+x^{2}+x+x^{0}

Step 3:Arrange the variable in descending order of their powersx

^{5}+x^{3}+x^{2}+x+x^{0}

Step 4:The largest power of the variable is the degree of the polynomialdeg(x

^{5}+x^{3}+x^{2}+x+x^{0}) = 5

**Zeros of a polynomial:**

Here we are going to see how to find zero of a polynomials.

If the value of a polynomial is zero for some value of the variable then that value is known as zero of the polynomial.

**Definition:**

Let p(x) be a polynomial in x. If p(x) = 0, then we say that a is a zero of the polynomial p(x).

Let us consider the example given below.**Zeros of linear polynomial****Example 1:****Find the zeros of the following linear polynomialp(x) = 2x + 3**

p(x) = 2x + 3

Now we have to think about the value of x, for which the given function will become zero.

For that let us factor out 2

p(x) = 2 (x + 3/2)

Instead of "x" , if we apply -3/2 p(x) will become zero.

Hence -3/2 is the zero of the given linear polynomial.

**Example 2: ****Find the zeros of the following linear polynomialp(x) = 4x - 1**

p(x) = 4x - 1

Now we have to think about the value of x, for which the given function will become zero.

For that let us factor out 4

p(x) = 4 (x - 1/4)

By applying the value 1/4 instead of x, the function p(x) will become zero.

Hence 1/4 is the zero of the given linear polynomial.

**Zeros of quadratic polynomial**

For a quadratic equation, there will be two zeros. In order to find those zeros, we may use the method called factoring.**Example 3: ****Find the zeros of the quadratic equation x² + 17 x + 60 by factoring.**

p(x) = x² + 17 x + 60

p(x) = x² + 12x + 5x + 60

p(x) = x (x + 12) + 5 (x + 12)

p(x) = (x + 5) (x + 12)

If x = -5

p(x) = (-5 + 5) (-5 + 12) = 0

If x = -12

p(x) = (-12 + 5) (-12 + 12) = 0

Hence the zeros are -5 and -12.

For a cubic equation, there will be three zeros. In order to find those zeros, we may use the methods

(i) Factor theorem

(ii) Synthetic division

Let p (x) = 4 x³ - 7 x + 3 x = 1

p (1) = 4 (1)³ -7 (1) + 3

= 4 - 7 + 3

= 7 - 7

= 0

So we can decide (x - 1) is a factor. To find other two factors, we have to use synthetic division.

So, the factors are (x - 1) and (4 x² - 4 x - 3). By factoring this quadratic equation we get (2 x +

3) (2 x - 1)

Hence the required three factors are (x - 1) (2 x + 3) (2 x - 1)

If a polynomial P(x) is divided by (x-a), the remainder is P(a).

Using division algorithm, we have

P(x) = Q(x)(x-a) + P(a)

Here, Q(x) is the quotient when P(x) is divided by (x-a).

**Factor theorem: **

A polynomial P(x) would have a factor (x-a), if and only if P(a) = 0.

**Remainder theorem and Factor theorem - Examples****Example 1: Using Remainder theorem, find the remainder when the polynomial 3x³- 2x² + 6x - 7 is divided by (x-2).****Solution : ****Step 1 :**

Let P(x) = 3x³- 2x² + 6x - 7.

If x - 2 = 0, then x = 2.**Step 2 :**

To know the remainder when P(x) divided by (x-2), aaaaa plug x = 2 in P(x).

Remainder = P(2)

Remainder = 3(2)³- 2(2)² + 6(2) - 7

Remainder = 24 - 8 + 12 - 7

Remainder = 21**Example 2: Using Remainder theorem, find the remainder when the polynomial 7x⁴ - x²- 3x + 9 is divided by (x-6).**

Let P(x) = 7x⁴ - x²- 3x + 9

If x - 6 = 0, then x = 6.

To know the remainder when P(x) divided by (x-6), aaaaa plug x = 6 in P(x).

Remainder = P(6)

Remainder = 7(6)⁴ - (6)² - 3(6) + 9

Remainder = 9072 - 36 - 18 + 9

Remainder = 9027

Let P(x) = x³+ 3x² - 5x + 2

If x + 5 = 0, then x = - 5.**Step 2 :**

To know the remainder when P(x) divided by (x+5), aaaaa plug x = - 5 in P(x).

Remainder = P(-5)

Remainder = (-5)³+ 3(-5)² - 5(-5) + 2

Remainder = -125 + 3(25) + 25 + 2

Remainder = -125 + 75 + 25 + 2

Remainder = -23**Example 4: Using Factor theorem, check whether (x-2) is a factor of the polynomial x⁴ - 3x³ + 2x² + 8x - 16.**

Let P(x) = x⁴ - 3x³ + 2x² + 8x - 16

If x - 2 = 0, then x = 2.

Using Factor theorem, to check whether (x-2) is factor of P(x), plug x = 2 in P(x).

P(2) = (2)⁴ - 3(2)³ + 2(2)² + 8(2) - 16

P(2) = 16 - 24 + 8 + 16 - 16

P(2) = 0

Because P(2) = 0, by Factor theorem, (x-2) is a factor of the polynomial P(x).

Let P(x) = x² - 8x + 16

If x + 4 = 0, then x = -4.

Using Factor theorem, to check whether (x+4) is factor of P(x), plug x = -4 in P(x).

P(-4) = (-4)² + 8(-4) + 16

P(-4) = 16 - 32 + 16

P(-4) = 0

Because P(-4) = 0, by Factor theorem, (x+4) is a factor of the polynomial P(x).

Let P(x) = x³ - 2x² + 5x + 6

If x - 3 = 0, then x = 3.

Using Factor theorem, to check whether (x-3) is factor of P(x), plug x = 3 in P(x).

P(3) = (3)³ - 2(3)² + 5(3) + 6

P(3) = 27 - 2(9) + 15 + 6

P(3) = 27 - 18 + 15 + 6

P(3) = 30 ≠ 0

Because P(3) ≠ 0, by Factor theorem, (x-3) is not a factor of the polynomial P(x).

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