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Polynomial

Polynomial- Polynomial is made up of two terms, namely Poly (meaning “many”) and Nominal (meaning “terms.”). A polynomial is defined as an expression which is composed of variables, constants and exponents, that are combined using mathematical operations such as addition, subtraction, multiplication and division (No division operation by a variable)
The polynomial function is denoted by P(x) where x represents the variable.
For example  P(x) = x²- 5x + 11
These are polynomials
8 is a polynomial because 8 can be written as 8x⁰ or 0x²+ 0x+  8, which represents the polynomial expression. Therefore, we can consider 8 as a polynomial.
x + x²  is a polynomial because the exponents of the variable x are 1 and 2. All the exponents are whole numbers. 
x + √3 and x are polynomials because the exponents of the variable x is 1. which is a whole number. 
x/2 is allowed, because you can divide by a constant
√2 is allowed, because it is a constant (= 1.4142...etc)
These are not polynomials
x  + 2 = 0  is a polynomial equation and not a polynomial expression.
3xy⁻² is not polynomial, because the exponent is "-2" (exponents can only be 0,1,2,...)
2/(x+2) is not polynomial, because dividing by a variable is not allowed
√x is not polynomial, because the exponent of √x  is "½" , it is not a whole number.
∛y + y²  is not polynomial, because the exponent of ∛y is 1/3 ,it is not a whole number.
Coefficient: A coefficient is a real number that is present along with variables.
Terms of a Polynomial
The terms of polynomials are the parts of the expression that are generally separated by “+” or “-” signs. 
Example
2x²  + 5x + 4
There are three Terms [2x²  , 5x and 4]  in this polynomial  
The roots or zero of a polynomial - The roots or zeros of polynomial are the real values of the variable for which the value of the polynomial would become equal to zero.
If p(x) is a polynomial in x, and if ‘k’ is any real number, then the value obtained by replacing ‘x’ by ‘k’ in p(x), is called the value of p(x) at x = k, and is denoted by p(k).
What is the value of p(x) = x²  –3x – 4  at x = –1 and 4
p(–1) = (–1)2 –{3 × (–1)} – 4 = 0 
p(4) = 4² – (3 x 4) – 4 = 0
        As p(–1) = 0 and p(4) = 0
         –1 and 4 are called the zeroes of the polynomial x²  – 3x – 4.
A real number k is said to be a zero of a polynomial p(x), if p(k) = 0.
Degree - The highest power of  x in polynomial p(x) of variable x  is called the degree of the polynomial
Example 
The degree of the polynomial  4x + 2  is 1
The degree of the polynomial 2y² – 3y + 4  is 2
The degree of the polynomial  5x³ – 4x²  + x – 2  is 3
The degree of the polynomial  √5 is 0
The degree of the polynomial 7u⁶ – ³/₂   + 4u²  + u -  8 is 6 
The degree of the polynomial  x³ + 6x²y⁴ + 3y²+5 is 6 
The degree of the polynomial  3xy is 2
If a polynomial has multiple variables, the degree of the polynomial can be found by adding the powers of different variables in any terms present in the polynomial expression. 

Polynomial Name 

Degree

Constant Polynomial      

0

Linear Polynomial          

1

Quadratic Polynomial     

2

Cubic Polynomial           

3

Quartic Polynomial         

4

1. Linear polynomial - A polynomial of degree 1 is called a linear polynomial.
In fact, the most general form of a Linear polynomial is  ax + b, where, a and b are real numbers and a≠ 0.
Example
2x – 3 ,   √3 x +  5, y + √2, x - 2/11, 3z + 4, 3/2  u - 1
2. Quadratic polynomial  - A polynomial of degree 2 is called a quadratic polynomial. The name ‘quadratic’ has been derived from the word ‘quadrate’, which means ‘square’.
In fact, the most general form of a cubic polynomial is ax²  + bx + c, where, a, b, c are real numbers and a  0.
Example
Y² – 2,2x²  + 3x  -  3/5,  2 – x²  +  c x, u/3  - 2u² + 5, √3v² - 2/3 v, 4z² + 1/7
3 Cubic polynomial – A Polynomial of degree 3 is called a cubic polynomial.
In fact, the most general form of a cubic polynomial is ax³ + bx²  + cx + d, where, a, b, c, d are real numbers and a ≠ 0.
Example
2 – x³ , x³ , √3 x³  , 3 – x²  + x³,  3x³ – 2x²  + x – 1
Geometrical Meaning of the Zeroes of a Polynomial
1. Geometrical Meaning of Zeroes of Linear Polynomial:
We know that the graph of  y = ax + b is a straight line
Example         y = 2x + 3
        When y = 0
          2x  + 3 = 0
          x = - 3/2

  x

0

1

2

-1

-2

-3/2

y = 2x +3

3

5

7

1

-1

    0




















From the graph, we can observe that graph y=2x+3 intersects the x-axis between x= -1 and x= -2
i.e., The straight line intersects the x-axis at the point (- 3/2, 0).
Therefore, - 3/2  is the zero of the polynomial y=2x+3.
In general, for a linear polynomial ax + b, a  0, the graph of y = ax + b is a straight line which intersects the x-axis at exactly one point, namely ( (-b)/a, 0). 
Therefore, the linear polynomial ax + b, a  0, has exactly one zero, namely, The zero of the linear polynomial is the x-coordinate of the point where the graph of y=ax+b intersects at the x-axis.
2. Geometrical Meaning of Zeroes of Quadratic Polynomial:
The zeroes of a quadratic polynomial ax2 + bx + c, a  0, are precisely the x-coordinates of the points where the parabola representing y = ax² + bx + c intersects the x-axis.
  y= x²  – 3x – 4
For the given quadratic equation, first find the coordinates (x, y), by taking a few values of x.

x

– 2

–1

0

1

2

3

4

5

y =x2–3x–4

6

0

–4

–6

–6

–4

0

6
























From the graph, we can observe that the two zeroes of the polynomial y = x² – 3x – 4 are -1 and 4
Generally, the graph of the quadratic polynomial, y=ax²   +bx   +   c, where a≠0, has two types of curves such as the parabolic curve open upwards or parabolic curve open downwards, depending on whether a>0 or a<0.
The quadratic equation has at most two zeroes, there exist three different cases. 
Case 1: The graph cuts the x-axis at two distinct points A and A'. In this case, the quadratic polynomial has two zeroes.






     Parabolic curve open downwards                       Parabolic curve open upwards
Case 2: The graph cuts the X-axis at exactly one point, A. In this case, there exists only one zero.

Parabolic curve open downwards              Parabolic curve open upwards
Case 3: The graph does not cut X-axis at any point. In this case, the curve for the given quadratic polynomial is completely above or below the x-axis. So, the quadratic polynomial has no zero in this case.

The graph of a quadratic polynomial is a parabola. It looks like a U, which either opens upwards or opens downwards depending on the value of ‘a’ in ax2+bx+c
If ‘a’ is positive, then parabola opens upwards and if ‘a’ is negative then it opens downwards
It can cut the x-axis at 0, 1 or two points
3. Geometrical Meaning of Zeroes of Cubic Polynomial:
A polynomial of the form ax³ + bx²+ cx+d, a≠0 is a cubic polynomial.
Consider the graph of y=x³− 4x.

x

–2

–1

0

1

2

y = x3 – 4x

0

3

0

–3

0



According to above table -2, 0 and 2 are zeroes of the cubic polynomial x³ – 4x. 
– 2, 0 and 2 are the x-coordinates of the points where the graph of y = x³ – 4x intersects the x-axis
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Consider the graph of  y= x³

x

–2

–1

0

1

2

y = x3

-8

-1

0

1

8

























From the graph we can observe that  0 is the only zero of the polynomial x³ . since the graph of y = x³ intersects the x-axis only at 0
Consider the graph of  y= x³ – x²

x

–2

–1

0

1

2

y = x- x2

4

-2

0

0

4






The polynomial x3 – x2 = x2(x – 1) has two zeroes, 0 and 1. From graph, we can see that the graph of y = x3 – x2 intersects the x-axis at 0 and 1
Relationship between Zeroes and Coefficients of a Polynomial
1. linear polynomial -The zero of a linear polynomial ax + b is  - b/a 
2. Quadratic polynomial - In general, if α and β are the zeroes of the quadratic polynomial p(x) = ax + bx + c, a≠ 0, then we know that x – α and x – β are the factors of p(x). Therefore,
        ax² + bx + c = k(x – α) (x – β), where k is a constant
                            = k[x² – ( α + β)x+ αβ]
                            = kx² –k( α + β)x+kαβ 
Comparing the coefficients of x2 , x and constant terms on both the sides, we get
                            a = k
                            b = – k(α + β)
                            c= kαβ
                            b = – k( α + β)
                      α + β = (-b)/k
                      α + β = (-b)/a            [ k = a]
                             c= kαβ
                           αβ = c/k
                           αβ = c/a                [ k = a]
Sum of zeroes = α + β = (-b)/a =   (- Coefficient of x)/ (Coefficient of x² )              
Product of zeroes = αβ  = c/a   = (Constant term)/(Coefficient of x^2 )              
3. Cubic polynomial - In general, it can be proved that if α, β, ϒ are the zeroes of the cubic polynomial ax³ + bx² + cx + d, then
α    + β    + ϒ    = (-b)/a   
αβ  + βϒ +αϒ   = c/a  
                  αβϒ = (-d)/a 

  1. Find the zeroes of the quadratic polynomial 3x2 + 5x – 2., and verify the relationship between the zeroes and the coefficients.

    p(x) = 3x2 + 5x – 2.

    By the method of splitting the middle term,

    3x2 + 5x – 2

    = 3x + 6x – x – 2

    = 3x(x + 2) –1(x + 2)

    = (3x – 1)(x + 2)

    Hence, the value of 3x + 5x – 2 is zero when either 3x – 1 = 0 or x + 2 = 0, i.e., when x =  1/3 or x = –2. So, the zeroes of 3x + 5x – 2 are  1/2 and – 2.

    Sum of zeroes = a + b

    Sum of zeroes = -b/a

                            =1/3+(-2)

                            = -5/3

                            = -(5)/3

                                              = -5/3

    Product of zeroes = ab

    Product of zeroes = c/a

                         =  1/3x (-2)

         = -2/3

                                  = -2/3

                                  = -2/3

  2. Find the zeroes of the quadratic polynomial 2x² – 8x + 6, and verify the relationship  between the zeroes and the coefficients.

    P(x) = 2x² – 8x + 6

    By the method of splitting the middle term,

    = 2x²– 6x – 2x + 6

    = 2x(x – 3) – 2(x – 3)

    = (2x – 2)(x – 3)

    = 2(x – 1)(x – 3)

    So, the value of p(x) = 2x² – 8x + 6 is zero when x – 1 = 0 or x – 3 = 0, i.e., when x = 1 or x = 3.

    So, the zeroes of 2x² – 8x + 6 are 1 and 3.

    Sum of zeroes = a + b

    Sum of zeroes =  -b/a

                             = 1+ 3

        = - (-8)/2

         = 4

                             = 4

    Product of zeroes = ab

    Product of zeroes =c/a

                                 =1x 3

                                  = 3

                                   = 6/2

                                   = 3

  3. Find the zeroes of the polynomial x²– 3 and verify the relationship between the zeroes and the coefficients.

    We know that

    a²  – b²  = (a – b)(a + b). Using it, we can write:

    x2 – 3 = (x - √3) (x + √3)                              

    (x - √3) = 0

    x = √3

    (x + √3) = 0

    x = -√3

    Hence, the value of x² – 3 is zero when either  (x - √3) Or (x +√3) i.e.,

    when x = √3 or x = –√3. So, the zeroes of x² – 3 are -2 and – 5.

    Sum of zeroes =α + β

    Sum of zeroes = √3 +(-√3) 

                            = 0

    Sum of zeroes = -b/a          

                            = -0/1

                            = 0

    Product of zeroes =α x β

                                = √3 x (-√3)

                                 = -3

    Sum of zeroes = c/a         

                            = -3/1    

                            = -3

  4. Find a quadratic polynomial, the sum and product of whose zeroes are 3 and 2, respectively.

    Let the quadratic polynomial be ax²  + bx + c, and its zeroes be a and b.

    We have

    Sum of zeroes = α + β =3   

    Product of zeroes  = αβ  = 2            

    If a =1 then b= 3 and c = 2       Because [ α + β =3= -b/a],  [αβ =2= c/a]

    So, one quadratic polynomial which fits the given conditions is  x2 + 3x + 2.

  5. If α, β are the zeroes of the polynomial x² – px + 36  and α² ²  = 9, then what is the value of p?

    Given polynomial  x² – px +36

    On comparing with the standard form of a quadratic polynomial ax² + bx + c, we get

    a = 1, b = -p, c = 36

    Here, α  and β  are the zeroes of  the polynomial.

      ⇒ α + β =  = -p

    and

         ⇒ α β= 36

    Now,

    α + β² = (α + β)²  - 2αβ  (x + y)²  = x² + y²  + 2xy                    

    ⇒ x² + y² = (x + y)²  - 2xy

    ⇒ 9 = p−2 × 36        [∵ α+ β = 9]

    ⇒ p² = 72 +9

    ⇒ p² = 81

    ⇒ p = 9  or  −9

  6. Find the value of 𝒌 such that the polynomial x² − (k + 6)x + 2(2k − 1) has sum of its zeroes equal to half  of their product.

    For the given polynomial x² - (k + 6)x + 2(2k -1)

    a = 1,    b =- (k + 6),   c = 2(2k -1) 

    Let a and b be the zeroes of the polynomial

    Sum of zeroes = α + β  = -b/c

                           = -(-k+6)/1

                           = k + 6                ..............(1)

    Product of zeroes  αβ  = c/a

                                        = 2(2k-1)/1

                                         =2(2k - 1)           ..............(2)

    It is given that

    Sum of zeroes = Product of zeroes

     a + β = 1/2× aβ

    Substituting the values from Equ. (1) and (2);

     k + 6 =1/2  x 2(2k - 1)

     k+ 6  = 2k - 1

     k - 2k = -6 -1

     k = 7

  7. Find the zeroes of the polynomial (x − 3)² – 4

    Given p(x) = (x-3)² - 4

    Factoraising    (x - 3)² - 4       (x – y)² = x² – 2xy + y

                            x² - 6x + 9 - 4

                            x² - 6x + 5

                             x² - 5x - x + 5

                               x(x – 5) – 1(x – 5)

                               (x - 5) (x - 1)

               (x -5 ) = 0              (x - 1) = 0

            ⇒    x = 5                     x = 1  

    Substituting x = 1 in the polynomial p(x) = (x-3)² – 4

                =  (1 – 3)² – 4

                 = 0

          p(1) = 0

    Substituting x = 5 in the polynomial p(x) = (x-3)² – 4

                = (5 - 3)² - 4

                = 0

          p(5) = 0

    Since p(1) = 0 and p(5) = 0, hence x = 1 and x = 1 are the zeroes of the given polynomial.

  8. What is the coefficient of  x in (x -5)3 ?

    Using (a - b)3 = a3  -   b3  - 3a2b  + 3ab2

    (x -5)3 = x3 – 53 – 3x2 5 + 3x 52

    = x3 – 125 – 15 x2 -75 x

    Hence, the coefficient of x  is 75.

  9. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. 4s² − 4s + 1


    Find the zeroes of the given polynomial

    The given polynomial 4s² − 4s + 1

    Now,

    = 4s² − 4s + 1

    = 4s² − 2s – 2s + 1

    = 2s(2s – 1) -1(2s -1)

    = (2s-1) (2s -1)

     (2s-1) = 0 or  (2s -1) = 0

     s = 1/2   or 1/2

    At s =1/2 , 1/2   are the zeroes of the given polynomial.

    Verify the relation of the sum of zeroes

    sum of zeroes = 1/2 + 1/2    = 1

    We know that

    sum of zeroes for as− bs + c

                            =  -b/a

                            = - (-4)/4

                            = 1

    Verity the relation of the product of the zeroes

    product of the zeroes = 1/2   x    1/2

                                       = 1/4

    We know that product of the zeroes

                                        = c/a

                                        = 1/4

  10. Find the zeroes of the quadratic polynomial x²  + 7x + 10, and verify the relationship between the zeroes and the coefficients.

    We have

        x² + 7x + 10        

        = x²  + 5x +2x + 10  [By the method of splitting the middle term]

        = x(x+5) + 2(x+5)

        = (x+2) (x+5)

        (x+2) = 0

        x = -2

    (x+5) = 0

    x = -5

    Hence, the value of x²  + 5x +2x + 10 is zero when either (x+2) Or (x+5), i.e.,

    when x = -2 or x = –5. So, the zeroes of x²  + 5x +2x + 10 are -2 and – 5.

    Sum of zeroes =α + β

    Sum of zeroes = -2 +(-5)

                            = -7

    Sum of zeroes = -b/a     

                            = - (7)/1 

                            = - 7      

    Product of zeroes =α x β

    Sum of zeroes = -2 X (-5)    

                              =10

    Sum of zeroes   = c/a                              = 10/1

                               = 10 

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