Polynomials
Mathematics Class Tenth
What is a Polynomial? NCERT Exercise 2.1 and 2.2
An expression that consists variables and coefficient and non-negative integer components involving operations of addition, subtraction, and multiplication only is called POLYNOMIAL.
Example:
x2+ 4x – 7, x3+2 x2y – y + 1, 3x, 5, etc.
On the other hand x– 2y, 1/x, 2/x + 1, x, etc. are not polynomials. Because a polynomial cannot have
(i) an exponent with negative sign , such as – 2, – 5, etc.
(ii) no any term in a polynomial can be dividing by a variable, such as 1/x
(iii) an exponent with fractional exponent, such as x as it is written as x1/2
A Polynomial can have constants, variables, and exponents.
Examples of
Constants: 3, 2, – 2, 1/4 etc.
Variables: x, yx, z, abc, etc.
Exponents: 0, 1, 2, 3, 4, etc.
Degree of Polynomial
If p(x) is a polynomial in x, the highest power of x in p(x) is called the DEGREE OF THE POLYNOMIAL p(x).
Linear Polynomial
Let a polynomial 4x + 2.
Variable x of this polynomial has power equal to one (1), thus the degree of this polynomial is one (1). Or this is called the Polynomial of degree 1.
A polynomial of degree 1 is known as LINEAR POLYNOMIAL also.
Quadratic Polynomial
Let a polynomial x2 + x + 2.
Variable x of this polynomial has the highest power equal to 2. Thus, this polynomial, the variable of which has the highest power equal to 2 is called a Polynomial of 2 degrees.
A polynomial of 2 degrees is known as QUADRATIC POLYNOMIAL also.
Cubic Polynomial
A polynomial of degree 3 is called a CUBIC POLYNOMIAL.
Example
x3 + 2x2 – x + 1, 2 – x3, 2x, etc.
The most general form of a cubic polynomial is
ax3 + bx2 + cx + d where a,b,c,d are real numbers and a ≠ 0.
Value of Polynomial
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).
Example:
Let a polynomial p(x) = x2 – 3x – 4
By putting x = 2, we get
p(2) = 22 – 3 × 2 – 4= – 6
Here, the value of 2 is called the value of polynomial p(x) at x = 2.
Zeroes of a Polynomial
A real number k is said to be a zero of a polynomial p(x), if p(k) = 0.
Example
Let a polynomial p(x) = x2 – 3x – 4
By putting x = – 1, we get
p( – 1) = ( – 1)2 – 3( – 1) – 4
⇒ p( – 1) = 1 + 3 – 4 = 0
By putting x = 4 in the polynomial in example, we get
p(4) = 42 – (3 × 4) – 4
⇒ p(4) = 16 – 12 – 4 = 0
Since, here p( – 1) = 0 and p(4) = 0
Thus, – 1 and 4 are called the zeros of the quadratic polynomial x2 – 3x – 4.
Zeroes of a Linear Polynomial
If k is a zero of p(x) = ax + b, then
p(k) = ak + b = 0
i.e. k = – b/a
Thus, zero of a LINEAR POLYNOMIAL ax + b is equal to – b/a
= – Constant term/Coefficient of x
Thus, the zero of a LINEAR POLYNOMIAL is related to its coefficients.
The zeroes of a polynomial p(x) are precisely the x– coordinates of the points, where the graph of y = p(x) intersects the x– axis.
NCERT Exercise 2.1 Solution Class ten Mathematics
Question (1) The graphs of y = p(x) are given in the figure below, for some polynomials p(x). Find the number of zeros of p(x), in each case.
(i) 
Answer: Since line of graph does not intersect x–axis, thus number of zeros = 0
(ii) 
Answer: Since line of graph intersects x–axis only once, thus number of zeros = 1
(iii) 
Answer: Since line of graph intersects x–axis thrice, thus number of zeros = 3
(iv) 
Answer: Since line of graph intersects x–axis twice, thus number of zeros = 2
(v) 
Answer: Since line of graph intersects x–axis three times, thus number of zeros = 3
(vi) 
Answer: Since line of graph intersects x–axis four times, thus number of zeros = 4
Relationship between Zeroes and Coefficients of a Polynomial
If α and β are the zeroes of the quadratic polynomial p(x) = ax2 + bx + c, where a ≠ 0, then x – α and x – β are the factors of p(x).
Therefore,
ax2 + bx + c =k(x – α )(x – β ), where k is constant.
= k [x2( α + β )x + α β ]
= k x2 – k( α + β )x + k α β
Now, by comparing the coefficients of x2, x and constant terms on both sides, we get
a = k -------(i)
b = – k( α + β ) ----------(ii)
c = k α β ------(iii)
Now,
∵ b = – k( α + β )
∴ α + β = – b/k
After replacing the value of k = a from equation (i) we get
α + β = – b/a ------(iv)
Now,
∵ c = k α β
∴ α β = c/k
After replacing the value of k = a from equation (i) we get
α β = c/a ---------(v)
Thus, sum of zeroes
= α + β = – b/a
= – Coefficient of x/Coefficient of x2
And Product of zeroes
= α β = c/a = Constant term/Coefficient of x2
NCERT Exercise 2.2 SolutionClass ten Mathematics
Question (1) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
Solution:
Given,
x2 – 2x – 8
After expanding middle term 2x, we get
= x2 – 4x+2x – 8
[∵ 2x = – 4x + 2x]
=x(x – 4) + 2(x – 4)
Now, after taking (x – 4) as common, we get
=(x – 4)(x + 2)
Thus, the value of x2 – 2x – 8 is zero, when x – 4 =0 or x + 2 = 0
Thus,
When x – 4 = 0
∴ x = 4
And, when x + 2 = 0
∴ x = – 2
Therefore,
The zeroes of x2 – 2x – 8 are 4 and – 2. Answer
Verification of relationship between zeroes and coefficients.
Now, Sum of zeroes
= 4 + ( – 2)
= – ( – 2/1)
= – Coefficient of x/Coefficient of x2
And Product of zeroes = 4 × ( – 2)
= – 8/1
= Constant term/Coefficient of x2
(ii) 4s2 – 4s + 1
Solution
Given, 4s2 – 4s + 1
= (2s)2 – 2 × 2 × s + 1
=(2s – 1)2
[∵ (a – b)2 = a2 – 2ab + b2]
=(2s – 1)(2s – 1)
Thus, value of 4s2 – 4s + 1 is equal to zero (0), when 2s – 1 = 0
Therefore, when, 2s – 1 = 0
⇒ 2s = 1
⇒ s = 1/2
Thus, the zeroes of 4s2 – 4s + 1 are 1/2 and 1/2
Verification:
Sum of zeroes = 1/2 + 1/2
= 1 = – – 4/4
= – Coefficient of s/Coefficient of s2
Product of zeroes = 1/2 × 1/2
= 1/4 = Constant term/Coefficient of s2
(iii) 6x2 – 3 – 7x
Solution
Given, 6x2 – 3 – 7x
= 6x2 – 7x – 3
After expanding middle term – 7x we get
= 6x2 – 9x + 2x – 3
[∵ – 9x + 2x = – 7x]
= 3x(2x – 3) + 1(2x – 3)
After taking 2x + 3 as common
= (2x – 3)(3x + 1)
Thus, value of 6x2 – 3 – 7x is equal to zero (0), when 2x – 3 = 0 or 3x + 1 = 0
Thus, when,
2x – 3 = 0
∴ 2x = 3
⇒ x = 3/2
And, when, 3x + 1 = 0
∴ 3x = – 1
⇒ x = – 1/3
Thus, zeroes of 6x2 – 3 – 7x are 3/2 and – 1/3
Verification:
Now,
Sum of zeroes = 3/2 + ( – 1/3)
= 3/2 – 1/3
= 9 – 2/6 = 7/6
= – ( – 7)/6 = – Coefficient of x/Coefficient of x2
And, product of zeroes
= 3/2 × – 1/3 = – 3/6
= Constant term/Coefficient of x2
(iv) 4u2 + 8u
Solution
Given, 4u2 + 8u
= 4u(u + 2)
Thus, the value of 4u2 + 8u is zero when, 4u = 0 or u + 2 = 0
Thus, when, 4u = 0
∴ u = 0
And, when u + 2 = 0
∴ u = – 2
Therefore, the zeroes of 4u2 + 8u are 0 and – 2
Verification:
Now, Sum of zeroes = u + ( – 2) = – 2
= – 8/4 = – Coefficient of u/Coefficient of u2
And, product of zeroes = 0 × ( – 2) = 0
= 0/4 = Constant term/Coefficient of u2
(v) t2 – 15
Solution:
Given, t2 – 15
=t2 – (15)2
= (t + 15)(t – 15)
[∵ a2 – b2 = (a + b)(a – b)]
Thus, the value of t2 – 15 is equal to zero, when t – 15 = 0 or t + 15 = 0
Therefore, when t – 15 = 0
∴ t = 15
And when, t + 15 = 0
∴ t = – 15
Thus, zeroes of t2 – 15 are 15 and – 15
Verification:
Now, sum of zeroes =15 + ( – 15)=0
= – 0/1 = – Coefficient of t/Coefficient of t2
And, product of zeroes = 15 × ( – 15)= – 15
= – 1/1 = Constant term/Coefficient of t2
(vi) 3x2 – x – 4
Solution:
Given, 3x2 – x – 4
By expanding middle term – x, we get
3x2 + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
After taking (x + 1) as common, we get
= (x + 1)(3x – 4)
Thus, the value of 3x2 – x – 4 is zero when x + 1 = 0 or 3x – 4 = 0
Thus, when, x + 1 = 0
Therefore, x = – 1
And when, 3x – 4 = 0
Therefore, 3x = 4
⇒ x = 4/3
Therefore, zeroes of 3x2 – x – 4 are – 1 and 4/3
Verification:
Sum of zeroes
= – 1 + 4/3 = – 3 + 4/3 = 1/3
= – – 1/3 = – Coefficient of x/Coefficient of x2
And, the product of zeroes, = – 1 × 3/4
= – 4/3 = Constant term/Coefficient of x2
Question (2) Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 1/4, – 1
Solution
Given, the Sum and product of zeroes of the polynomial are 1/4 and – 1 respectively.
Let, the polynomial be ax2 + bx + c ------(i)
And its zeroes are α and β
Thus, according to question,
α + β = 1/4 = – b/a
Thus, a = 4 and b = – 1
α β = -1 = – 4/4 = c/a
Therefore, a = 4, b = – 1 and c = – 4
Thus, after substituting the values of a, b, and c in equation (i), we get
4x2 – 1 × x – 4
= 4x2 – x – 4
Therefore, the quadratic polynomial for the given values is 4x2 – x – 4 Answer
(ii) 2, 1/3
Solution
Given, the sum and product of the quadratic polynomial are 2, and 1/3
Let, the polynomial be ax2 + bx + c ------(i)
And its zeroes are α and β
Thus, according to the question,
Product of zeroes
= α β = 1/3 = c/a
Therefore, c = 1 and a = 3
Sum of zeroes
= α + β = 2 = – b/a
⇒ 32/3 = – b/a
[∵ a = 3]
Thus, a = 3, b = – 3 2 and c = 1
After substituting the values of a,b and c in equation (i), we get
3x2 – 3 x + 1
Therefore, the quadratic polynomial for the given values is 3x2 – 3x + 1 Answer
(iii) 0, 5
Solution
Given, the sum and product of the quadratic polynomial are 0, and 5
Let, the polynomial be ax2 + bx + c ------(i)
And its zeroes are α and β
Thus, according to the question,
Sum of zeroes
= α + β = 0 = 0/1 = – b/a
And the product of zeroes
= α β = 5 = 5/1 = c/a
Thus, a = 1, b = – 0 and c = 5
After substituting the values of a, b and c in equation (i), we get
x2 + (– 0 × x) + 5
= x2 + 5
Therefore, the quadratic polynomial for the given values is = x2 + 5 Answer
(iv) 1, 1
Solution
Given, the sum and product of the quadratic polynomial are 1, and 1
Let, the polynomial be ax2 + bx + c ------(i)
And its zeroes are α and β
Thus, according to the question,
Sum of zeroes
= α + β = 1 = 1/1 = – b/a
And, the product of zeroes
= α β = 1 = 1/1 = c/a
Thus, a = 1, b = – 1 and c = 1
After substituting the values of a, b and c in equation (i), we get
x2 – x + 1
Therefore, the quadratic polynomial for the given values is x2 – x + 1 Answer.
(v) – 1/4, 1/4
Solution
Given, the sum and product of the quadratic polynomial are – 1/4, and 1/4
Let, the polynomial be ax2 + bx + c ------(i)
And its zeroes are α and β
Thus, according to the question,
Sum of zeroes
= α + β = – 1/4 = – b/a
And Product of zeroes,
= α β = 1/4 = c/a
Thus, a = 4, b = 1 and c = 1
After substituting the values of a,b and c in equation (i), we get
4x2 + x + 1
Therefore, the quadratic polynomial for the given values is 4x2 + x + 1 Answer
(vi) 4, 1
Solution
Given, the sum and product of the quadratic polynomial are 4, and 1
Let, the polynomial be ax2 + bx + c ------(i)
And its zeroes are α and β
Thus, according to the question,
Sum of zeroes
= α + β = 4 = 4/1 = – b/a
And product of zeroes,
= α β = 1 = 1/1 = c/a
Thus, a = 1, b = – 4 and c = 1
After substituting the values of a, b and c in equation (i), we get
x2 – 4x + 1
Therefore, the quadratic polynomial for the given values is x2 – 4x + 1 Answer
Reference: