The Polynomials are fundamental algebraic expressions consisting of variables raised to the whole-number powers and combined using addition, subtraction, and multiplication. They form the basis of many concepts in algebra including solving equations, graphing functions, and mathematical modeling. In this chapter, we will explore polynomials in detail and solve various problems to understand their properties and applications better. Exercise 2.1 | Set 1 focuses on solving the fundamental problems to build a strong foundation in polynomials.
Polynomials
A polynomial is an algebraic expression that involves a sum of powers of one or more variables multiplied by the coefficients. For example: 3x2+2xā5 is a polynomial in the variable x. The degree of a polynomial is the highest power of the variable in the expression. Polynomials can be classified into different types based on their degree and number of terms such as monomials, binomials, and trinomials. They are crucial for solving algebraic equations and analyzing mathematical relationships.
Question 1. Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:
(i) f(x) = x2 ā 2x ā 8
Solution:
Given that,
f(x) = x2 ā 2x ā 8
To find the zeros of the equation, put f(x) = 0
= x2 ā 2x ā 8 = 0
= x2 ā 4x + 2x ā 8 = 0
= x(x ā 4) + 2(x ā 4) = 0
= (x ā 4)(x + 2) = 0
x = 4 and x = -2
Hence, the zeros of the quadratic equation are 4 and -2.
Now, Verification
As we know that,
Sum of zeros = ā coefficient of x / coefficient of x^2
4 + (-2)= ā (-2) / 1
2 = 2
Product of roots = constant / coefficient of x^2
4 x (-2) = (-8) / 1
-8 = -8
Hence the relationship between zeros and their coefficients are verified.
(ii) g(s) = 4s2 ā 4s + 1
Solution:
Given that,
g(s) = 4s2 ā 4s + 1
To find the zeros of the equation, put g(s) = 0
= 4s2 ā 4s + 1 = 0
= 4s2 ā 2s ā 2s + 1= 0
= 2s(2s ā 1) ā (2s ā 1) = 0
= (2s ā 1)(2s ā 1) = 0
s = 1/2 and s = 1/2
Hence, the zeros of the quadratic equation are 1/2 and 1/2.
Now, Verification
As we know that,
Sum of zeros = ā coefficient of s / coefficient of s2
1/2 + 1/2 = ā (-4) / 4
1 = 1
Product of roots = constant / coefficient of s2
1/2 x 1/2 = 1/4
1/4 = 1/4
Hence the relationship between zeros and their coefficients are verified.
(iii) h(t)=t2 ā 15
Solution:
Given that,
h(t) = t2 ā 15 = t2 +(0)t ā 15
To find the zeros of the equation, put h(t) = 0
= t2 ā 15 = 0
= (t + ā15)(t ā ā15)= 0
t = ā15 and t = -ā15
Hence, the zeros of the quadratic equation are ā15 and -ā15.
Now, Verification
As we know that,
Sum of zeros = ā coefficient of t / coefficient of t2
ā15 + (-ā15) = ā (0) / 1
0 = 0
Product of roots = constant / coefficient of t2
ā15 x (-ā15) = -15/1
-15 = -15
Hence the relationship between zeros and their coefficients are verified.
(iv) f(x) = 6x2 ā 3 ā 7x
Solution:
Given that,
f(x) = 6x2 ā 3 ā 7x
To find the zeros of the equation, we put f(x) = 0
= 6x2 ā 3 ā 7x = 0
= 6x2 ā 9x + 2x ā 3 = 0
= 3x(2x ā 3) + 1(2x ā 3) = 0
= (2x ā 3)(3x + 1) = 0
x = 3/2 and x = -1/3
Hence, the zeros of the quadratic equation are 3/2 and -1/3.
Now, Verification
As we know that,
Sum of zeros = ā coefficient of x / coefficient of x2
3/2 + (-1/3) = ā (-7) / 6
7/6 = 7/6
Product of roots = constant / coefficient of x2
3/2 x (-1/3) = (-3) / 6
-1/2 = -1/2
Hence the relationship between zeros and their coefficients are verified.
(v) p(x) = x2 + 2ā2x ā 6
Solution:
Given that,
p(x) = x2 + 2ā2x ā 6
To find the zeros of the equation, put p(x) = 0
= x2 + 2ā2x ā 6 = 0
= x2 + 3ā2x ā ā2x ā 6 = 0
= x(x + 3ā2) ā ā2 (x + 3ā2) = 0
= (x ā ā2)(x + 3ā2) = 0
x = ā2 and x = -3ā2
Hence, the zeros of the quadratic equation are ā2 and -3ā2.
Now, Verification
As we know that,
Sum of zeros = ā coefficient of x / coefficient of x2
ā2 + (-3ā2) = ā (2ā2) / 1
-2ā2 = -2ā2
Product of roots = constant / coefficient of x2
ā2 x (-3ā2) = (-6) / 2ā2
-3 x 2 = -6/1
-6 = -6
Hence the relationship between zeros and their coefficients are verified.
(vi) q(x)=ā3x2 + 10x + 7ā3
Solution:
Given that,
q(x) = ā3x2 + 10x + 7ā3
To find the zeros of the equation, put q(x) = 0
= ā3x2 + 10x + 7ā3 = 0
= ā3x2 + 3x +7x + 7ā3x = 0
= ā3x(x + ā3) + 7 (x + ā3) = 0
= (x + ā3)(ā3x + 7) = 0
x = -ā3 and x = -7/ā3
Hence, the zeros of the quadratic equation are -ā3 and -7/ā3.
Now, Verification
As we know that,
Sum of zeros = ā coefficient of x / coefficient of x2
-ā3 + (-7/ā3) = ā (10) /ā3
(-3-7)/ ā3 = -10/ā3
-10/ ā3 = -10/ā3
Product of roots = constant / coefficient of x2
(-ā3) x (-7/ā3) = (7ā3) / ā3
7 = 7
Hence the relationship between zeros and their coefficients are verified.
(vii) f(x) = x2 ā (ā3 + 1)x + ā3
Solution:
Given that,
f(x) = x2 ā (ā3 + 1)x + ā3
To find the zeros of the equation, put f(x) = 0
= x2 ā (ā3 + 1)x + ā3 = 0
= x2 ā ā3x ā x + ā3 = 0
= x(x ā ā3) ā 1 (x ā ā3) = 0
= (x ā ā3)(x ā 1) = 0
x = ā3 and x = 1
Hence, the zeros of the quadratic equation are ā3 and 1.
Now, Verification
Sum of zeros = ā coefficient of x / coefficient of x2
ā3 + 1 = ā (-(ā3 +1)) / 1
ā3 + 1 = ā3 +1
Product of roots = constant / coefficient of x2
1 x ā3 = ā3 / 1
ā3 = ā3
Hence the relationship between zeros and their coefficients are verified.
(viii) g(x) = a(x2+1)āx(a2+1)
Solution:
Given that,
g(x) = a(x2+1)āx(a2+1)
To find the zeros of the equation put g(x) = 0
= a(x2+1)āx(a2+1) = 0
= ax2 + a ā a2x ā x = 0
= ax2 ā a2x ā x + a = 0
= ax(x ā a) ā 1(x ā a) = 0
= (x ā a)(ax ā 1) = 0
x = a and x = 1/a
Hence, the zeros of the quadratic equation are a and 1/a.
Now, Verification :
As we know that,
Sum of zeros = ā coefficient of x / coefficient of x2
a + 1/a = ā (-(a2 + 1)) / a
(a^2 + 1)/a = (a2 + 1)/a
Product of roots = constant / coefficient of x2
a x 1/a = a / a
1 = 1
Hence the relationship between zeros and their coefficients are verified.
(ix) h(s) = 2s2 ā (1 + 2ā2)s + ā2
Solution:
Given that,
h(s) = 2s2 ā (1 + 2ā2)s + ā2
To find the zeros of the equation put h(s) = 0
= 2s2 ā (1 + 2ā2)s + ā2 = 0
= 2s2 ā 2ā2s ā s + ā2 = 0
= 2s(s ā ā2) -1(s ā ā2) = 0
= (2s ā 1)(s ā ā2) = 0
x = ā2 and x = 1/2
Hence, the zeros of the quadratic equation are ā3 and 1.
Now, Verification
As we know that,
Sum of zeros = ā coefficient of s / coefficient of s2
ā2 + 1/2 = ā (-(1 + 2ā2)) / 2
(2ā2 + 1)/2 = (2ā2 +1)/2
Product of roots = constant / coefficient of s2
1/2 x ā2 = ā2 / 2
ā2 / 2 = ā2 / 2
Hence the relationship between zeros and their coefficients are verified.
(x) f(v) = v2 + 4ā3v ā 15
Solution:
Given that,
f(v) = v2 + 4ā3v ā 15
To find the zeros of the equation put f(v) = 0
= v2 + 4ā3v ā 15 = 0
= v2 + 5ā3v ā ā3v ā 15 = 0
= v(v + 5ā3) ā ā3 (v + 5ā3) = 0
= (v ā ā3)(v + 5ā3) = 0
v = ā3 and v = -5ā3
Hence, the zeros of the quadratic equation are ā3 and -5ā3.
Now, for verification
Sum of zeros = ā coefficient of v / coefficient of v2
ā3 + (-5ā3) = ā (4ā3) / 1
-4ā3 = -4ā3
Product of roots = constant / coefficient of v2
ā3 x (-5ā3) = (-15) / 1
-5 x 3 = -15
-15 = -15
Hence the relationship between zeros and their coefficients are verified.
(xi) p(y) = y2 + (3ā5/2)y ā 5
Solution:
Given that,
p(y) = y2 + (3ā5/2)y ā 5
To find the zeros of the equation put f(v) = 0
= y2 + (3ā5/2)y ā 5 = 0
= y2 ā ā5/2 y + 2ā5y ā 5 = 0
= y(y ā ā5/2) + 2ā5 (y ā ā5/2) = 0
= (y + 2ā5)(y ā ā5/2) = 0
This gives us 2 zeros,
y = ā5/2 and y = -2ā5
Hence, the zeros of the quadratic equation are ā5/2 and -2ā5.
Now, Verification
As we know that,
Sum of zeros = ā coefficient of y / coefficient of y2
ā5/2 + (-2ā5) = ā (3ā5/2) / 1
-3ā5/2 = -3ā5/2
Product of roots = constant / coefficient of y2
ā5/2 x (-2ā5) = (-5) / 1
ā (ā5)2 = -5
-5 = -5
Hence the relationship between zeros and their coefficients are verified.
(xii) q(y) = 7y2 ā (11/3)y ā 2/3
Solution:
Given that,
q(y) = 7y2 ā (11/3)y ā 2/3
To find the zeros of the equation put q(y) = 0
= 7y2 ā (11/3)y ā 2/3 = 0
= (21y2 ā 11y -2)/3 = 0
= 21y2 ā 11y ā 2 = 0
= 21y2 ā 14y + 3y ā 2 = 0
= 7y(3y ā 2) ā 1(3y + 2) = 0
= (3y ā 2)(7y + 1) = 0
y = 2/3 and y = -1/7
Hence, the zeros of the quadratic equation are 2/3 and -1/7.
Now, Verification
As we know that,
Sum of zeros = ā coefficient of y / coefficient of y2
2/3 + (-1/7) = ā (-11/3) / 7
-11/21 = -11/21
Product of roots = constant / coefficient of y2
2/3 x (-1/7) = (-2/3) / 7
ā 2/21 = -2/21
Hence the relationship between zeros and their coefficients are verified.
Question 2. For each of the following, find a quadratic polynomial whose sum and product respectively of the zeros are as given. Also, find the zeros of these polynomials by factorization.
(i) -8/3, 4/3
Solution:
As we know that the quadratic polynomial formed for the given sum and product of zeros is given by : f(x) = x2 + -(sum of zeros) x + (product of roots)
The sum of zeros = -8/3 and
Product of zero = 4/3
Therefore,
Required polynomial f(x) is,
= x2 ā (-8/3)x + (4/3)
= x2 + 8/3x + (4/3)
To find the zeros we put f(x) = 0
= x2 + 8/3x + (4/3) = 0
= 3x2 + 8x + 4 = 0
= 3x2 + 6x + 2x + 4 = 0
= 3x(x + 2) + 2(x + 2) = 0
= (x + 2) (3x + 2) = 0
= (x + 2) = 0 and, or (3x + 2) = 0
Hence, the two zeros are -2 and -2/3.
(ii) 21/8, 5/16
Solution:
As we know that the quadratic polynomial formed for the given sum and product of zeros is given by : f(x) = x2 + -(sum of zeros) x + (product of roots)
The sum of zeros = 21/8 and
Product of zero = 5/16
Therefore,
The required polynomial f(x) is,
= x2 ā (21/8)x + (5/16)
= x2 ā 21/8x + 5/16
To find the zeros we put f(x) = 0
= x2 ā 21/8x + 5/16 = 0
= 16x2 ā 42x + 5 = 0
= 16x2 ā 40x ā 2x + 5 = 0
= 8x(2x ā 5) ā 1(2x ā 5) = 0
= (2x ā 5) (8x ā 1) = 0
= (2x ā 5) = 0 and, or (8x ā 1) = 0
Hence, the two zeros are 5/2 and 1/8.
(iii) -2ā3, -9
Solution:
As we know that the quadratic polynomial formed for the given sum and product of zeros is given by : f(x) = x2 + -(sum of zeros) x + (product of roots)
The sum of zeros = -2ā3 and
Product of zero = -9
Therefore,
The required polynomial f(x) is,
= x2 ā (-2ā3)x + (-9)
= x2 + 2ā3x ā 9
To find the zeros we put f(x) = 0
= x2 + 2ā3x ā 9 = 0
= x2 + 3ā3x ā ā3x ā 9 = 0
= x(x + 3ā3) ā ā3(x + 3ā3) = 0
= (x + 3ā3) (x ā ā3) = 0
= (x + 3ā3) = 0 and, or (x ā ā3) = 0
Hence, the two zeros are -3ā3and ā3.
(iv) -3/2ā5, -1/2
Solution:
As we know that the quadratic polynomial formed for the given sum and product of zeros is given by : f(x) = x2 + -(sum of zeros) x + (product of roots)
The sum of zeros = -3/2ā5 and
Product of zero = -1/2
Therefore,
The required polynomial f(x) is,
= x2 ā (-3/2ā5)x + (-1/2)
= x2 + 3/2ā5x ā 1/2
To find the zeros we put f(x) = 0
= x2 + 3/2ā5x ā 1/2 = 0
= 2ā5x2 + 3x ā ā5 = 0
= 2ā5x2 + 5x ā 2x ā ā5 = 0
= ā5x(2x + ā5) ā 1(2x + ā5) = 0
= (2x + ā5) (ā5x ā 1) = 0
= (2x + ā5) = 0 and, or (ā5x ā 1) = 0
Hence, the two zeros are -ā5/2 and 1/ā5.
Question 3. If Ī± and Ī² are the zeros of the quadratic polynomial f(x) = x2 ā 5x + 4, find the value of 1/Ī± + 1/Ī² ā 2Ī±Ī².
Solution:
Given that,
Ī± and Ī² are the roots of the quadratic polynomial f(x) where a = 1, b = -5 and c = 4
Using these values we can find,
Sum of the roots = Ī±+Ī² = -b/a = ā (-5)/1 = 5,
Product of the roots = Ī±Ī² = c/a = 4/1 = 4
We have to find 1/Ī± +1/Ī² ā 2Ī±Ī²
= [(Ī± +Ī²)/ Ī±Ī²] ā 2Ī±Ī²
= (5)/ 4 ā 2(4) = 5/4 ā 8 = -27/ 4
Question 4. If Ī± and Ī² are the zeros of the quadratic polynomial p(y) = 5y2 ā 7y + 1, find the value of 1/Ī±+1/Ī².
Solution:
Given that,
Ī± and Ī² are the roots of the quadratic polynomial f(x) where a =5, b = -7 and c = 1,
Using these values we can find,
Sum of the roots = Ī±+Ī² = -b/a = ā (-7)/5 = 7/5
Product of the roots = Ī±Ī² = c/a = 1/5
We have to find 1/Ī± +1/Ī²
= (Ī± +Ī²)/ Ī±Ī²
= (7/5)/ (1/5) = 7
Question 5. If Ī± and Ī² are the zeros of the quadratic polynomial f(x)=x2 ā x ā 4, find the value of 1/Ī±+1/Ī²āĪ±Ī².
Solution:
Given that,
Ī± and Ī² are the roots of the quadratic polynomial f(x) where a = 1, b = -1 and c = ā 4
So, we can find,
Sum of the roots = Ī±+Ī² = -b/a = ā (-1)/1 = 1
Product of the roots = Ī±Ī² = c/a = -4 /1 = ā 4
We have to find, 1/Ī± +1/Ī² ā Ī±Ī²
= [(Ī± +Ī²)/ Ī±Ī²] ā Ī±Ī²
= [(1)/ (-4)] ā (-4) = -1/4 + 4 = 15/ 4
Question 6. If Ī± and Ī² are the zeroes of the quadratic polynomial f(x) = x2 + x ā 2, find the value of 1/Ī± ā 1/Ī².
Solution:
Given that:
Ī± and Ī² are the roots of the quadratic polynomial f(x) where a = 1, b = 1 and c = ā 2
So, we can find
Sum of the roots = Ī±+Ī² = -b/a = ā (1)/1 = -1,
Product of the roots = Ī±Ī² = c/a = -2 /1 = ā 2
We have to find, 1/Ī± ā 1/Ī²
= [(Ī² ā Ī±)/ Ī±Ī²] = [Ī²-Ī±]/(Ī±Ī²) x (Ī±-Ī²)/Ī±Ī² = (ā(Ī±+Ī²)2 -4Ī±Ī²) / Ī±Ī² = ā(1+8) / 2 = 3/2
Question 7. If one of the zero of the quadratic polynomial f(x) = 4x2 ā 8kx ā 9 is negative of the other, then find the value of k.
Solution:
Given that,
The quadratic polynomial f(x) where a = 4, b = -8k and c = ā 9
And, for roots to be negative of each other, let us assume that the roots Ī± and ā Ī±.
Using these values we can find,
Sum of the roots = Ī± ā Ī± = -b/a = ā (-8k)/1 = 8k = 0 [āµ Ī± ā Ī± = 0]
= k = 0
Question 8. If the sum of the zeroes of the quadratic polynomial f(t)=kt2 + 2t + 3k is equal to their product, then find the value of k.
Solution:
Given that,
The quadratic polynomial f(t)=kt2 + 2t + 3k, where a = k, b = 2 and c = 3k ,
Sum of the roots = Product of the roots
= (-b/a) = (c/a)
= (-2/k) = (3k/k)
= (-2/k) = 3
Hence k = -2/3
Question 9. If Ī± and Ī² are the zeros of the quadratic polynomial p(x) = 4x2 ā 5x ā 1, find the value of Ī±2 Ī²+Ī± Ī²2
Solution:
Given that,
Ī± and Ī² are the roots of the quadratic polynomial p(x) where a = 4, b = -5 and c = -1
Using these values we can find,
Sum of the roots = Ī±+Ī² = -b/a = ā (-5)/4 = 5/4
Product of the roots = Ī±Ī² = c/a = -1/4
We have to find, Ī±^2 Ī²+Ī± Ī²^2
= Ī±Ī²(Ī± +Ī²)
= (-1/4)(5/4) = -5/16
Question 10. If Ī± and Ī² are the zeros of the quadratic polynomial f(t)=t2ā 4t + 3, find the value of Ī±4 Ī²3+Ī±3 Ī²4.
Solution:
Given that,
Ī± and Ī² are the roots of the quadratic polynomial f(t) where a = 1, b = -4 and c = 3
Using these values we can find,
Sum of the roots = Ī±+Ī² = -b/a = ā (-4)/1 = 4 ,
Product of the roots = Ī±Ī² = c/a = 3/1 = 3
We have to find, Ī±4 Ī²3 + Ī±3 Ī²4
= Ī±3 Ī²3 (Ī± +Ī²)
= (Ī±Ī²)3 (Ī± +Ī²)
= (3)3 (4) = 27 x 4 = 108
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Summary
Exercise 2.1 | Set 1 of RD Sharma's Class 10 Mathematics textbook introduces the basic concepts of polynomials. This section covers the definition of polynomials, their classification based on degree and number of terms, identification of polynomials, and basic operations with polynomials. Students learn to recognize polynomials in different forms, determine their degrees, identify coefficients and constants, and understand the concept of zeros of polynomials. The exercise also includes problems on evaluating polynomials for given values and finding the values of variables for which a polynomial equals zero.
