MathsClass 10 MathsChapter 2: Polynomials
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Important Questions Class 10 Maths Chapter 2 Polynomials
Important questions for Class 10 Maths Chapter 2 Polynomials are provided here as per the NCERT book. These questions are provided with answers that will help the students prepare and score well in the CBSE Class 10 Maths exam. The questions here cover the latest syllabus as prescribed by the board. Also, they are based on the latest exam pattern. Students can practice the important questions for all the chapters of 10th standard Maths subject and prepare well for the board exam 2021-22. Along with the important questions, students will also find detailed solutions. In case they get stuck while solving the questions, then they can refer to the solutions.
We have also provided some additional important questions from the examination point of view for Chapter 2 Polynomials of Class 10 Maths so that students can revise the chapter by solving these problems instantly. This chapter deals with expressions carrying unknown variables with different degrees. The solutions of these expressions are termed zeros of the polynomial. The polynomial with degree 1 is the linear polynomial, with degree 2 is a quadratic polynomial, with degree 3 is a cubic polynomial.
Students can also get access to Class 10 Maths Chapter 2 Polynomials MCQs here.
Also Check:
Important 2 Marks Questions for CBSE 10th Maths
Important 3 Marks Questions for CBSE 10th Maths
Important 4 Marks Questions for CBSE 10th Maths
Class 10 Maths Chapter 2 Important Questions With Solutions
A few important Class 10 polynomials questions are provided below with solutions. These questions include both short and long answer questions to let the students get acquainted with the in-depth concepts.
Q.1: Find the value of “p” from the polynomial x2 + 3x + p, if one of the zeroes of the polynomial is 2.
Solution:
As 2 is the zero of the polynomial.
We know that if α is a zero of the polynomial p(x), then p(α) = 0
Substituting x = 2 in x2 + 3x + p,
⇒ 22 + 3(2) + p = 0
⇒ 4 + 6 + p = 0
⇒ 10 + p = 0
⇒ p = -10
Q.2: Does the polynomial a4 + 4a2 + 5 have real zeroes?
Solution:
In the aforementioned polynomial, let a2 = x.
Now, the polynomial becomes,
x2 + 4x + 5
Comparing with ax2 + bx + c,
Here, b2 – 4ac = 42 – 4(1)(5) = 16 – 20 = -4
So, D = b2 – 4ac < 0
As the discriminant (D) is negative, the given polynomial does not have real roots or zeroes.
Q.3: Compute the zeroes of the polynomial 4×2 – 4x – 8. Also, establish a relationship between the zeroes and coefficients.
Solution:
Let the given polynomial be p(x) = 4×2 – 4x – 8
To find the zeroes, take p(x) = 0
Now, factorise the equation 4×2 – 4x – 8 = 0
4×2 – 4x – 8 = 0
4(x2 – x – 2) = 0
x2 – x – 2 = 0
x2 – 2x + x – 2 = 0
x(x – 2) + 1(x – 2) = 0
(x – 2)(x + 1) = 0
x = 2, x = -1
So, the roots of 4×2 – 4x – 8 are -1 and 2.
Relation between the sum of zeroes and coefficients:
-1 + 2 = 1 = -(-4)/4 i.e. (- coefficient of x/ coefficient of x2)
Relation between the product of zeroes and coefficients:
(-1) × 2 = -2 = -8/4 i.e (constant/coefficient of x2)
Q.4: Find the quadratic polynomial if its zeroes are 0, √5.
Solution:
A quadratic polynomial can be written using the sum and product of its zeroes as:
x2 – (α + β)x + αβ
Where α and β are the roots of the polynomial.
Here, α = 0 and β = √5
So, the polynomial will be:
x2 – (0 + √5)x + 0(√5)
= x2 – √5x
Q.5: Find the value of “x” in the polynomial 2a2 + 2xa + 5a + 10 if (a + x) is one of its factors.
Solution:
Let f(a) = 2a2 + 2xa + 5a + 10
Since, (a + x) is a factor of 2a2 + 2xa + 5a + 10, f(-x) = 0
So, f(-x) = 2×2 – 2×2 – 5x + 10 = 0
-5x + 10 = 0
5x = 10
x = 10/5
Therefore, x = 2
Q.6: How many zeros does the polynomial (x – 3)2 – 4 have? Also, find its zeroes.
Solution:
Given polynomial is (x – 3)2 – 4
Now, expand this expression.
=> x2 + 9 – 6x – 4
= x2 – 6x + 5
As the polynomial has a degree of 2, the number of zeroes will be 2.
Now, solve x2 – 6x + 5 = 0 to get the roots.
So, x2 – x – 5x + 5 = 0
=> x(x – 1) -5(x – 1) = 0
=> (x – 1)(x – 5) = 0
x = 1, x = 5
So, the roots are 1 and 5.
Q.7: α and β are zeroes of the quadratic polynomial x2 – 6x + y. Find the value of ‘y’ if 3α + 2β = 20.
Solution:
Let, f(x) = x² – 6x + y
From the given,
3α + 2β = 20———————(i)
From f(x),
α + β = 6———————(ii)
And,
αβ = y———————(iii)
Multiply equation (ii) by 2. Then, subtract the whole equation from equation (i),
=> α = 20 – 12 = 8
Now, substitute this value in equation (ii),
=> β = 6 – 8 = -2
Substitute the values of α and β in equation (iii) to get the value of y, such as;
y = αβ = (8)(-2) = -16
Q.8: If the zeroes of the polynomial x3 – 3×2 + x + 1 are a – b, a, a + b, then find the value of a and b.
Solution:
Let the given polynomial be:
p(x) = x3 – 3×2 + x + 1
Given,
The zeroes of the p(x) are a – b, a, and a + b.
Now, compare the given polynomial equation with general expression.
px3 + qx2 + rx + s = x3 – 3×2 + x + 1
Here, p = 1, q = -3, r = 1 and s = 1
For sum of zeroes:
Sum of zeroes will be = a – b + a + a + b
-q/p = 3a
Substitute the values q and p.
-(-3)/1 = 3a
a = 1
So, the zeroes are 1 – b, 1, 1 + b.
For the product of zeroes:
Product of zeroes = 1(1 – b)(1 + b)
-s/p = 1 – 𝑏2
=> -1/1 = 1 – 𝑏2
Or, 𝑏2 = 1 + 1 =2
So, b = √2
Thus, 1 – √2, 1, 1 + √2 are the zeroes of equation 𝑥3 − 3𝑥2 + 𝑥 + 1.
Q.9: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes, respectively.
(i) 1/4, -1
(ii) 1, 1
(iii) 4, 1
Solution:
(i) From the formulas of sum and product of zeroes, we know,
Sum of zeroes = α + β
Product of zeroes = αβ
Given,
Sum of zeroes = 1/4
Product of zeroes = -1
Therefore, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:-
x2 – (α + β)x + αβ
= x2 – (1/4)x + (-1)
= 4×2 – x – 4
Thus, 4×2 – x – 4 is the required quadratic polynomial.
(ii) Given,
Sum of zeroes = 1 = α + β
Product of zeroes = 1 = αβ
Therefore, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:-
x2 – (α + β)x + αβ
= x2 – x + 1
Thus, x2 – x + 1 is the quadratic polynomial.
(iii) Given,
Sum of zeroes, α + β = 4
Product of zeroes, αβ = 1
Therefore, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:-
x2 – (α + β)x + αβ
= x2 – 4x + 1
Thus, x2 – 4x +1 is the quadratic polynomial.
Q.10: Obtain all other zeroes of 3×4 + 6×3 – 2×2 – 10x – 5, if two of its zeroes are √(5/3) and-√(5/3).
Solution: Since this is a polynomial of degree 4, hence there will be a total of 4 roots.
√(5/3) and-√(5/3) are zeroes of polynomial f(x).
∴ [x-√(5/3)] [x+√(5/3)] = x2-(5/3)
Important Questions class 10 chapter 2 polynomials Q10
Therefore, 3×2 + 6x + 3 = 3x(x + 1) +3 (x + 1)
= (3x + 3)(x + 1)
= 3(x + 1)(x + 1)
= 3(x + 1)(x + 1)
Hence, x + 1 = 0 i.e. x = – 1 , – 1 is a zero of p(x).
So, its zeroes are given by: x = −1 and x = −1.
Therefore, all four zeroes of the given polynomial are:
√(5/3) and-√(5/3), −1 and −1.
Q.11: Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial f(x) = ax2 + bx + c, a ≠ 0, c ≠ 0.
Solution:
Let α and β be the zeroes of the polynomial f(x) = ax2 + bx + c.
So, α + β = -b/a
αβ = c/a
According to the given, 1/α and 1/β are the zeroes of the required quadratic polynomial.
Now, the sum of zeroes = (1/α) + (1/β)
= (α + β)/αβ
= (-b/a)/ (c/a)
= -b/c
Product of two zeroes = (1/α) (1/β)
= 1/αβ
= 1/(c/a)
= a/c
The required quadratic polynomial = k[x2 – (sum of zeroes)x + (product of zeroes)]
= k[x2 – (-b/c)x + (a/c)]
= k[x2 + (b/c) + (a/c)]
Q.12: Divide the polynomial f(x) = 3×2 – x3 – 3x + 5 by the polynomial g(x) = x – 1 – x2 and verify the division algorithm.
Solution:
Given,
f(x) = 3×2 – x3 – 3x + 5
g(x) = x – 1 – x2
Dividing f(x) = 3×2 – x3 – 3x + 5 by g(x) = x – 1 – x2
Important Questions Class 10 Maths Chapter 2 Polynomials 12A
Here,
Quotient = q(x) = x – 2
Remainder = r(x) = 3
By division algorithm of polynomials,
Dividend = (Quotient × Divisor) + Remainder
So,
[q(x) × g(x)] + r(x) = (x – 2)(x – 1 – x2) + 3
= x2 – x – x3 – 2x + 2 + 2×2 + 3
= 3×2 – x3 – 3x + 5
= f(x)
Hence, the division algorithm is verified.
Q.13: For what value of k, is the polynomial f(x) = 3×4 – 9×3 + x2 + 15x + k completely divisible by 3×2 – 5?
Solution:
Given,
f(x) = 3×4 – 9×3 + x2 + 15x + k
g(x) = 3×2 – 5
Dividing f(x) by g(x),
Important Questions Class 10 Maths Chapter 2 Polynomials 13A
Given that f(x) is completely divisible by 3×2 – 5.
So, the remainder = 0
k + 10 = 0
k = -10
Q.14: If 4 is a zero of the cubic polynomial x3 – 3×2 – 10x + 24, find its other two zeroes.
Solution:
Given cubic polynomial is p(x) = x3 – 3×2 – 10x + 24
4 is a zero of p(x).
So, (x – 4) is the factor of p(x).
Let us divide the given polynomial by (x – 4).
Important Questions Class 10 Maths Chapter 2 Polynomials 14A
Here, the quotient = x2 + x – 6
= x2 + 3x – 2x – 6
= x(x + 3) – 2(x + 3)
= (x – 2)(x + 3)
Therefore, the other two zeroes of the given cubic polynomial are 2 and -3.
Practice Questions for Class 10 Maths Chapter 2 Polynomials
Find the zeroes of the quadratic polynomial 7y2 – (11/3)y – (2/3) and verify the relationship between the zeroes and the coefficients.
Find all zeroes of the polynomial (2×4 – 9×3 + 5×2 + 3x – 1) if two of its zeroes are (2 + √3) and (2 – √3).
If the remainder on division of x3 + 2×2 + kx +3 by x – 3 is 21, find the quotient and the value of k. Hence, find the zeroes of the cubic polynomial x3 + 2×2 + kx – 18.
For which values of a and b, are the zeroes of q(x) = x3 + 2×2 + a also the zeroes of the polynomial p(x) = x5 – x4 – 4×3 + 3×2 + 3x + b? Which zeroes of p(x) are not the zeroes of q(x)?
Verify that 3, –1, -1/3 are the zeroes of the cubic polynomial p(x) = 3×3 – 5×2 – 11x – 3, and then verify the relationship between the zeroes and the coefficients.
Find a quadratic polynomial, the sum and product of whose zeroes are √2 and -3/2, respectively. Also find its zeroes.
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21 Comments
Sakkhya Sinha APRIL 19, 2020 AT 2:14 PM
nice and helpful questions
REPLY
kavita MAY 24, 2020 AT 11:52 PM
my doubt is of relationship between zeros ans coffieients
REPLY
lavanya MAY 29, 2020 AT 8:27 PM
Please visit: https://byjus.com/maths/relation-between-zeros-and-coefficients/
REPLY
Dr Myst NOVEMBER 15, 2020 AT 11:34 AM
relationship between zeroes and coefficients means that you have to show that alpha + beta = -b/a and alpha x beta = c/a.
REPLY
Kanishq Garg JUNE 22, 2020 AT 1:50 AM
good questions
REPLY
Tanishka JUNE 23, 2020 AT 9:01 PM
It was really useful
REPLY
Manasvishnoi JULY 1, 2020 AT 9:08 AM
Please tell me some tough questions???
These were too easy.
REPLY
Riddhi JULY 22, 2020 AT 1:45 PM
Ya! It is useful , good questions .
REPLY
bhumi JULY 25, 2020 AT 1:58 PM
this all are so easy and nice thanks to byju’s
REPLY
aneetta gitto AUGUST 9, 2020 AT 3:16 PM
happy with these questions .. will be more happy if got hard questions and i need some difficult questions on relationship between zeroes and coefficient
REPLY
shivani AUGUST 13, 2020 AT 11:06 AM
its easy
REPLY
mith jain SEPTEMBER 15, 2020 AT 8:28 PM
it is very helpfull questions my sons douts are all cleared . thank u
REPLY
Nancy SEPTEMBER 16, 2020 AT 9:19 AM
very helpful thank’s
REPLY
Peehu SEPTEMBER 26, 2020 AT 1:44 PM
These questions were useful.
REPLY
Aaryan singh OCTOBER 14, 2020 AT 8:16 AM
I need some hard question from this chapter
REPLY
Akash NOVEMBER 10, 2020 AT 6:48 PM
Thank you for the questions, these really helped me clear my doubts.
REPLY
Dr Myst NOVEMBER 15, 2020 AT 11:35 AM
Very useful questions for revision
REPLY
Maths slayer MAY 14, 2021 AT 1:09 PM
Very easy but good question thanks byjus
REPLY
Jiyu JULY 5, 2021 AT 8:08 PM
Veryy amazing I just love it I hope byjus go through the sky 😃😃😃😃😃😃
REPLY
Mohammed Abdul Raees JULY 29, 2021 AT 7:14 PM
It’s very useful before examination for us like 10 std so thank you for posting questions like this every week
REPLY
ISHANT SHARMA APRIL 13, 2022 AT 5:21 PM
It is very very helpful for students .
REPLY
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