Important questions of Chapter 2 Polynomials for Class 10 Maths are provided here as per the NCERT book. These questions with answers 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. Along with the important questions, students will also find detailed solutions. In case they get stuck while practising the questions, then they can refer to the solutions.
We have also provided some additional questions so that students can revise the chapter by solving these problems in an instant. This chapter deals with expressions carrying unknown variables with varying degrees. The solutions of these expressions are termed as 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.
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.
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
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.
3. Compute the zeroes of the polynomial 4x2 – 4x – 8. Also, establish a relationship between the zeroes and coefficients.
Solution:
Let the given polynomial be p(x) = 4x2 – 4x – 8
To find the zeroes, take p(x) = 0
Now, factorise the equation 4x2 – 4x – 8 = 0
4x2 – 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 4x2 – 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)
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 is:
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
5. Find the value of “x” in the polynomial 2a2 + 2xa + 5a + 10 if (a + x) is one of its factors.
Solution:
Say, f(a) = 2a2 + 2xa + 5a + 10
Since, (a + x) is a factor of 2a2 + 2xa + 5a + 10, f(-x) = 0
So, f(-x) = 2x2 – 2x2 – 5x + 10 = 0
Or, -5x + 10 = 0
Thus, x = 2
6. How many zeros does the polynomial (x – 3)2 – 4 can have? Also, find its zeroes.
Solution:
Given equation is (x – 3)2 – 4
Now, expand this equation.
=> x2 + 9 – 6x – 4
= x2 – 6x + 5
As the equation 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.
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
put the value of α and β in equation (iii) to get the value of y, such as;
y = αβ = (8)(-2) = -16
8. If the zeroes of the polynomial x3 – 3x2 + 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 – 3x2 + 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 – 3x2 + 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
Or, 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.
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)
4x2 – x – 4
Thus, 4x2 – 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.
10. Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are √(5/3) and-√(5/3).
Solution: Do YourSellf