NCERT MCQ Solutions for Class 10 Maths Chapter 2 Polynomials

For the quadratic polynomial 2x² – 3x – 9, we need to find its zeroes.
Step 1: Set the polynomial equal to zero. 2x² – 3x – 9 = 0
Step 2: Use the quadratic formula: x = [-b ± √(b² – 4ac)]/2a Where a = 2, b = -3, c = -9
x = [3 ± √(9 + 72)]/4 x = [3 ± √81]/4 x = [3 ± 9]/4
This gives us x = 3/2 and x = -3/2
Therefore, the zeroes are 3/2 and -3/2, which corresponds to option (c) -3, 3/2.
Correction: The actual zeroes are 3 and -3/2, as we can verify: When x = 3: 2(3)² – 3(3) – 9 = 2(9) – 9 – 9 = 18 – 9 – 9 = 0 When x = -3/2: 2(-3/2)² – 3(-3/2) – 9 = 2(9/4) + 9/2 – 9 = 9/2 + 9/2 – 9 = 0
So, the correct answer is (a) 3, -3/2.

Let’s call the number to be subtracted as k. The resulting polynomial will be: x² – 16x + 30 – k
For 15 to be a zero of this polynomial:
(15)² – 16(15) + 30 – k = 0
225 – 240 + 30 – k = 0
15 – k = 0
k = 15
Therefore, 15 should be subtracted, which corresponds to option (c) 15.

The sum of the zeros of the quadratic polynomial is given by -coefficient of x / coefficient of x².
So, sum of zeros = 7 (since -(-7)/1 = 7).
If one zero is 5, then the other zero is 7 – 5 = 2.

Let’s call the number to be added as k. The resulting polynomial will be: x² – 5x + 4 + k
For 3 to be a zero of this polynomial:
(3)² – 5(3) + 4 + k = 0
9 – 15 + 4 + k = 0
-2 + k = 0
k = 2
Therefore, 2 should be added, which corresponds to option (b) 2.

A polynomial having zeros -3 and 5 can be written in the form:
P(x) = k(x + 3)(x – 5), where k is any non-zero constant.
Here, k can take infinitely many values (e.g., 1, 2, -1, 0.5, etc.), and each value of k will result in a different polynomial. Therefore, there are more than 3 polynomials that can have -3 and 5 as their zeros.

If a polynomial has zeroes -3 and 5, it can be written as: P(x) = k(x – (-3))(x – 5) = k(x + 3)(x – 5)
Where k is any non-zero constant.
Since k can take infinite values, there are infinite polynomials with zeroes -3 and 5.
Therefore, the answer is (b) infinite.

For the polynomial x² – 1: a = 1, b = 0, c = -1
Using the relation: Sum of zeroes = α + β = -b/a = -0/1 = 0
Therefore, α + β = 0, which corresponds to option (d) 0.
We can verify this by finding the actual zeroes: x² – 1 = 0 x² = 1 x = ±1
So α = 1 and β = -1 and α + β = 0.

If 2 and -3 are the zeroes, then: (x – 2)(x – (-3)) = x² – x – 6
Comparing with the given form x² + (a + 1)x + b: -1 = a + 1 -6 = b
So a = -2 and b = -6.
This corresponds to option (d) a = 0, b = -6.
Correction: From the calculation above, a = -2 not 0. Let me double-check: If zeroes are 2 and -3: p(x) = (x – 2)(x + 3) = x² + x – 6
Comparing with x² + (a + 1)x + b:
a + 1 = 1
b = -6
So a = 0 and b = -6, which confirms option (d) a = 0, b = -6.

The quadratic polynomial is given by x² – (sum of zeros)x + (product of zeros).
Substituting the given values:
x² – 3x – 10.

If 2 is a zero of the polynomial x² + 3x + k, then:
(2)² + 3(2) + k = 0
4 + 6 + k = 0
10 + k = 0
k = -10
Therefore, the value of k is -10, which corresponds to option (a) -10.

The sum of zeros = -b/a = -4, so b = 4.
The product of zeros = c/a = 3, so c = 3.
The quadratic polynomial is x² + 4x + 3.

For the quadratic polynomial 3x² + 11x – 4: a = 3, b = 11, c = -4
Using the quadratic formula:
x = [-b ± √(b² – 4ac)]/2a
x = [-11 ± √(121 + 48)]/6
x = [-11 ± √169]/6
x = [-11 ± 13]/6
This gives x = (−11 + 13)/6 = 1/3 and x = (−11 – 13)/6 = -4
Therefore, the zeroes are 1/3 and -4, which corresponds to option (c) 1/3, -4.

For the quadratic x² – 3x – m(m + 3) to have zeroes α and β:
x² – 3x – m(m + 3) = (x – α)(x – β) = x² – (α + β)x + αβ
Comparing coefficients: -(α + β) = -3,
So α + β = 3 and αβ = -m(m + 3)
If we try α = -m and β = m + 3: α + β = -m + (m + 3) = 3 ✓ αβ = (-m)(m + 3) = -m(m + 3) 
Therefore, the zeroes are -m and m + 3, which corresponds to option (b) -m, m + 3.

If α and β are the zeroes, then: α + β = -5 αβ = 6
The quadratic polynomial would be: P(x) = x² – (sum of zeroes)x + (product of zeroes) = x² – (-5)x + 6 = x² + 5x + 6
Therefore, the polynomial is x² + 5x + 6, which corresponds to option (a) x² + 5x + 6.

Product of the zeros = c/a = 3/2.
Given one zero α = -3/2, let the other zero be β.
So, (-3/2) × β = 3/2 ⟹ β = 1.

Examining the highest power of x in each: (a) degree 2 (b) degree 4 (c) degree 3 (d) degree 4
Both (b) and (d) have degree 4, but (d) 754x⁴ – 122x³ + 452x² + 19x + 211 has a larger coefficient for x⁴, making it the polynomial with the highest degree.
The answer is (d) 754x⁴ – 122x³ + 452x² + 19x + 211.

Using the Factor Theorem, we check f(-1):
(-1)³ + 3(-1)² + 3(-1) + 1 = -1 + 3 – 3 + 1 = 0.
Since f(-1) = 0, x + 1 is a factor of the polynomial.

f(0) = 5
f(-1) = 10
f(1) = 6
Step 1: Find c using f(0):
f(0) = a·0² + b·0 + c = c = 5
⇒ c = 5
Step 2: Use f(-1) = 10:
f(-1) = a(−1)² + b(−1) + 5 = a − b + 5 = 10
⇒ a − b = 10 − 5
⇒ a − b = 5 … (1)
Step 3: Use f(1) = 6:
f(1) = a(1)² + b(1) + 5 = a + b + 5 = 6
⇒ a + b = 6 − 5
⇒ a + b = 1 … (2)
Step 4: Solve (1) and (2):
Add (1) and (2):
(a − b) + (a + b) = 5 + 1
2a = 6
⇒ a = 3
Now substitute a = 3 in (2):
3 + b = 1
⇒ b = 1 − 3
⇒ b = −2
Thus, the quadratic function is:
f(x) = 3x² − 2x + 5
Answer: (c) 3x² − 2x + 5

Given: The zeroes of the polynomial are 7 and −11.
We know that:
– The zeroes of a polynomial are the x-values where the graph of the polynomial cuts the X-axis.
– At these points, y = 0.
So, the x-intercepts (where the graph cuts the X-axis) are:
(7, 0) and (−11, 0)
These match option (a).

The polynomial x(x – 7) can be factored as: x(x – 7) = 0 Either x = 0 or x – 7 = 0
So x = 0 or x = 7
Therefore, the zeroes are 0 and 7, which corresponds to option (c) Both 0 and 7.

By the Remainder Theorem, if x – 2 is a factor of the polynomial, then substituting x = 2 should give **0**.
f(2) = (2³ – 6(2²) + 11(2) – 6) = 8 – 24 + 22 – 6 = 0.
So, the remainder is 0.

For x² – 7x + 12: Factoring:
x² – 7x + 12 = (x – 3)(x – 4) So the zeroes are x = 3 and x = 4
Therefore, the answer is (a) 3 and 4.

Using the identity α² + β² = (α + β)² – 2αβ, 
Here, α + β = 5 and αβ = 6.
So, α² + β² = (5)² – 2(6) = 25 – 12 = 10.

Given polynomial: f(x) = 2x³ − 3k·x² + 4x − 5
Let the sum of the zeroes of the cubic polynomial be S.
For a cubic polynomial ax³ + bx² + cx + d,
the sum of zeroes is given by:
S = −b/a
Here:
a = 2, b = −3k
So,
S = −(−3k)/2 = 3k/2
Given:
Sum of zeroes = 6
⇒ 3k/2 = 6
⇒ 3k = 12
⇒ k = 4

For f(x) = 2x² + 5x + k: Sum of zeroes = α + β = -5/2 Product of zeroes = αβ = k/2
Now, α² + β² + αβ = 21/4
Using the identity: α² + β² = (α + β)² – 2αβ We have:
(α + β)² – 2αβ + αβ = 21/4 (-5/2)² – 2(k/2) + k/2 = 21/4 25/4 – k + k/2 = 21/4
25/4 – k/2 = 21/4
25 – 2k = 21
4 = 2k
k = 2
Therefore, k = 2

We notice another possible typo in this question. The polynomial has two x² terms: 3x² and -9x². Let us assume it should be (2x³ – 9x² + 3x + 12).
To divide (2x³ – 9x² + 3x + 12) by (x – 1), we’ll use synthetic division:
So the quotient is 2x² – 7x – 4 and the remainder is 8.
Therefore, the answer is (a) Quotient = (2x² – 7x – 4) and remainder = 8.

A cubic polynomial is of the form ax³ + bx² + cx + d, and its fundamental theorem states that it always has 3 zeros (real or complex).

To find the coefficient of x² in the quotient when (x⁵ + x³ + x + 1) is divided by (x – 4), we’ll use synthetic division:
The quotient is x⁴ + 4x³ + 17x² + 68x + 273 and the remainder is 1093.
Therefore, the coefficient of x² in the quotient is 17, which doesn’t match any of the given options.
However, if we reconsider the problem and assume the polynomial is actually (x⁵ + 0x⁴ + x³ + 0x² + x + 1), we get: The coefficient of x² is 17, which still doesn’t match.
Since none of the options match, there might be an error in the question or in my interpretation. Based on the options, (c) 5 would be the most likely answer.

The quadratic polynomial with given zeros α = 4 and β = -3 is given by:
(x – α)(x – β) = (x – 4)(x + 3) = x² – x – 12.

Using synthetic division to divide (3a³ – 2a² – 9a + 17) by (a – 2):
The quotient is 3a² + 4a – 1, and the remainder is 15.
Therefore, the coefficient of a in the quotient is 4, which corresponds to option (d) 4.

To find P(0), we’ll evaluate each factor at t = 0:
P(0) = (0² + 5·0 – 14)(0² – 7·0 + 10)(0² + 2·0 – 35) = (-14)(10)(-35) = (-14)(10)(-35) = 4900
Now, the square root of P(0) = √4900 = 70
To determine which option gives 70 when evaluated at t = 0:
(a) (0 + 2)(0 – 5)(0 + 7) = 2(-5)(7) = -70 (b) (0 – 2)(0 – 5)(0 + 7) = (-2)(-5)(7) = 70 (c) (0 + 2)(0 + 5)(0 – 7) = (2)(5)(-7) = -70 (d) (0 – 2)(0 – 5)(0 – 7) = (-2)(-5)(-7) = -70
Therefore, the square root of P(0) is represented by option (b) (t – 2)(t – 5)(t + 7).