Zeroes of a Polynomial Questions
Chapter – Polynomials
Topic – Zeroes of a Polynomial
Question 1. What is a zero of a polynomial? Explain with an example.
Solution:
A zero of a polynomial p(x) is a value of x for which p(x) = 0.
For example:
p(x) = x − 5
Putting x = 5:
p(5) = 5 − 5
= 0
Therefore, 5 is a zero of the polynomial x − 5.
Answer: A zero of a polynomial is a value of the variable that makes the polynomial equal to zero.
Question 2. Find the zero of the polynomial p(x) = 3x − 12.
Solution:
Given:
p(x) = 3x − 12
To find the zero, put p(x) = 0.
3x − 12 = 0
3x = 12
x = 12/3
x = 4
Answer: The zero of the polynomial is 4.
Question 3. Check whether x = 2 is a zero of p(x) = x² − 5x + 6.
Solution:
Given:
p(x) = x² − 5x + 6
Substitute x = 2:
p(2) = (2)² − 5(2) + 6
= 4 − 10 + 6
= 0
Since p(2) = 0, x = 2 is a zero of the polynomial.
Answer: Yes, 2 is a zero.
Question 4. Find the zeroes of the polynomial p(x) = x² − 9.
Solution:
Given:
p(x) = x² − 9
Put p(x) = 0.
x² − 9 = 0
Using the identity:
a² − b² = (a − b)(a + b)
We get:
(x − 3)(x + 3) = 0
Therefore:
x − 3 = 0 or x + 3 = 0
x = 3 or x = −3
Answer: The zeroes are 3 and −3.
Question 5. Find the zeroes of the polynomial p(x) = x² − 7x + 12.
Solution:
Given:
p(x) = x² − 7x + 12
Put p(x) = 0.
x² − 7x + 12 = 0
Split the middle term:
x² − 3x − 4x + 12 = 0
Take common factors:
x(x − 3) − 4(x − 3) = 0
(x − 3)(x − 4) = 0
Therefore:
x = 3 or x = 4
Answer: The zeroes are 3 and 4.
Question 6. Find the zeroes of the polynomial p(x) = x² + 5x + 6.
Solution:
Given:
p(x) = x² + 5x + 6
Put p(x) = 0.
x² + 5x + 6 = 0
Split the middle term:
x² + 2x + 3x + 6 = 0
x(x + 2) + 3(x + 2) = 0
(x + 2)(x + 3) = 0
Therefore:
x + 2 = 0 or x + 3 = 0
x = −2 or x = −3
Answer: The zeroes are −2 and −3.
Question 7. Find the zeroes of the polynomial p(x) = 2x² − 8x.
Solution:
Given:
p(x) = 2x² − 8x
Put p(x) = 0.
2x² − 8x = 0
Take 2x common:
2x(x − 4) = 0
Therefore:
2x = 0 or x − 4 = 0
x = 0 or x = 4
Answer: The zeroes are 0 and 4.
Question 8. Find the value of k if x = 2 is a zero of p(x) = x² + kx − 6.
Solution:
Given:
p(x) = x² + kx − 6
Since x = 2 is a zero:
p(2) = 0
Substitute x = 2:
(2)² + k(2) − 6 = 0
4 + 2k − 6 = 0
2k − 2 = 0
2k = 2
k = 1
Answer: k = 1.
Question 9. Find the zeroes of the polynomial p(x) = x³ − 4x.
Solution:
Given:
p(x) = x³ − 4x
Put p(x) = 0.
x³ − 4x = 0
Take x common:
x(x² − 4) = 0
Using the difference of squares:
x(x − 2)(x + 2) = 0
Therefore:
x = 0
x − 2 = 0, so x = 2
x + 2 = 0, so x = −2
Answer: The zeroes are 0, 2 and −2.
Question 10. Find the number of zeroes of the polynomial p(x) = x² + 4 over the real numbers.
Solution:
Given:
p(x) = x² + 4
To find the zeroes, put:
x² + 4 = 0
x² = −4
For every real number x, we know that:
x² ≥ 0
Therefore, x² cannot equal −4.
Hence, there is no real value of x for which p(x) = 0.
Answer: The polynomial x² + 4 has no real zeroes.