Practice Test: Polynomials - Class 10 MCQ

Explanation: Sum of zeroes = α + β =√2
Product of zeroes = αβ = 1/3
∴ If α and β are zeroes of any quadratic polynomial, then the polynomial is;
x² – (α+β)x + αβ
= x² – (√2)x + (1/3)
= 3x² - 3√2x +1

x² – 2x – 8 = x² – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x - 4)(x + 2)
Therefore, x = 4, -2.

For equal roots, discriminant will be equal to zero.
b² - 4ac = 0
b² = 4ac
ac = b²/4
ac > 0 (as square of any number cannot be negative)

If p(x) is a polynomial of degree one, it can be written as:
p(x) = mx + c
where m and c are constants, and m ≠ 0.
Given that p(a) = 0, substituting x = a into p(x):
p(a) = m(a) + c = 0
This equation implies that a is a root or zero of the polynomial p(x), as it satisfies p(a) = 0.
Hence, the correct answer is:
A: Zero of p(x)

Maximum number of zeroes of a polynomial = Degree of the polynomial

The polynomials x² -3x -10, 2x² - 6x - 20, (1/2)x² -(3/2)x - 5, 3x² - 9x - 30, have zeroes as - 2 and 5.

x² - 27 = 0

x² = 27

x = √27

x = ±3√3

Given that 2 is the zero of the quadratic polynomial x² + 3x + k.

⇒ (2)² + 3(2) + k = 0

⇒ 4 + 6 + k = 0

⇒ k = -10

Let the given zeroes be α = -3 and β = 4.

Sum of zeroes, α + β= -3 + 4 = 1 

Product of Zeroes, αβ = -3 × 4 = -12 

Therefore, the quadratic polynomial = x² – (sum of zeroes)x + (product of zeroes) 

= x² – (α + β)x + (αβ) 

= x² – (1)x + (-12) 

= x² – x – 12

Dividing by 2,

= (x²/2) – (x/2) – 6

Given quadratic polynomial is x² + 99x + 127.

By comparing with the standard form, we get;

a = 1, b = 99 and c = 127

a > 0, b > 0 and c > 0

We know that in any quadratic polynomial, if all the coefficients have the same sign, then the zeroes of that polynomial will be negative.

Therefore, the zeroes of the given quadratic polynomial are negative.