Polynomials MCQ Test — Class 10 CBSE

Solution:

By definition, a number 'k' is called a zero of a polynomial P(x) if P(k) = 0.

This means that when 'k' is substituted for 'x' in the polynomial, the entire expression evaluates to zero.

Solution:

A quadratic polynomial is a polynomial of degree 2. This means the highest power of the variable in the expression must be 2.

Option A (2x + 5) is a linear polynomial (degree 1). — Degree of 2x + 5 is 1

Option B (3x² - 4x + 1) has the highest power of x as 2, so it is a quadratic polynomial. — Degree of 3x² - 4x + 1 is 2

Option C (x³ + 2x - 7) is a cubic polynomial (degree 3). — Degree of x³ + 2x - 7 is 3

Option D (4x⁴) is a polynomial of degree 4. — Degree of 4x⁴ is 4

Solution:

The zeros of a polynomial P(x) are the x-values for which P(x) = 0. Geometrically, these are the points where the graph of y = P(x) intersects or touches the x-axis.

If the graph intersects the x-axis at exactly three distinct points, it means there are three distinct x-values for which P(x) = 0.

Therefore, P(x) has exactly three real zeros.

Solution:

For a quadratic polynomial ax² + bx + c, the sum of its zeros (α + β) is given by the formula -b/a.

In the given polynomial x² - 7x + 10, we have a = 1, b = -7, and c = 10.

Substituting these values into the formula, the sum of zeros = -(-7)/1 = 7/1 = 7.

Ravi's statement that the sum of zeros is 7 is correct, and the reasoning for option D correctly explains why.

Solution:

For a quadratic polynomial, if α and β are its zeros, then the polynomial can be written in the form x² - (sum of zeros)x + (product of zeros).

Given: Sum of zeros (α + β) = -5.

Given: Product of zeros (αβ) = 6.

Substitute these values into the general form: x² - (-5)x + (6) = x² + 5x + 6.

Solution:

The Division Algorithm states that for any polynomial P(x) and any non-zero polynomial G(x), there exist unique polynomials Q(x) (quotient) and R(x) (remainder) such that P(x) = G(x) × Q(x) + R(x).

A crucial condition for the remainder is that its degree must be less than the degree of the divisor G(x), or the remainder R(x) must be the zero polynomial (in which case its degree is undefined or considered -∞, which is certainly less than any positive degree).

Solution:

Let the zeros be α = 2 and β = -3.

Calculate the sum of zeros: α + β = 2 + (-3) = -1.

Calculate the product of zeros: αβ = 2 × (-3) = -6.

The quadratic polynomial can be formed as x² - (α + β)x + αβ. — x² - (-1)x + (-6) = x² + x - 6

Thus, the polynomial is x² + x - 6.

Solution:

We need to find the value of 1/α + 1/β. First, combine the fractions:

1/α + 1/β = (β + α) / (αβ)

For a quadratic polynomial P(x) = ax² + bx + c, we know the relationships:

Sum of zeros: α + β = -b/a

Product of zeros: αβ = c/a

Substitute these into the combined expression: — (β + α) / (αβ) = (-b/a) / (c/a)

Simplify the expression: — (-b/a) × (a/c) = -b/c

Solution:

Polynomials are classified by their degree, which is the highest power of the variable.

A polynomial with degree 1, such as P(x) = ax + b (where a ≠ 0), is called a linear polynomial because its graph is a straight line.

Solution:

A zero of the polynomial h(t) occurs when h(t) = 0.

In this context, h(t) represents the height of the ball. So, h(t) = 0 means the ball is at ground level.

Therefore, the value of 't' for which h(t) = 0 (i.e., the zero of the polynomial) represents the time when the ball hits the ground.

Since time 't' cannot be negative in this physical context, we are interested in the positive zero.