Polynomials - Roots and Factor Theorem

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Q1. If f(x) = x^2 - 4, what are the roots of the polynomial?

Solution:

The polynomial can be factored as (x - 2)(x + 2). Setting each factor to zero gives us x - 2 = 0 or x + 2 = 0, so the roots are x = -2 and x = 2.

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Q2. What are the roots of the polynomial x^2 + 6x + 9?

Solution:

The polynomial can be factored as (x + 3)(x + 3) or (x + 3)^2. Therefore, the roots are x = -3 and x = -3.

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Q3. If x^2 - 6x + 9 = 0, what is the repeated root?

Solution:

The polynomial can be factored as (x - 3)(x - 3) = 0, giving a repeated root of x = 3.

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Q4. Which of the following is a solution to the equation x^2 - 7x + 10 = 0?

Solution:

Factoring gives (x - 5)(x - 2) = 0, so the solutions are x = 5 and x = 2.

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Q5. What are the roots of the polynomial x^2 - 4?

Solution:

x^2 - 4 can be factored as (x - 2)(x + 2) = 0. Therefore, the roots are -2 and 2.

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Q6. What is the solution set for the inequality 2x - 3 < 5?

Solution:

First, add 3 to both sides: 2x < 8. Then divide by 2: x < 4.

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Q7. Which polynomial has a root at x = -1?

Solution:

The polynomial x^2 + 2x + 1 can be factored as (x + 1)(x + 1), indicating that -1 is a root.

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Q8. What is the value of k if x^2 + kx + 16 has roots 4 and -4?

Solution:

Using Vieta's formulas, the sum of the roots is -k = 4 + (-4) = 0, so k = 0.

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Q9. Which of the following represents the roots of the equation 2x^2 - 8 = 0?

Solution:

First, simplify the equation: 2x^2 = 8, so x^2 = 4. The roots are x = ±2.

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Q10. What is the value of k if the polynomial x^2 + kx + 16 has roots that are both 4?

Solution:

Using the fact that the sum of the roots is -k, we have 4 + 4 = -k, so k = -8.