Polynomials - Roots and Factor Theorem

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Q1. What is the value of k if x^2 - kx + 12 has roots 3 and 4?

Solution:

Using the sum of roots, 3 + 4 = k, we find k = 7.

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Q2. What is the value of k if the polynomial x^2 + kx + 16 has roots 4 and -4?

Solution:

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

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Q3. What is the solution set for the inequality 2x - 4 < 0?

Solution:

Solving the inequality: 2x < 4, thus x < 2.

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Q4. If x^2 + 6x + 9 = 0, what are the roots?

Solution:

The polynomial can be factored as (x + 3)(x + 3) = 0, giving the root x = -3.

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Q5. Which of the following is a root of the polynomial x^3 - 6x^2 + 11x - 6?

Solution:

Using the Rational Root Theorem, we can test x = 1, 2, and 3. Testing x = 3 gives 3^3 - 6(3^2) + 11(3) - 6 = 0, so x = 3 is a root.

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Q6. 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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Q7. What is the factored form of the polynomial x^2 - 10x + 24?

Solution:

To factor, we look for two numbers that multiply to 24 and add to -10. The numbers -4 and -6 work, so the factored form is (x - 4)(x - 6).

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Q8. 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.

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Q9. 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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Q10. 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.