Question

one root of (f(x) = x^3 - 9x^2 + 26x - 24) is (x = 2). what are all the roots of the function? use the remainder theorem.

• (x = 2, x = 3, \text{ or } x = 4)
• (x = -2, x = -3, \text{ or } x = -4)
• (x = 1, x = 2, x = 3, \text{ or } x = 13)
• (x = -1, x = -2, x = -3, \text{ or } x = -13)

Explanation:

⚡ Using what you learned: Zeros and Multiplicity · 🆕 New Concept: Polynomial Division and Remainder Theorem
Using division to break down polynomials into simpler factors

Step 1: Set up synthetic division

Since \( x = 2 \) is a root, we divide the polynomial \( f(x) = x^3 - 9x^2 + 26x - 24 \) by \( x - 2 \) using synthetic division:

$$
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$$

Step 2: Write the depressed polynomial

The quotient is a quadratic polynomial:

$$ q(x) = x^2 - 7x + 12 $$

Step 3: Factor the quadratic equation

To find the remaining roots, set \( q(x) = 0 \):

$$ x^2 - 7x + 12 = 0 $$

$$ (x - 3)(x - 4) = 0 $$

$$ x = 3 \quad \text{or} \quad x = 4 $$

Answer:

\( x = 2, x = 3, \text{or } x = 4 \)