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one root of (f(x) = x^3 - 4x^2 - 20x + 48) is (x = 6). what are all the…

Question

one root of (f(x) = x^3 - 4x^2 - 20x + 48) is (x = 6). what are all the factors of the function? use the remainder theorem.

(a) ((x + 6)(x + 8))
(b) ((x - 6)(x - 8))
(c) ((x - 2)(x + 4)(x - 6))
(d) ((x + 2)(x - 4)(x + 6))

Explanation:

🆕 New Concept Discovered: Polynomial Factor Theorem
Connecting roots of a polynomial to its linear factors.

Step 1: Identify the first factor from the given root

If \( x = c \) is a root of a polynomial function \( f(x) \), then by the Factor Theorem (a consequence of the Remainder Theorem), \( (x - c) \) is a factor of the function.

Given the root:

$$ x = 6 $$

The corresponding factor is:

$$ (x - 6) $$

Step 2: Divide the polynomial to find the remaining quadratic factor

We divide \( f(x) = x^3 - 4x^2 - 20x + 48 \) by \( (x - 6) \) using synthetic division:

$$ LATEXBLOCK0 $$

The quotient is the quadratic expression:

$$ x^2 + 2x - 8 $$

Step 3: Factor the remaining quadratic expression

We find two numbers that multiply to \(-8\) and add to \(2\). These numbers are \(4\) and \(-2\).

$$ x^2 + 2x - 8 = (x + 4)(x - 2) $$

Step 4: Combine all factors

Combining the initial factor with the factored quadratic gives:

$$ f(x) = (x - 2)(x + 4)(x - 6) $$

Answer:

$$ (x - 2)(x + 4)(x - 6) $$