Question

select all of the factors of (x^3 + 5x^2 + 2x - 8).

(x + 5)
(x - 3)
(x + 4)
(x - 1)
(x + 3)
(x + 2)

Explanation:

🆕 New Concept Discovered: Factoring Polynomials with the Factor Theorem
Using roots to find linear factors of a polynomial.

Step 1: Apply the Factor Theorem

The Factor Theorem states that a linear binomial \( x - c \) is a factor of a polynomial \( p(x) \) if and only if \( p(c) = 0 \).

Let our polynomial be:

$$ p(x) = x^3 + 5x^2 + 2x - 8 $$

We will test the values of \( c \) corresponding to each given option:

• For \( x + 5 \), test \( c = -5 \)
• For \( x - 3 \), test \( c = 3 \)
• For \( x + 4 \), test \( c = -4 \)
• For \( x - 1 \), test \( c = 1 \)
• For \( x + 3 \), test \( c = -3 \)
• For \( x + 2 \), test \( c = -2 \)

Step 2: Evaluate the polynomial at each test value

• Test \( c = 1 \) (for \( x - 1 \)):

$$ p(1) = (1)^3 + 5(1)^2 + 2(1) - 8 = 1 + 5 + 2 - 8 = 0 $$

Since \( p(1) = 0 \), \( x - 1 \) is a factor.

• Test \( c = -4 \) (for \( x + 4 \multi-line \)):

$$ p(-4) = (-4)^3 + 5(-4)^2 + 2(-4) - 8 = -64 + 5(16) - 8 - 8 $$

$$ p(-4) = -64 + 80 - 16 = 0 $$

Since \( p(-4) = 0 \), \( x + 4 \) is a factor.

• Test \( c = -2 \) (for \( x + 2 \)):

$$ p(-2) = (-2)^3 + 5(-2)^2 + 2(-2) - 8 = -8 + 5(4) - 4 - 8 $$

$$ p(-2) = -8 + 20 - 12 = 0 $$

Since \( p(-2) = 0 \), \( x + 2 \) is a factor.

Step 3: Verify remaining options

Since a cubic polynomial can have at most 3 linear factors, and we have already found three:

$$ p(x) = (x - 1)(x + 4)(x + 2) $$

We can double-check one of the incorrect options to verify, for example \( x + 5 \) (\( c = -5 \)):

$$ p(-5) = (-5)^3 + 5(-5)^2 + 2(-5) - 8 = -125 + 125 - 10 - 8 = -18 eq 0 $$

Thus, \( x + 5 \) is not a factor.

Answer:

The factors of the polynomial are:

• \( x + 4 \)
• \( x - 1 \)
• \( x + 2 \)