Let's solve the three given polynomial equations by finding all rational zeros and writing them in factored form:

1. $P(x) = x^3 + 3x^2 - 4x - 12$

Using the Rational Root Theorem, the potential rational roots are factors of $-12$ (constant term) divided by factors of $1$ (leading coefficient): $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$

Testing these values:

• $P(2) = 2^3 + 3(2^2) - 4(2) - 12 = 8 + 12 - 8 - 12 = 0$

Thus, $x = 2$ is a root. Using synthetic division to divide $P(x)$ by $(x - 2)$:

$x^3 + 3x^2 - 4x - 12 \div (x - 2) = x^2 + 5x + 6$

Factoring $x^2 + 5x + 6$:

$x^2 + 5x + 6 = (x + 2)(x + 3)$

The fully factored form is: $P(x) = (x - 2)(x + 2)(x + 3)$

2. $P(x) = x^4 - 6x^2 + 5$

Let $y = x^2$, so the equation becomes $y^2 - 6y + 5$: $y^2 - 6y + 5 = (y - 1)(y - 5)$

Substituting back $y = x^2$: $(x^2 - 1)(x^2 - 5)$

Factoring further: $x^2 - 1 = (x - 1)(x + 1)$

Thus: $P(x) = (x - 1)(x + 1)(x^2 - 5)$

(Note: $x^2 - 5$ cannot be factored further over the rationals.)

3. $P(x) = x^4 - 5x^3 + 5x^2 + 5x - 6$

Using the Rational Root Theorem, the potential rational roots are factors of $-6$ divided by factors of $1$: $\pm 1, \pm 2, \pm 3, \pm 6$

Testing these values:

• $P(1) = 1^4 - 5(1^3) + 5(1^2) + 5(1) - 6 = 1 - 5 + 5 + 5 - 6 = 0$

Thus, $x = 1$ is a root. Using synthetic division to divide $P(x)$ by $(x - 1)$:

$x^4 - 5x^3 + 5x^2 + 5x - 6 \div (x - 1) = x^3 - 4x^2 + x + 6$

Factoring $x^3 - 4x^2 + x + 6$:

Testing roots for $x^3 - 4x^2 + x + 6$:

• $P(-2) = (-2)^3 - 4(-2)^2 + (-2) + 6 = -8 - 16 - 2 + 6 = 0$

Thus, $x = -2$ is a root. Using synthetic division: $x^3 - 4x^2 + x + 6 \div (x + 2) = x^2 - 6x + 3$

Factoring $x^2 - 6x + 3$ (using the quadratic formula): $x = \frac{6 \pm \sqrt{36 - 12}}{2} = \frac{6 \pm \sqrt{24}}{2} = \frac{6 \pm 2\sqrt{6}}{2} = 3 \pm \sqrt{6}$

The fully factored form is: $P(x) = (x - 1)(x + 2)(x - (3 + \sqrt{6}))(x - (3 - \sqrt{6}))$

Summary of Factorizations:

1. $P(x) = (x - 2)(x + 2)(x + 3)$
2. $P(x) = (x - 1)(x + 1)(x^2 - 5)$
3. $P(x) = (x - 1)(x + 2)(x - (3 + \sqrt{6}))(x - (3 - \sqrt{6}))$

Would you like further details or have questions on any of these steps?

Expanding Questions:

1. How is the Rational Root Theorem applied to determine potential roots?
2. Why does $x^2 - 5$ remain unfactored over rationals?
3. What is the significance of synthetic division in factoring polynomials?
4. How can you confirm the roots of a polynomial using substitution?
5. Can irrational or complex roots appear in similar problems?

Tip: Always verify potential roots by substitution into the original polynomial!