Let's go through each equation and solve them by factoring.

Problem (a)

Given: $x^4 - 16x^2 + 75 = 2x^2 - 6$ Rearrange the equation: $x^4 - 18x^2 + 81 = 0$ Let $y = x^2$, then the equation becomes: $y^2 - 18y + 81 = 0$ Factor this as: $(y - 9)^2 = 0 \Rightarrow y = 9$ Since $y = x^2$, we have: $x^2 = 9 \Rightarrow x = \pm 3$ Solution for (a): $x = 3$ or $x = -3$.

Problem (b)

Given: $2x^2 + 4x - 1 = x + 1$ Rearrange: $2x^2 + 3x - 2 = 0$ To factor, find two numbers that multiply to $-4$ and add to $3$: $(2x - 1)(x + 2) = 0$ Set each factor equal to zero: $2x - 1 = 0 \Rightarrow x = \frac{1}{2}$ $x + 2 = 0 \Rightarrow x = -2$ Solution for (b): $x = \frac{1}{2}$ or $x = -2$.

Problem (c)

Given: $4x^3 - x^2 - 2x + 2 = 3x^3 - 2(x^2 - 1)$ Expand and simplify the right side: $4x^3 - x^2 - 2x + 2 = 3x^3 - 2x^2 + 2$ Rearrange: $x^3 + x^2 - 2x = 0$ Factor out $x$: $x(x^2 + x - 2) = 0$ Factor further: $x(x + 2)(x - 1) = 0$ Solution for (c): $x = 0$, $x = -2$, or $x = 1$.

Problem (d)

Given: $-2x^2 + x - 6 = -x^3 + 2x - 8$ Rearrange: $x^3 - 2x^2 - x + 2 = 0$ Factor by grouping: $x^2(x - 2) - 1(x - 2) = 0$ $(x^2 - 1)(x - 2) = 0$ Factor further: $(x - 1)(x + 1)(x - 2) = 0$ Solution for (d): $x = 1$, $x = -1$, or $x = 2$.

Would you like more details on any of these steps?

Here are some related questions:

1. How do you approach solving higher-degree polynomial equations?
2. What is factoring by grouping, and when is it useful?
3. How can we use substitution to simplify polynomial equations?
4. How do you check your solutions in polynomial equations?
5. Why is factoring important in solving polynomial equations?

Tip: Always start by simplifying equations as much as possible before factoring; this can reveal easier paths to the solution.