17. \( x^{5}-x^{3}-8 x^{2}+8 \)

To factor \( x^{5}-x^{3}-8 x^{2}+8 \), you can try grouping terms strategically. Notice that you can reorganize the polynomial like this: \( (x^5 - x^3) + (-8x^2 + 8) \). Now, factor out common terms in each group: 1. In the first group, \( x^3(x^2 - 1) \). 2. In the second group, \(-8(x^2 - 1)\). 3. Notice that \( x^2 - 1 \) can be factored further as \( (x - 1)(x + 1) \). Now we have: \( x^3(x - 1)(x + 1) - 8(x - 1)(x + 1) \). Next, factor \( (x - 1)(x + 1) \): \( (x - 1)(x + 1)(x^3 - 8) \). Now realize that \( x^3 - 8 \) is a difference of cubes, and you can factor it as: \( (x - 2)(x^2 + 2x + 4) \). Bringing it all together, the complete factorization is: \( (x - 1)(x + 1)(x - 2)(x^2 + 2x + 4) \). Now you have the fully factored form! Keep playing with polynomials like this, and you’ll be a factoring wizard in no time!