1) Factor xReturn to the tutorial^{3}+ 1 SOLUTION: A factor of the form (x - k) exists only if k divides (+1) exactly. Therefore, k must equal +1, or -1. This means that (x - 1) and/or (x + 1) are factors. See if (x - 1) is a factor by long division (remember that if it is, the remainder must equal zero): **Note: when using long division, you must include x terms whose coefficients are zero!! x^{2}+ x + 1 ----------------- (x - 1)| x^{3}+ 0x^{2}+ 0x + 1 x^{3}- x^{2}------- x^{2}+ 0x x^{2}- x ------- x + 1 x - 1 ----- 2 The remainder is NOT zero and thus (x - 1) is NOT a factor. This means that (x + 1) probably is (if there is any factor of the form (x - k)) but we will use long division to see if it actually is. x^{2}- x + 1 ----------------- (x + 1)| x^{3}+ 0x^{2}+ 0x + 1 x^{3}+ x^{2}------- -x^{2}+ 0x -x^{2}- x ------- x + 1 x + 1 ----- 0 The remainder is zero, thus (x + 1) is a factor and (x^{2}- x + 1) is the other factor. That is: x^{3}+ 1 = (x + 1)((x^{2}- x + 1) 2) Factor 2x^{3}- x^{2}- 13x + 5 SOLUTION: Since the coefficient of the leading term (highest order term) is 2, we need to find a factor of the form (hx - k) such that h divides exactly into 2, and k divides exactly into 5. Factors of 5 are: 1, -1, 5, and -5 (possible k values) Factors of 2 are: 1, -1, 2, and -2 (possible h values) Thus, possible factors of the original polynomial needing to be factored are: if h = 1: (x - 1), (x + 1), (x - 5), (x + 5) if h = -1: (-x -1), (-x + 1), (-x - 5), (-x + 5) if h = 2: (2x - 1), (2x + 1), (2x - 5), (2x + 5) if h = -2 (-2x -1), (-2x + 1), (-2x - 5), (-2x + 5) Note that: (x - 1) = -(-x + 1), and (2x - 1) = -(-2x + 1). Continue on with factoring out a negative from every possiblity in the 1st and 3rd lines above and see that you get the factors listed in rows 2 and 4 above. This means that every factor above appears twice (because the negative of a factor is also a factor), and so you only have to test half of these to see which of them are factors. Thus, we need to check: (x - 1), (x + 1), (x - 5), (x + 5), (2x - 1), (2x + 1), (2x - 5), (2x + 5) **The best way to get only those possible factors that you want to check is to always only use the positive values for h but use both positive and negative values of k. It is a lot of work but if you perform long division on all these factors, you will see that all of them result in remainders not equal to zero, except (2x + 5). x^{2}- 3x + 1 --------------------- (2x + 5)| 2x^{3}- x^{2}- 13x + 5 2x^{3}+ 5x^{2}-------- -6x^{2}- 13x -6x^{2}- 15x ----------- 2x + 5 2x + 5 ------ 0 Thus, the factors of 2x^{3}- x^{2}- 13x + 5 are (2x + 5) and (x^{2}- 3x + 1)