 Quandaries and Queries Hello I am having a really hard time with the factor theorem. Can you please help me with the following? (a'3 means a to the exponent 3) a)8a'3+64b'3c'6 b)8x'4-12x'3-44x'2+24x c)x'4-21x'2-100 Thank you very much Marcin Hi Marcin, In many instances factoring is recognizing patterns. For example in your first problem I see that 8, a3, 64, b3 and c6 are all cubes.Hence I can write 8a3 + 64b3c6 = (2a)3 + (4bc2)3 Can you factor it now as a sum of cubes? In b), all the coefficients are divisible by 4 and there is an x common factor in each term, thus 8x4 - 12x3 - 44x2 + 24x = 4x(x3 -3x2 -11x + 6) Hence you are left with the cubic x3 -3x2 -11x + 6 which is a candidate for the factor theorem. To use the factor theorem you need to find a root, that is a number x for which x3 -3x2 -11x + 6 = 0. Usually we look for a root that is an integer. Such an integer must divide the constant term (6 in this case). Thus the only candidates for x are 1,2,3,6,-1,-2,-3 and -6. If you try these numbers (start with the small numbers to make the calculation easy) you will find that x=-2 is a root. The factor theorem then implies that (x+2) is a factor of the polynomial. Now divide (x+2) into x3 -3x2 -11x + 6. to get x3 -3x2 -11x + 6 = (x + 2) (a quadratic) In this example you can also factor the quadratic. Cheers, Penny Go to Math Central