For the first, divide 2x^{3}  5x^{2} 28x + 15 by (x  5). You should get 2x^{2} + 5x  3. Now you can factor that: look for combinations of 1, 2, and 3 that can add to 5 (for 5x) and multiply to get 3. Let's try (2x  1) and (x + 3),
since (1)(+3) = 3 and (2x)(+3) + (x)(1) = +6x  x = +5x. So now we have 3 factors of the original problem: (x  5)(x + 3)(2x  1).
For the second: It's asking what values of k in (x + k) would give you a remainder of +3 after you divide (x^{2} + 5x + 7) by that (x + k). So let's subtract that 3 from (x^{2} + 5x + 7) and see if what remains can be factored. (x^{2} + 5x + 7)  3 = (x^{2} + 5x + 4). Looking for factors that will multiply to a product of 4 and also add to 5 (for 5x), let's try 4 and 1: (x + 4)(x + 1) = (x2 + 5x + 4), leaving our remainder of +3. So our k values are +1 and +4, or (x + 1) and (x + 4).
1/3/2013

Bill F.
Comments
Oh my goodness..Thank you soooooo much!!! You totally helped me out!(: