Inactive Tutor answered 01/27/13
I believe, that this formula will help you too in the future: "ax^2+bx+c=a(x-x‚)(x-x„), x‚ and x„ are roots of equation ax^2+bx+c=0
Raygen B.
asked 01/03/13Im learning about the factor theorem and i cant figure out these two things...
Given a polynomial and one of its factors, find the remaining factors of the polynomials. Some factors may not be binomials.
1.) 2x^3-5x^2-28x+15; x-5
Find values of K ao that each remainder is 3
2.) (x^2+5x+7)/(x+k)
Inactive Tutor answered 01/27/13
I believe, that this formula will help you too in the future: "ax^2+bx+c=a(x-x‚)(x-x„), x‚ and x„ are roots of equation ax^2+bx+c=0
Inactive Tutor answered 01/03/13
For the first, divide 2x3 - 5x2 -28x + 15 by (x - 5). You should get 2x2 + 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 (x2 + 5x + 7) by that (x + k). So let's subtract that 3 from (x2 + 5x + 7) and see if what remains can be factored. (x2 + 5x + 7) - 3 = (x2 + 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).
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Raygen B.
Oh my goodness..Thank you soooooo much!!! You totally helped me out!(:
01/03/13