Question 719665: Show that (x-2) and (x-3) are factors of F(x)=2x^4+7x^3-4x^2-27x-18. Then find the remaining factors of f(x).
Answer by swincher4391(1107)   (Show Source):
You can put this solution on YOUR website! By the factor theorem, if x-2 is a factor of f(x): 2x^4+7x^3-4x^2-27x-18, then f(2) = 0. That is to say, if we were to plug in 2, we would result in 0.
2(2^4) + 7(2^3) - 4(2^2)-27(2) - 18
32 + 56 - 16 - 54 - 18 = 0 (check).
I do believe that there is a typo in your question. It should read (x+3).
Plugging in -3 yields 0.
If you have not learned the factor theorem, disregard the previous work.
Alternative method (long division):
(x-2) | 2x^4 +7x^3 - 4x^2 -27x - 18
Going through this we get 2x^3+11x^2+18x+9, with no remainder. Since there is no remainder, x-2 factors into it evenly. You would repeat the process for x+3 into the remaining 2x^3 + 11x^2 + 18x + 9 leaving you 2x^2 + 5x + 3. This factors into (2x+3)(x+1) [remaining factors]
|
|
|
