questions
| hey guys, for 4.2 i am not sure how to do number 23, and 45. thanks for ur help! melissa |
posted by melishappy at 12:26 AM
Blog for students in Mr. Karafiol's Period 4 Precalculus BC class to talk about math, learning math, etc.
| hey guys, for 4.2 i am not sure how to do number 23, and 45. thanks for ur help! melissa |
posted by melishappy at 12:26 AM
1 Comments:
Well since you said you want to know HOW to do them so to do number 45, you're given that it is a polynomial to the 5th degree and are given a graph of it showing its 0's. Since you know the zeros you just set it up like (x-(one of the zeros)*(x-(another zero) etc etc and that gives you (x-3)*(x-2)*(x-1)*(x+1)*(x-0) and that simplifies to x^5-5x^4+5x^3+5x^2-6x. For number 23, you know that when f(x) is divided by x-c (in this case c is = to 1) the remainder is f(c) so you just plug in 1 into x^10+x^8 so that gives you a remainder of 2. Hope that helps!
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