6.5 and 6.6 notes.
Divide x3 -2x2 -13x - 10 by x + 1 and find all solutions.
Now what if you weren't told what to divide by?
There has to be a way to figure out what numbers will work.
The rational root theorem says "all possible solutions are in the
form p/q where p is all factors of the constant and p/q is all
factors of the leading coefficient."
P(x) = x3 -2x2 -13x - 10
You will ALWAYS have q of 1.
Makes it easier.
x3 -2x2 -13x - 10
possible solutions are +/- 1, 2, 5, 10
now you just have to find them.
List all possible solutions for P(x)
Find all solutions for P(x)
x3 + x2 -3x - 3 = 0
List all possible solutions for P(x)
Find all solutions for P(x)
x3 + 2x2 - 8x - 16 = 0
Fundamental Theorem of Algebra.
Every polynomial has at least one
complex solution.
remember complex numbers are just real
numbers in disguise.
a + bi
3 + 0i
4 - 5i
List all possible solutions for P(x)
Find all solutions for P(x)
P(x) = x3 + x2 -x + 2
Homework pg. 343 # 9-16
on your test you are expected to find all the
POSSIBLE solutions. If you can find the ACTUAL
solutions, you get extra credit.