Posted by Alyssa on Saturday, July 25, 2009 at 10:29pm.
values I would try are
±1, ±3, ±1/2, and ±3/2
b)
on the second try
f(-1) = -2-3+8-3 = 0
so x+1 is a factor.
by synthetic division I got
2x^3 – 3x^2 – 8x – 3 = (x+1)(2x^2 - 5x -3)
with a couple trial and error stabs, I factored the quadratic into (x-3)(2x+1)
so the factors are
(x+1)(x-3)(2x+1)
of course I could have continued with the above values of a) and tried
f(3), f(-3), f(1/2) etc
and would have found
f(3) and f(-1/2) also to result in zero.
f(x)=0
F(x) = (2x+1) (x+1) (x-3)
x=-1/2
x=-1
x=3
F(x) =0
F(x) = (2x+1) (x+1) (x-3)
(2x+1) (x+1) (x-3) = 0
Is this a good way to show the procedure?
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