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.
Consider the polynomial:
f(x) = 2x^3 – 3x^2 – 8x – 3.
(a) By using the Rational Zero Theorem, list all possible rational zeros of the given polynomial.
(b) Find all of the zeros of the given polynomial. Be sure to show work, explaining how you have found them.
2 answers
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?
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?