This is probably going to be straightforward, but given
$P(x)=2x^3−3x^2+4x+3$
Which values of $x$ will give rise to possible factors?
I have the following alternatives (More than one can be correct):
a) −1
b) 1
c) −3
d) 3
e) $\frac{−1}{2}$
f) $\frac{1}{2}$
g) $\frac{-3}{2}$
h) $\frac{-3}{2}$
Using the factor theorem, I've decided to use $\frac{P(x)}{2}$.
Hence, my $d$ will become $\frac{d}{2}$ and I chose what would be the factors of $d$ do make tests, for instance: $\{\frac{-1}{2}, \frac{1}{2}, \frac{-3}{2}, \frac{3}{2}\}$
I've found one solution with $x=\frac{-1}{2}$, which yields the factor $(x + \frac{1}{2})$
Which gives me
$(x + \frac{1}{2})(2x^2 -4x +6)$
Then I can transform that to $P(x)$ to get me back to
$(2x+1)(x^2 -2x +3)$
It has two factors: one linear in the form $(ax+b)$, both integers, and one quadratic. The quadratic factor cannot be further factorised.
Now I don't understand why none of the alternatives above is working for me? For me the correct answer should be $e)$ only as no other value brings me zero. But perhaps I am missing something obvious?
The question is asking you to select which one out of all the answers is correct. So as you've stated the correct choice is e).
It's a multiple choice question! Good job :)