We can use the factor theorem, which basically says that if a polynomial has a zero at \(x=c\), then \(f(c)=0\). In other words, if we plug in the \(x\)-value of the zeros, we
should obtain an output of zero.
$$ P(1)=0 $$
$$ 2(1)^2+3(1)+k=0 $$
$$ 2+3+k=0 $$
$$ k=\boxed{-5} $$
Since \(x=1\) is a zero, \((x-1)\) is a factor. We can use long division or synthetic division to obtain the other factor.
$$ \begin{array}{r}
2x+5\phantom{-10} \\
x-1{\overline{\smash{\big)}\,2x^2+3x+k}} \\
\underline{-2x^2+2x} \hphantom{-10}\\
5x+k\phantom{} \\
\underline{-5x+5} \phantom{} \\
\end{array} $$
$$ \text{ Remainder}=k+5 $$
The other possible factor for our quadratic equation is \(2x+5\). It is only a factor if there is no remainder, so \(k=\boxed{-5}\)
Since \((x-1)\) is a factor, we can anticipate what the other factor is to construct our quadratic \(P(x)=2x^2+3x+k \).
$$ (x-1)(ax+b)=2x^2+3x+k $$
Since we need to obtain \(2x^2\), \(a=2\).
$$ (x-1)(2x+b)=2x^2+3x+k $$
Expanding the left side allows us to identify the correct value for \(b\).
$$ 2x^2+\colorbox{aqua}{$bx-2x$}-b=2x^2+\colorbox{aqua}{$3x$}+k $$
$$ bx-2x=3x $$
$$ b=5 $$
We can substitute the values of \(a\) and \(b\) and expand to find the value of \(k\).
$$ (x-1)(ax+b)=2x^2+3x+k $$
$$ (x-1)(2x+5)=2x^2+3x+k $$
$$ 2x^2+3x-5=2x^2+3x+k $$
$$ k=\boxed{-5 }$$