$$P(x)=2x^2+3x+k$$

In the function above, $k$ is a constant. If $1$ is a zero of the function, what is the value of $k$?

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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 $k$.

$$(x-1)(ax+b)=2x^2+3x+k$$ $$(x-1)(2x+5)=2x^2+3x+k$$ $$2x^2+3x-5=2x^2+3x+k$$ $$k=\boxed{-5 }$$