Answer
Zeros: $-\dfrac{3}{2},2$
$x$-intercepts: $-\dfrac{3}{2},2$
Work Step by Step
To find the zeros of a function $f$, solve the equation $f(x)=0$
The zeros of the function are also the $x-$intercepts.
Let $P(x)=0$:
$$2(x+1)^2-5(x+1)-3=0$$
Let $u=x+1$, the original equation becomes
$$2u^2-5u-3=0$$
By factoring
$$(2u+1)(u-3) = 0$$
Use the Zero-Product Property by equating each factor to zero, then solve ach equation to obtain:
\begin{align*}
2u+1& =0 &\text{ or }& &u-3=0\\
2u& =-1 &\text{ or }& &u=3\\
u &= -\dfrac{1}{2} &\text{ or }& &u=3\\
\end{align*}
To solve for $x$, we use $u=x+1$
$$\because u = x+1$$
$$\therefore x =u-1$$
For $u=-\dfrac{1}{2}$
$$x=-\dfrac{1}{2}-1 $$
$$x= -\dfrac{3}{2}$$
For $u=3$
$$x=3-1$$
$$x= 2$$
$\therefore x =-\dfrac{3}{2},2$
