Answer
Zeros: $2,\dfrac{5}{3}$
$x$-intercepts: $2,\dfrac{5}{3}$
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 $H(x)=0$:
$$3(1-x)^2+5(1-x)+2=0$$
Let $u=1-x$, the original equation becomes
$$3u^2+5u+2=0$$
By factoring
$$(u+1)(3u+2) = 0$$
Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain:
\begin{align*}
u +1&=0 &\text{ or }& &3u+2=0\\
u &=-1 &\text{ or }& &3u=-2\\
u &= -1 &\text{ or }& &u=-\dfrac{2}{3}\\
\end{align*}
To solve for $x$, we use $u=1-x$
$$\because u = 1-x$$
$$\therefore x =1-u$$
For $u=-1$
$$x=1-(-1) $$
$$x= 2$$
For $u=-\dfrac{2}{3}$
$$x=1-(-\dfrac{2}{3})$$
$$x= \dfrac{5}{3}$$
$\therefore x =2,\dfrac{5}{3}$
