91Ó°ÊÓ

Assume the first integer to be x.

Next consecutive even integer$=x+2$

From the question,

localid="1644678835234" $$\begin{gathered} x\left(x+2\right)=168......\left(i\right) \\ {x}^2+2x=168 \\ {x}^2+2x-168=0 \end{gathered}$$

On factoring,

$$\begin{gathered} x^2+14x-12x-168=0 \\ x\left(x+14\right)-12\left(x+14\right)=0 \\ \left(x+14\right)\left(x-12\right)=0 \end{gathered}$$

Use the zero product property,

$$\begin{gathered} x+14=0 \\ x=-14 \end{gathered}$$

Then,

$$\begin{gathered} x-12=0 \\ x=12 \end{gathered}$$

If the first integer $x=-14$, then the next even integer,

$$\begin{gathered} =x+2 \\ =-14+2 \\ =-12 \end{gathered}$$

Therefore, the sets of two consecutive even integers is $-14,-12$.

If the first integer $x=12$, then the next even integer,

$$\begin{gathered} =n+2 \\ =12+2 \\ =14 \end{gathered}$$

Therefore, the sets of two consecutive even integers is$12,14$.

Substitute the consecutive integers $-14,-12$in equation (i),

$$\begin{gathered} \left(-14\right)\left(-12\right)=168 \\ 168=168 \end{gathered}$$

This is true.

Substitute the consecutive integers $12,14$in equation (i),

$$\begin{gathered} 12×14=168 \\ 168=168 \end{gathered}$$

This is also true.

Hence, the sets of consecutive integer are$\{-14,-12\},\{12,14\}$.