x2−x−6=(x+2)(x−3) Case 1: x2−x−6<0 i.e. (x+2)(x−3)<0 $⇒-2 Therefore, |x2−x−6|=x+2 =−(x+2)(x−3)=x+2 ⇒(x−3)=−1⇒x=2 Case 2:x2−x−6≥0 i.e. (x+2)(x−3)≥0 ⇒x≤−2 or x≥3 Checking for boundary conditions: For x=−2,|x2−x−6|=x+2, therefore, x=−2 is also the root. But for x=3,|x2−x−6|≠x+2. Hence x=3 is NOT the root. And for the interval x<−2 or x>3 the expression |x2−x−6|=x2−x−6 Therefore, |x2−x−6|=x+2 =(x+2)(x−3)=x+2 ⇒(x−3)=1⇒x=4 Therefore, the root are −2,2, and 4 . So the required product =(2)(−2)(4)=−16